Ganesh - Engineering Mathematics II

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ISBN (13) : 978-81-224-2882-7

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Preface This Textbook has been prepared as per the syllabus for the Engineering Mathematics Second semester B.E classes of Visveswaraiah Technological University. The book contains eight chapters, and each chapter corresponds to one unit of the syllabus. The topics covered are: Unit I and II— Differential Calculus, Unit III and IV—Integral Calculus and Vector Integration, Unit V and VI— Differential Equations and Unit VII and VIII—Laplace Transforms. It gives us a great pleasure in presenting this book. In this edition, the modifications have been dictated by the changes in the VTU syllabus. The main consideration in writing the book was to present the considerable requirements of the syllabus in as simple manner as possible. This will help students gain confidence in problem-solving. Each unit treated in a systematic and logical presentation of solved examples is followed by an exercise section and includes latest model question papers with answers from an integral part of the text in which students will get enough questions for practice. The book is designed as self-contained, comprehensive and friendly from students’ point of view. Both theory and problems have been explained by using elegant diagrams wherever necessary. We are grateful to New Age International (P) Limited, Publishers and the editorial department for their commitment and encouragement in bringing out this book within a short span of period. AUTHORS

Acknowledgement It gives us a great pleasure to present this book ENGINEERING MATHEMATICS-II as per the latest syllabus and question pattern of VTU effective from 2008-2009. Let us take this opportunity to thank one and all who have actually given me all kinds of support directly and indirectly for bringing up my textbook. We whole heartedly thank our Chairman Mr. S. Narasaraju Garu, Executive Director, Mr. S. Ramesh Raju Garu, Director Prof. Basavaraju, Principal Dr. T. Krishnan, HOD Dr. K. Mallikarjun, Dr. P.V.K. Perumal, Dr. M. Surekha, The Oxford College of Engineering, Bangalore. We would like to thank the other members of our Department, Prof. K. Bharathi, Prof. G. Padhmasudha, Mr. Ravikumar, Mr. Sivashankar and other staffs of The Oxford College of Engineering, Bangalore for the assistance they provided at all levels for bringing out this textbook successfully. We must acknowledge Prof. M. Govindaiah, Principal, Prof. K.V. Narayana, Reader, Department of Mathematics, Vivekananda First Grade Degree College, Bangalore are the ones who truly made a difference in our life and inspired us a lot. We must acknowledge HOD Prof. K. Rangasamy, Mr. C. Rangaraju, Dr. S. Murthy, Department of Mathematics, Govt. Arts College (Men), Krishnagiri. We are also grateful to Dr. A.V. Satyanarayana, Vice-Principal of R.L. Jalappa Institute of Technology, Doddaballapur, Prof. A.S. Hariprasad, Sai Vidya Institute of Technology, Prof. V.K. Ravi, Mr. T. Saravanan, Bangalore college of Engineering and Technology, Prof. L. Satish, Raja Rajeshwari College of Engineering, Prof. M.R. Ramesh, S.S.E.T., Bangalore. A very special thanks goes out to Mr. K.R. Venkataraj and Bros., our well wisher friend Mr. N. Aswathanarayana Setty, Mr. D. Srinivas Murthy without whose motivation and encouragement this could not have been completed. We express our sincere gratitude to Managing Director, New Age International (P) Limited, and Bangalore Division Marketing Manager Mr. Sudharshan for their suggestions and provisions of the font materials evaluated in this study. We would also like to thank our friends and students for exchanges of knowledge, skills during our course of time writing this book. AUTHORS

Dedicated to my dear parents, Shiridi Sai Baba, my dear loving son Monish Sri Sai G and my wife and best friend S. Mamatha — A . Ganesh

& my dear parents, and my wife S. Geetha — G. Balasubramanian


QUESTION PAPER LAYOUT Engineering Mathematics-II O6MAT21 Units-1, 2, 3, 4

PART-A

4 Qns.

1 Qn. from each unit

PART-B

4 Qns.

Units-5, 6, 7, 8

1 Qn. from each unit

To answer fivefull questions choosing at leasttwo questions from each part

Time: 3 Hrs.

Unit/Qn. No. 1.

Max. Marks: 100

Topics DIFFERENTIAL CALCULUS-I

Unit/Qn. No. 5.

Topics DIFFERENTIAL EQUATIONS-I Linear differential equation with constant coefficients, Solution of homogeneous and non homogeneous linear D.E., Inverse differential operator and the Particular Integral (P.I.)

Radius of Curvature: Cartesian curve Parametric curve, Pedal curve, Polar curve and some fundamental theorems.

Method of undetermined coefficients. 2.

DIFFERENTIAL CALCULUS-II

6.

Taylor’s, Maclaurin’s Maxima and Minima for a function of two variables.

3.

INTEGRAL CALCULUS-II

Method of variation of parameters, Solutions of Cauchy's homogeneous linear equation and Legendre’s linear equation, Solution of initial and Boundary value problems. 7.

VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

LAPLACE TRANSFORMS Periodic function, Unit step function (Heaviside function), Unit impulses function.

Double and triple integral, Beta and Gamma functions. 4.

DIFFERENTIAL EQUATIONS-II

8.

INVERSE LAPLACE TRANSFORMS Applications of Laplace transforms.


Contents PREFACE

(v)

ACKNOWLEDGEMENT

(vi)

UNIT I Differential CalculusI

160

1.1 Introduction 1.2 Radius of Curvature 1.2.1 Radius of Curvature in Cartesian Form 1.2.2 Radius of Curvature in Parametric Form Worked Out Examples Exercise 1.1 1.2.3 Radius of Curvature in Pedal Form 1.2.4 Radius of Curvature in Polar Form Worked Out Examples Exercise 1.2 1.3 Some Fundamental Theorem 1.3.1 Rolle’s Theorem 1.3.2 Lagrange’s Mean Value Theorem 1.3.3 Cauchy’s Mean Value Theorem 1.3.4 Taylor’s Theorem Worked Out Examples Exercise 1.3 Additional Problems (from Previous Years VTU Exams.) Objective Questions

UNIT II Differential CalculusII 2.1 Indeterminate Forms 2.1.1 Indeterminate Form

1 1 2 3 4 18 19 19 21 26 27 27 27 28 29 30 50 52 57

61104 61

0 0

Worked Out Examples Exercise 2.1 2.1.2 Indeterminate Forms ∞ – ∞ and 0 × ∞ Worked Out Examples Exercises 2.2 2.1.3 Indeterminate Forms 00, 1∞, ∞0, 0∞ Worked Out Examples Exercise 2.3 2.2 Taylor’s Theorem for Functions of Two Variables

61 62 68 69 69 74 74 74 77 78

(xii)

Worked Out Examples Exercise 2.4 2.3 Maxima and Minima of Functions of Two Variables 2.3.1 Necessary and Sufficient Conditions for Maxima and Minima Worked Out Examples Exercise 2.5 2.4 Lagrange’s Method of Undetermined Multipliers Working Rules Worked Out Examples Exercise 2.6 Additional Problems (from Previous Years VTU Exams.) Objective Questions

UNIT III Integral Calculus

78 82 83 83 84 89 89 90 91 94 94 101

105165

3.1 Introduction 3.2 Multiple Integrals 3.3 Double Integrals Worked Out Examples Exercise 3.1 3.3.1 Evaluation of a Double Integral by Changing the Order of Integration 3.3.2 Evaluation of a Double Integral by Change of Variables 3.3.3 Applications to Area and Volume Worked Out Examples Type 1. Evaluation over a given region Type 2. Evaluation of a double integral by changing the order of integration Type 3. Evaluation by changing into polars Type 4. Applications of double and triple integrals Exercise 3.2 3.4 Beta and Gamma Functions 3.4.1 Definitions 3.4.2 Properties of Beta and Gamma Functions 3.4.3 Relationship between Beta and Gamma functions Worked Out Examples Exercise 3.3 Additional Problems (From Previous Years VTU Exams.) Objective Questions

UNIT IV Vector Integration and Orthogonal Curvilinear Coordinates 4.1 Introduction 4.2 Vector Integration 4.2.1 Vector Line Integral Worked Out Examples Exercise 4.1

105 105 105 106 112 113 113 113 114 114 119 122 124 128 129 129 129 133 135 156 159 162

166213 166 166 166 166 172

(xiii)

4.3 Integral Theorem 4.3.1. Green’s Theorem in a Plane 4.3.2. Surface Integral and Volume Integral 4.3.3. Stoke’s Theorem 4.3.4. Gauss Divergence Theorem Worked Out Examples Exercise 4.2 4.4 Orthogonal Curvilinear Coordinates 4.4.1 Definition 4.4.2 Unit Tangent and Unit Normal Vectors 4.4.3. The Differential Operators Worked Out Examples Exercise 4.3 4.4.4. Divergence of a Vector Worked out Examples Exercise 4.4 4.4.5. Curl of a Vector Worked Out Examples Exercise 4.5 4.4.6. Expression for Laplacian ∇2ψ 4.4.7. Particular Coordinate System Worked Out Examples Exercise 4.6 Additional Problems (From Previous Years VTU Exams.) Objective Questions

UNIT V Differential EquationsI 5.1 Introduction 5.2 Linear Differential Equations of Second and Higher Order with Constant Coefficients 5.3 Solution of a Homogeneous Second Order Linear Differential Equation Worked Out Examples Exercise 5.1 5.4 Inverse Differential Operator and Particular Integral 5.5 Special Forms of x Worked Out Examples Exercise 5.2 Exercise 5.3 Exercise 5.4 5.6 Method of Undetermined Coefficients Worked Out Examples Exercise 5.5

173 173 173 174 174 174 188 189 189 189 191 192 194 194 195 196 196 197 198 199 199 203 208 208 210

214279 214 214 215 215 219 220 221 224 236 241 251 251 252 262

(xiv)

5.7 Solution of Simultaneous Differential equations Worked Out Examples Exercise 5.6 Additional Problems (From Previous Years VTU Exams.) Objective Questions

UNIT VI Differential EquationsI

280320

6.1 Method of Variation of Parameters Worked Out Examples Exercise 6.1 6.2 Solution of Cauchy’s Homogeneous Linear Equation and Lengendre’s Linear Equation Worked Out Examples Exercise 6.2 6.3 Solution of Initial and Boundary Value Problems Worked Out Examples Exercise 6.3 Additional Problems (From Previous Years VTU Exams.) Objective Questions

UNIT VII Laplace Transforms

280 281 291 292 294 306 308 308 310 310 318

321368

7.1 Introduction 7.2 Definition 7.3 Properties of Laplace Transforms 7.3.1 Laplace Transforms of Some Standard Functions Worked out Examples Exercise 7.1 7.3.2 Laplace Transforms of the form eat f (t) Worked Out Examples Exercise 7.2 7.3.3 Laplace Transforms of the form tn f (t) where n is a positive integer 7.3.4 Laplace Transforms of

264 265 267 268 277

=B

f t t

Worked out Examples Exercise 7.3 7.4 Laplace Transforms of Periodic Functions Worked Out Examples Exercise 7.3 7.5 Laplace Transforms of Unit Step Function and Unit Impulse Function Unit Step Function (Heaviside Function) 7.5.1 Properties Associated with the Unit Step Function

321 321 321 322 325 330 331 332 335 336 337 337 345 346 347 351 352 352 352

(xv)

7.5.2 Laplace Transform of the Unit Impulse Function Exercise 7.4 Additional Problems (From Previous Years VTU Exams.) Objective Questions

354 359 360 365

UNIT VIII Inverse Laplace Transforms 8.1 Introduction 8.2 Inverse Laplace Transforms of Some Standard Functions Worked Out Examples 8.3 Inverse Laplace Transforms using Partial Fractions Exercise 8.1 8.4 Inverse Laplace Transforms of the Functions of the Form

8.5

8.6 8.7

8.8

Worked Out Example Exercise 8.2 Convolution Theorem Worked Out Examples Exercise 8.3 Laplace Transforms of the Derivatives Solution of Linear Differential Equations Worked Out Examples Solution of Simultaneous Differential Equations Exercise 8.4 Applications of Laplace Transforms Worked Out Examples Exercise 8.5 Additional Problems (From Previous Years VTU Exams.) Objective Questions

369426

=B

Fs s

369 369 372 376 382 384 384 387 388 389 396 397 398 398 406 410 411 412 416 417 422

Model Question PaperI

427440

Model Question PaperII

441447


UNIT

1

Differential Calculus—I 1.1 INTRODUCTION In many practical situations engineers and scientists come across problems which involve quantities of varying nature. Calculus in general, and differential calculus in particular, provide the analyst with several mathematical tools and techniques in studying how the functions involved in the problem behave. The student may recall at this stage that the derivative, obtained through the basic operation of calculus, called differentiation, measures the rate of change of the functions (dependent variable) with respect to the independent variable. In this chapter we examine how the concept of the derivative can be adopted in the study of curvedness or bending of curves.

1.2 RADIUS OF CURVATURE Let P be any point on the curve C. Draw the tangent at P to the circle. The circle having the same curvature as the curve at P touching the curve at P, is called the circle of curvature. It is also called the osculating circle. The centre of the circle of the curvature is called the centre of curvature. The radius of the circle of curvature is called the radius of curvature and is denoted by ‘ρ’.

Y

C O P

Note : 1. If k (> 0) is the curvature of a curve at P, then the radius

1 . This follows from the definition k of radius of curvature and the result that the curvature of a circle is the reciprocal of its radius.

O

of curvature of the curve of ρ is

Note : 2. If for an arc of a curve, ψ decreases as s increases, then But the radius of a circle is non-negative. So to take ρ = i.e., k =

dψ . ds 1

X

Fig. 1.1

dψ is negative, i.e., k is negative. ds

1 ds = some authors regard k also as non-negative dψ k

2

ENGINEERING MATHEMATICS—II

dψ indicates the convexity and concavity of the curve in the neighbourhood of ds ds the point. Many authors take ρ = and discard negative sign if computed value is negative. dψ The sign of

∴ Radius of curvature ρ =

1 · k

1.2.1 Radius of Curvature in Cartesian Form Suppose the Cartesian equation of the curve C is given by y = f (x) and A be a fixed point on it. Let P(x, y) be a given point on C such that arc AP = s. Then we know that

dy = tan ψ dx where ψ is the angle made by the tangent to the curve C at P with the x-axis and ds = dx

R|1 + FG dy IJ S| H dx K T

2

Differentiating (1) w.r.t x, we get

U| V| W

...(1)

1 2

...(2)

dψ d2y sec 2 ψ ⋅ 2 = dx dx

d

2 = 1 + tan ψ

i ddsψ ⋅ dxds

LM1 + FG dy IJ OP 1 LM1 + FG dy IJ OP MN H dx K PQ ρ MN H dx K PQ 1 R| F dy I U| S1 + G J V ρ |T H dx K |W R|1 + FG dy IJ U| S| H dx K V| T W

1 2 2

2

=

2

=

2

Therefore,

where y1 =

ρ=

dy d2y and y2 = . dx dx 2

d2y dx 2

[By using the (1) and (2)]

3 2

3 2

...(3)

3

DIFFERENTIAL CALCULUS—I

Equation (3) becomes, ρ=

3 2 2 1

o1 + y t y2

This is the Cartesian form of the radius of curvature of the curve y = f (x) at P (x, y) on it.

1.2.2 Radius of Curvature in Parametric Form Let x = f (t) and y = g (t) be the Parametric equations of a curve C and P (x, y) be a given point on it. Then

dy dt dy = d x dt dx

...(4)

RS T

UV W

d2y d dy / dt dt ⋅ 2 = dt dx / dt dx dx

and

dx d 2 y dy d 2 x ⋅ − ⋅ dt dt 2 dt dt 2 ⋅ 1 = 2 dx dx dt dt

FG IJ H K

dx d 2 y dy d 2 x – ⋅ ⋅ d2y dt dt 2 dt dt 2 = 3 dx 2 dx dt

FG IJ H K

Substituting the values of

...(5)

dy d2y and in the Cartesian form of the radius of curvature of the dx dx 2

curve y = f (x) [Eqn. (3)]

∴

ρ=

=

R|1 + FG dy IJ o1 + y t = S|T H dx K 3 2 2 1

d2y dx 2

y2

|RS1 + FG dy / dt IJ |T H dx / dt K 2

2

2

3 2

3 2

3

2

R|FG dx IJ + FG dy IJ U| S|H dt K H dt K V| T W 2

ρ=

U| V| W

|UV |W RS dx ⋅ d y – dy ⋅ d x UV / FG dx IJ T dt dt dt dt W H dt K 2

∴

2

2

3 2

dx d 2 y dy d 2 x ⋅ − ⋅ dt dt 2 dt dt 2

...(6)

4

ENGINEERING MATHEMATICS—II 2 2 dx dy , y′ = , x″ = d x , y″ = d y 2 dt dt dt dt 2

where x′ =

ρ=

o

x′ 2 + y′ 2

t

3 2

x ′ y ″ – y′ x ″

This is the cartesian form of the radius of curvature in parametric form.

WORKED OUT EXAMPLES 1. Find the radius of curvature at any point on the curve y = a log sec Solution Radius of curvature ρ = Here,

o

t

1 + y12

3 2

y2

y = a log sec

FG x IJ H aK

1

y1 = a ×

sec

FG x IJ H aK

⋅ sec

FG x IJ tan FG x IJ ⋅ 1 H aK H aK a

FG x IJ H aK F xI 1 = sec G J ⋅ H aK a

y 1 = tan y2

2

RS1 + tan FG x IJ UV H aKW T ρ = 1 F xI sec G J H aK a RSsec FG x IJ UV T H aKW = = 1 F xI sec G J H aK a 3 2

2

Hence

2

32

2

2

∴

Radius of curvature = a sec

FG x IJ H aK

FG x IJ H aK F xI sec G J H aK F xI = a sec G J H aK a sec 3 2

FG x IJ . H aK

5


FG x IJ , show that ρ = H cK d1 + y i ρ =

2. For the curve y = c cos h

y2 · c

3 2 2 1

Solution

y2

FG x IJ H cK F xI 1 F xI y = c sin h G J × = sin h G J H cK c H cK F xI 1 y = cos h G J × H cK c RS1 + sin h FG x IJ UV c FG cos h x IJ H cKW = H cK T ρ = x 1 F xI cos h cos h G J H cK c c F xII F xI 1 F = c cos h G J = G c cos h G J J H cKK H cK c H y = c cos h

Here,

1

and

2

3 2

2

2

3 2

2

2

=

1 2 ⋅y c

y2 ρ = · c

∴

Hence proved.

3. Find the radius of curvature at (1, –1) on the curve y = x2 – 3x + 1.

d1 + y i ρ =

3 2 2 1

Solution. Where

y2

Here, Now,

( y1)(1, ( y2)(1,

∴

ρ(1,

at (1, – 1)

y = x2 – 3x + 1 y 1 = 2x – 3, y2 = 2 –1) = – 1 –1) = 2

–1)

b1 + 1g =

3 2

2

=

=

2 2 2

2

4. Find the radius of curvature at (a, 0) on y = x3 (x – a).

Solution. We have

ρ =

d

1 + y12 y2

i

3 2

at (a, 0)

6


y = x3(x – a) = x4 – x3a y 1 = 4x3 – 3ax2 y 2 = 12x2 – 6ax ( y1)(a, 0) = 4a3 – 3a3 = a3 ( y2)(a, 0) = 12a2 – 6a2 = 6a2

Here, and Now

RS1 + da i UV T W = 2

3

∴

ρ(a,

0)

6a 2

o

1 + a6

=

6a

2

t

3 2

d

1 + y12

Solution. We have

ρ =

Here

y = a sec y1 y1

and

y2

i

· πa on y = a sec 4

5. Find the radius of curvature at x =

∴

3 2

3 2

at x =

y2

πa 4

FG x IJ H aK F xI F xI 1 = a sec G J ⋅ tan G J × H aK H aK a F xI F xI = sec G J tan G J H aK H aK x 1 F xI F xI 1 × + sec G J ⋅ tan G J ⋅ = sec H aK H aK a a a 1L F xI F xI F xIO = Msec G J + sec G J tan G J P H aK H aK H aKQ aN 3

2

3

and y2 =

2

πa π π , y = sec ⋅ tan = 2 1 4 4 4

At x =

e

j

1 3 2 2 2+ 2 = a a

RS1 + e 2 j UV T W 2

∴

ρ

x=

πa 4

FG x IJ . H aK

=

=

3 2 a

3 a. 2

3 2

=

3 3 3 2

⋅a

7


FG x IJ . H 2K

π on y = 2 log sin 3

6. Find ρ at x =

Solution. We have ρ =

d1 + y i

3 2 2 1

y2

y = 2 log sin

The curve is

at x =

FG x IJ H 2K

1

× cos

sin

y2

At

x=

π ,y 3 1

and

2

y2 =

–1 π cosec 2 = – 2 2 6

=

RS1 + e 3j UV T W

=

b1 + 3g

2

∴

ρ

x=

π 3

Solution. We have ρ =

3 2

–2 3 2

–2

=

o

1 + y12

t

y2

4×2 = – 4. –2

FG 3a , 3a IJ H 2 2K

7. Find the radius of curvature at

x3

FG x IJ × 1 H 2K 2

FG x IJ H 2K F xI = cot G J H 2K F xI 1 = – cosec G J × H 2K 2 F πI = cot G J = 3 H 6K

y1 = 2 ⋅

and

π 3

3 2

at

on x3 + y3 = 3axy.

FG 3a , 3a IJ . H2 2K

y3

+ = 3axy Here, Differentiating with respect to x 3x2 + 3y2 y1 = 3a (xy1 + y) 3 ( y2 – ax) y1 = 3 (ay – x2) ⇒

y1 =

ay – x 2 y 2 − ax

...(1)

8


Again differentiating w.r.t x. ⇒

y2 =

Now, from (1), at

dy

2

ib

g d

− ax ⋅ ay1 − 2 x − ay − x 2

FG 3a , 3a IJ H 2 2K F 3a I F 3a I aG J −G J H 2K H 2K y = FG 3a IJ − a FG 3a IJ H 2K H 2K

dy

2

− ax

i

i b2 yy

1

−a

g

2

2

2

1

=

6a 2 − 9a 2 9a 2 − 6a 2

d

– 9a 2 − 6a 2 =

From (2), at

FG 3a , 3a IJ H 2 2K y2 =

d9 a

F 9a GH 2

2

− 6a

−

2

3a 2 2

i

i

= –1

I b– a − 3ag − F 3a JK GH 2 F 9a − 3a I GH 4 2 JK

2

2

–

3 2 3a 2 a × 4a – × 4a 4 4

F 3a I GH 4 JK

=

=

2

2

2

– 6a 3 – 32 = 3a 9a 4 16

Using these

ρ F 3a

GH 2 , 32a IJK

{1 + b– 1g } FG − 32 IJ H 3a K

3 2 2

=

= – ∴ Radius of curvature at

2 2 × 3a − 3a = 32 8 2


is

3a · 8 2

2

2

−

9a 2 4

I b– 3a − ag JK

...(2)

9


8. Find the radius of curvature of b2x2 + a2y2 = a2b2 at its point of intersection with the y-axis. Solution. We have ρ = Here,

o1 + t

3 2 2 y1

at x = 0

y2

b2x2 + a2y2 = a2b2

When x = 0,

a2y2 = a2b2 y2 = b 2

⇒

y =±b

i.e., the point is (0, b) or (0, – b) The curve is b2x2 + a2y2 = a2b2. Differentiating w.r. to x 2b2x + 2a2yy1 = 0

b2 x a2 y Differentiating again w.r. to x y1 = –

y2 =

F y − xy I GH y JK – b b0g =0 a bbg – b F b − 0I G J a H b K – b2 a2

1

2

2

Now at (0, b),

y1 =

2

2

y2 =

and

2

2

–b a2 i.e., Radius of curvature at (0, b) is =

∴

ρ(0,

b1 + 0g FG – b IJ Ha K

3 2

b)

=

=

– a2 b

2

∴ Radius of curvature is

a2 b

Next consider (0, – b), y1 =

– b2 0 × =0 2 –b a

y2 =

– b2 a2

FG – b – 0IJ = a H b K b

2

2

10


ρ(0,

– b)

=

b1 + 0g FG b IJ Ha K

3 2

=

a2 b

2

∴ Radius of curvature of (0, – b) is

a2 . b

9. Show that at any point P on the rectangular hyperbola xy = c2, ρ =

r3 where r is the 2c 2

distance of the point from the origin. Solution. The curve is xy = c2 Differentiating w.r. to x xy1 + y = 0

y x

y1 = – Again differentiating w.r.t. x

=

RS xy − y UV T x W R| – xy − y U| 2y –S x = V x |W x |T R|1 + FG y IJ U| S| H x K V| T W

=

d

=

dx

1

y2 = –

=

ρ=

where

x2

∴

+

y2

=

r2

2

2

2

d1 + i

and

2

3 2 2 y1

y2

xy =

3 2

2y x2

x2 + y2

i

3 2

i

3 2

2y x3 × 2 x 2

+ y2 2 xy

c2.

ρ =

10. Show that, for the ellipse

r3 . 2c2 x2 y2 a 2b 2 + 2 = 1, ρ = where p is the length of the perpen2 p3 a b

dicular from the centre upon the tangent at (x, y) to the ellipse.

11


Solution. The ellipse is

x2 y2 + = 1 a 2 b2

Differentiating w.r.t. x

2 x 2 yy1 + 2 = 0 a2 b ⇒

y1 =

Again Differentiating w.r. to x y2 =

b2 x a2 y

–

LM y − xy OP a N y Q LM y + b ⋅ x OP –b M y P a a M MN y PPQ b Ly x O – + P M a y Nb a Q LM3 x –b a y N a – b2

1

2

2

2

2

=

2

2

2

4

=

y2 =

ρ =

Now,

2

2

2

3

2

2

2

2

4

2

3

2

o1 + t

3 y12 2

y2

+

F1 + b x I GH a x JK F– b I GH a y JK 4 2

=

ρ =

da y –

4

4

4

da y 4

Taking magnitude

ρ =

The tangent at (x0, y0) to the ellipse

x x0 y y0 + 2 = 1 a2 b

2

+ b4 x 2

2

a6 y3 2

3 2

4 2

=

2

da y −

OP Q

y2 =1 b2

i

3 2

i

3 2

+ b4 x 2

×

3

a2 y3 b4

a 4b 4

+ b4 x 2

a 4b 4

i

3 2

x2 y2 + = 1 is a 2 b2

...(1)

12


Length of perpendicular from (0, 0) upon this tangent

1

=

FG x IJ + FG y IJ Ha K Hb K 2

0 2

2

0 2

a 2b2

=

a 4 y02 + b 4 x02 So, the length of perpendicular from the origin upon the tangent at (x, y) is a 2b 2

p =

a 4 y02 + b 4 x02

By replacing x0 by x and y0 by y

a 2b 2

p =

a4 y2 + b4 x2

Reciprocal and cube on both sides, we get, ⇒

1 = p3

=

da y 4

+ b4 x 2

2

i

3 2

i

3 2

a 6b 6

d

a 4 y 2 + b4 x2 4 4

a b

×

1 a b

2 2

By using eq. (1), we get ρ 1 2 2 3 = a b p

⇒

ρ =

a 2b 2 · p3

ax , 11. Show that, for the curve y = a+x Solution. Here,

y =

FG 2ρ IJ HaK

2 3

F xI F yI = G J + GH JK H yK x 2

ax a+x

Differentiating w.r.t. x

LM ba + xgb1g − x OP a MN ba + xg PQ = ba + xg 2

y1 = a Again Differentiating w.r.t. x

2 y2 = a

2

–2

=

– 2a 2

b a + x g ba + x g 3

3

2

2

·

13


ρ =

Now,

d1 + y i

3 2 2 1

y2

Substituting y1 and y2, we get

R|1 + a U| S| ba + xg V| T W R| − 2a U| S| ba + xg V| W T {ba + xg + a } – 2a b a + x g FG x IJ + FG y IJ H yK H xK R| |S 2 {ba + xg + a } || 2a ba + xg T ba + x g + a a ba + x g ba + x g + a a ba + x g 3 2

4

4

=

2

3

3 4 2

4

ρ =

To show that

L.H.S.

FG 2ρIJ HaK

23

FG 2ρ IJ HaK

2

=

2 3

2

4

=

4

=

=

using (1)

2

2

FG x IJ + FG y IJ H yK H xK

R| x U| S| FG ax IJ V| TH a + xK W ba + x g + a2

∴ L.H.S. = R.H.S. using (2) and (3).

...(2)

2

2

=

2 3

2

2

2

R.H.S.

U| |V || W

4

2

=

3 2

3

3

4

=

...(1)

3

2

2

2

R| F ax I U| S GH a + x JK V| + | T x W a2

ba + x g

2

2

...(3)

14


12. Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ). Solution. Here x = a (θ + sinθ), y = a (1 – cosθ) Differentiating w.r.t. θ

dx dy = a (1 + cos θ), = a sin θ dθ dθ dy dy a sin θ = dθ = y1 = dx dx a 1 + cos θ dθ

b

g

θ θ cos 2 2 2 θ 2 cos 2

2 sin =

y 1 = tan Again differentiating w.r.t. θ y2 = = =

θ 2

FG IJ H K d F G tan θ2 IJK × ddxθ dθ H 1 F θI 1 sec G J × × H 2 K 2 a b1 + cos θg d θ tan dx 2

2

θ 2

sec 2 =

2a × 2 cos2 y2 =

ρ =

1 4a cos4

θ 2

o1 + y t

3 2 2 1

y2

RS1 + tan θ UV 2W T R| U| 1 S| θV 4a cos | 2W T 2

=

θ 2

4

3 2

15


= =

RSsec FG θ IJ UV T H 2K W 2

3 2

1

× 4a cos θI F cos G J H 2K F θI 4 a cos G J · H 2K 3

ρ =

FG θ IJ H 2K FG θ IJ H 2K

× 4a cos4 4

13. Find the radius of curvature at the point ′θ′ on the curve x = a log sec θ, y = a (tan θ – θ). Solution x = a log sec θ, y = a (tan θ – θ) Differentiating w.r.t. θ dy dx 1 ⋅ sec θ ⋅ tan θ , = a = a (sec2 θ – 1) dθ dθ sec θ = a tan θ = a tan2 θ

∴

dy dy = dθ y1 = dx dx dθ a tan 2 θ = a tan θ y 1 = tan θ y2 = =

b g

d2y d tanθ = dx 2 dx

b g

d dθ tan θ ⋅ dθ dx

2 = sec θ ×

=

Now,

ρ =

sec 2 θ a tan θ

o1 + y t

3 2 2 1

y2

d1 + tan θi F sec θ I GH tan θ JK 2

=

1 a tan θ

2

3 2

16


sec 3 θ × a tan θ sec 2 θ ρ = a sec θ tan θ. =

14. For the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), show that the radius of curvature at ′θ′ varies as θ. Solution x = a (cos θ + θ sin θ)

dx = a (– sin θ + θ cos θ + sin θ) = a θ cos θ dθ y = a (sin θ – θ cos θ)

⇒

dy = a (cos θ + θ sin θ – cos θ) = a θ sin θ dθ

⇒

y1 =

dy a θ sin θ = = tan θ dx a θ cos θ

y2 =

d dy d = tan θ dx dx dx

=

FG IJ b H K d dθ tan θg ⋅ b dθ dx

2 = sec θ ×

=

ρ =

Now,

=

g

1 a θ cos θ

1 a θ cos3 θ

d

1 + y12

i

3 2

y2

d i F 1 I GH a θ cos θ JK 1 + tan 2 θ

3 2

3

= sec3 θ × a θ cos3 θ = aθ ρ ∝ θ.

i.e.,

15. If ρ1 and ρ2 are the radii of curvatures at the extremities of a focal chord of the parabola

b g . Solution. If dat , 2at i and dat , 2at i are the extremities of a focal chord of the parabola

y2 = 4ax, then show that ρ 1 2 1

y2 = 4ax. Then

1

–2 3

+ ρ2

–2 3

= 2a

2 2

t1 · t2 = – 1

2

−2 3

17


The parametric equations to x x′ x″

∴

the parabola are y = 2at = at2, = 2at y′= 2a = 2a y″ = 0

ρ =

ox ′

2

+ y′2

t

3 2

x′ y ″ – x″ y′

{b2at g + b2ag } d – 4a i

3 2 2

2

=

2

= –

d

8a 3 1 + t 2 4a

2

d

i

2 ρ = − 2a 1 + t

ρ

–2 3

–2 ρ3

= =

i

3 2

3 2

b2ag d1 + t i –2 3

1

2

−1

1

×

b2ag d1 + t i 2 3

2

Let t1 and t2 be extremities of a focal chord. Then t2 = – –2

–2

ρ1 3

Now,

3 = ρt = t1 =

–2

ρ 23

= ρ

= –2

Adding

–2

ρ1 3 + ρ23

=

=

i.e.,

–2 3 1

ρ

–2

+ ρ 23

=

–2 3 t=

–1 t1

1

1

×

1

b2ag d1 + t i 2 3

=

×

1

2 1

1

×

b2ag d1 + t i 2 3

2 1

t12

b2ag d1 + t i 1 R 1 ST1 + t + 1 +t t UVW b2 a g b2ag × d11 ++ tt i d i 2 3

2 1

2 1

2 3

–2 3

b2 a g

–2 3

2 1

2 1

2 1 2 1

· Hence proved.

1 · t1

18


EXERCISE 1.1 1. Find ρ at any point on y = log sin x. 2. Find ρ at x = 1 on y =

Ans. cosec x

log x . x

FG x IJ · H 2K

π 3. Find the radius of curvature at x = on y = log tan 4

x2 y2 + 4. Find the radius of curvature of (3, 4) on = 2. 9 16

5. Find the radius of curvature at (x1, y1) on b2x2 – a2y2 = a2b2.

LMAns. N

OP 2 2Q LM O 3I P F MNAns. 2 × GH 2 JK PQ LMAns. 125OP N 12 Q LM b x + a y OP MNAns. e a b j PQ 3 2

2 2 1

Ans. 4 2

d4 x

7. Show that ρ at any point on 2xy = a2 is

4

+ a4

i

3 2

8a 2 x 3

8. Show that ρ at (0, 0) on y2 = 12x is 6. 9. Find radius of curvature at x = 2 on y2 =

b

·

g·

LMAns. 1 OP N 3Q LMAns. a OP 2Q N LMAns. 5 5 a OP 6 PQ MN

x x−2 x −5

10. Find ρ at (a, a) on x3 + y3 = 2a3.

11. Find the radius of curvature at (a, 2a) on x2y = a(x2 + a2). 12. Find the radius of curvature at (–2a, 2a) on x2y = a (x2 + y2). 13. Show that ρ at (a

a

sin3θ)

on

3 4 2 2 1

4 4

6. Find ρ at (4, 2) on y2 = 4 (x – 3).

cos3θ,

3

2 x3

+

2 y3

=

2 a3

Ans. 2a

is 3a sin θ cos θ.

π on x = a sin θ, y = b cos 2θ. 14. Find radius of curvature at θ = 3

LM a MNAns. e

15. Find ρ for x = t – sin ht cos ht, y = 2cos ht.

Ans. 2 cos h 2 t sin ht

2

+ 12b 2 4ab

O j PP Q 3 2

19


1.2.3 Radius of Curvature in Pedal Form Let polar form of the equation of a curve be r = f (θ) and P(r, θ) be a given point on it. Let the tangent to the curve at P subtend an angle ψ with the initial side. If the angle between the radius vector OP and the tangent at P is φ then we have ψ = θ + φ (see figure). Let p denote the length of the perpendicular from the pole O to the tangent at P. Then from the figure,

r = f (G)

OM p = OP r p = r sin φ

B

r

O

P (r, G)

O

G

X

sin φ = Hence, ∴

dψ dθ dφ dθ dφ dr 1 + = + ⋅ = = ds ds ds ds dr ds ρ

dθ ds dr ds

i.e.,

Hence,

sin φ = r ⋅

From (2),

dr ds

LM N

1 dφ sin φ + r cos φ ⋅ r dr

b

1 d ⋅ r sin φ r dr r sin φ = p

=

Since,

dθ ds

sin φ dφ 1 + cos φ ⋅ = r dr ρ

=

g

OP Q

dr dp This is the Pedal form of the radius of curvature.

Therefore,

M

r⋅

sin φ = cos φ

cos φ =

...(2)

p

Fig. 1.2

dθ dr

We know that tan φ = r ⋅

and

...(1)

ρ = r⋅

...(3)

1.2.4 Radius of Curvature in Polar Form Let r = f (θ) be the equation of a curve in the polar form and p(r, θ) be a point on it. Then we know that 1 1 1 Differentiating w.r.t. r, we2 get = 2 + 4 p r r

FG dr IJ H dθ K

2

...(4)

20


FG IJ d– 4r i + 1 ⋅ 2 ⋅ dr ⋅ d FG dr IJ H K dθ dr H dθ K r – 2 4 F dr I 2 dr d r dθ ⋅ ⋅ ⋅ − .G J + r r H dθ K r dθ dθ dr – 2 4 F dr I 2 d r + ⋅ − G J H K r r dθ r dθ L 1 2 F dr I – 1 d r OP p M + MN r r GH dθ JK r dθ PQ 2

– 2 dp –2 dr ⋅ + = 3 3 p dr dθ r

–5

4

2

=

3

2

5

4

2

2

=

Hence, Now,

dp = dr

3

2

5

4

2

2

2

3

3

ρ = r⋅

5

dr = dp

=

r2 + 2

4

2

r

p3

R| 1 + 2 FG dr IJ S| r r H dθ K T

r6 ⋅

1 p3

3

FG dr IJ H dθ K

5

2

−r⋅

2

−

1 d 2r r 4 dθ 2

U| V| W

d 2r dθ 2

By using equation (4),

R| 1 1 FG dr IJ U| r ⋅S + V T| r r H dθ K W| F dr I d r r +2G J –r⋅ H dθ K dθ

3 2

2

6

2

=

4

2

2

2

2

R|r + FG dr IJ U| S| H dθ K V| T W F dr I d r +2G J −r⋅ H dθ K dθ 2

2

ρ=

2

r

where

∴

r1 =

ρ=

2

3 2

2

2

dr d 2r , r = · dθ 2 dθ 2

o

r 2 + r12

r + 2

2r12

t

3 2

− r r2

This is the formula for the radius of curvature in the polar form.

...(5)

21


WORKED OUT EXAMPLES 1. Find the radius of the curvature of each of the following curves: (i) r3 = 2ap2 (Cardiod) (ii) p2 = ar

1 1 1 r2 = + − (Ellipse). p 2 a 2 b 2 a 2b 2

(iii)

Solution. (i) Here r3 = 2ap2 Differentiating w.r.t. p, we get

3r 2 ⋅

dr = 4ap dp 4ap dr = dp 3r 2

⇒

ρ = r⋅

Hence,

Fr I where p = G J H 2a K 3

dr 4ap 4ap =r⋅ 2 = dp 3r 3r

1 2

Fr I 4a ⋅ G J H 2a K 3

ρ =

1 2

3r 3

4a r 2 2 2ar = = 3 3r 2a (ii) Here

p2 = ar

Differentiating w.r.t. p, we get

dr dp

Then

2p = a ⋅

⇒

2p dr = a dp

where p =

ar · 3

dr 2 ⋅ ar 2r 2 =r⋅ = ρ = r dp a a (iii) Given

1 p2

=

1 1 r2 + – a2 b 2 a 2b 2

Differentiating w.r.t. p, we get –2 –1 dr = 2 2 2r ⋅ p3 dp a b

22


Hence Therefore,

dr a 2b 2 = dp p 3r ρ = r.

a 2b 2 a 2b 2 dr =r. 3 = 3 dp p r p

2. Find the radius of curvature of the cardiod r = a (1 + cos θ) at any point (r, θ) on it. Also

ρ2 is a constant. r Solution. Given r = a (1 + cos θ) Differentiating w.r.t. θ dr = – a sin θ r1 = dθ

prove that

d 2r = – a cos θ and r2 = dθ 2 ∴ The radius of curvature in the polar form ρ =

or

2

r + 2

+

t

3 2 2 r1

2r12

− r r2

g + a sin θ} { b a b1 + cos θg + 2a sin θ − ab1 + cos θgb – a cos θg a o1 + 2 cos θ + cos θ + sin θt a o1 + 2 cos θ + cos θ + 2 sin θ + cos θ + cos θt a m2 b1 + cos θgr 3 b1 + cos θg 2 2 a b1 + cos θg a 2 1 + cos θ

=

2

2

2

2

2

2

3 2

2

2

3

=

2

2

2

3 2

2

2

3 2

=

1 2

=

3 2

=

F θI 2 a G 2 cos J H 2K 2

3 4 θ a cos ρ = 3 2 Squaring on both sides, we get ρ2 =

16 2 θ a cos2 9 2

b

8 a2 1 + cos θ = 9

1 2

LM where cos θ = 1 b1 + cos θgOP 2 2 N Q LM where 1 + cos θ = r OP aQ N 2

g

23


ρ2

8a 2 r 8ar ⋅ = = 9 a 9

8a ρ2 = which is constant. 9 r

Hence,

3. Show that for the curve rn = an cos nθ the radius of curvature is rn

an

= cos nθ Solution. Here Taking logarithms on both sides, we get n log r = n log a + log cos nθ Differentiating w.r.t. θ, we have

an · n + 1 rn –1

b g

n dr n sin nθ = 0– r dθ cos nθ

dr = – r tan nθ dθ Differentiating w.r.t. θ again, we obtain r1 =

r2 =

RS T – onr sec

dr d 2r − rn sec 2 nθ + tan nθ ⋅ 2 = dθ dθ

=

t

nθ − r tan 2 nθ

2

= r tan2 nθ – nr sec2 nθ Using the polar form of ρ, we get ρ =

=

=

= = =

o

r 2 + r12

r + 2

2r12

t

3 2

− rr2

or

b

2

r 2 + 2 – r tan nθ

t

+ r 2 tan 2 nθ

g

2

d

3 2

− r r tan 2 nθ − nr sec 2 nθ

i

r 3 sec 3 nθ r 2 1 + 2 tan 2 nθ − tan 2 nθ + n sec 2 nθ

r sec 3 nθ n + 1 sec 2 nθ

b g

r n + 1 cos nθ

b g

r

bn + 1g FGH ar IJK n

n

=

UV W

an · n + 1 rn–1

b g

LMwhere cos nθ = r OP a Q N n

n

24


4. Find the radii of curvature of the following curves: (ii) r (1 + cos θ) = a (i) r = aeθ cot α (iii) θ =

r2 − a2

– cos –1

FG a IJ · H rK

a Solution. (i) Here r = aeθ Differentiating w.r.t θ

cot α

dr = aeθ cot α · cot α dθ = r · cot α

So that,

tan φ =

=

r dr dθ r = tan α r cot α

Hence, φ = α, since p = r sin φ We get, p = r sin α. This is the Pedal equation of the given curve. From which, we get

dr 1 = dp sin α Hence,

p = r⋅

dr = r cosec α. dp

(ii) Given equation of the curve is r (1 + cos θ) = a Differentiating w.r.t. θ, we get r (– sin θ) + (1 + cos θ) ·

dr =0 dθ

dr r sin θ = dθ 1 + cos θ

or

We have,

1 p2

1 1 = 2 + 4 r r

=

FG dr IJ H dθ K

2

1 1 r 2 sin 2 θ + ⋅ r 2 r 4 1 + cos θ 2

b

g 1 L sin θ O M P 1+ r M b1 + cos θg P N Q 1 L b1 + cos θg + sin M r M N b1 + cos θg 2

=

2

2

2

=

2

2

2

θ

OP PQ

25


= = where 1 + cos θ =

LM 2 b1 + cos θg OP MN b1 + cos θg PQ

1 r2

r

2

2 1 + cos θ

b

2

g

a r 1 p2

=

2 a r2 ⋅ r

=

2 ar

ar which is the pedal equation of the curve. 2 Differentiating w.r.t. p, we get Hence,

p2 =

2p =

a dr ⋅ 2 dp

⇒

4p dr = a dp

∴

ρ = r⋅

dr dp

= r⋅

4p where p = a

= r.

4 a

ar 2

ar 2 3

a r2 .

= 2 2 (iii) Here,

Then,

θ =

r2 − a2

− cos–1

a

dθ 2r + = dr a ⋅ 2 ⋅ r2 − a2

=

=

r a r −a 2

2

–

r2 − a2 ar r 2 − a 2

FG a IJ H rK 1

F1 − a GH r a

r r − a2 2

2 2

FG – a IJ I HrK JK 1 2

26


so that

dθ = dr

r2 − a2 ar

dr = dθ

ar

r2 − a2 We have the Pedal equation, we get 1 p2

1 p2

FG dr IJ H dθ K

2

=

1 1 + 4 2 r r

=

1 1 a 2r 2 + ⋅ r2 r4 r2 − a2

=

1 a2 1 + r2 r2 − a2

=

1 r − a2

RS T

d

UV W

i

2

p2 = r2 – a2

Hence From this we get

p dr = r dp ρ = r⋅

∴

p = p = r2 − a2 ⋅ r

EXERCISE 1.2 1. Find the radius curvature at the point ( p, r) on each of the following curves:

LMAns. r OP N aQ LMAns. a OP N 3r Q LM Ans. a OP MN bn + 1g r PQ LM a + r OP MMAns. d r + 2ai PP MN PQ 3

(i) pr =

a2

(ii)

r3

a2p

(iii)

pan

(Hyperbola)

2

2

=

(Lemniscate)

n

(iv) p =

=

r n+1

(Sine spiral)

r4 (Archimedian spiral) r2 + a2

n −1

2

3 2 2

2

2

27


2. Find the radius of curvature at (r, θ) on each of the following curves:

LM r a + r OP MM Ans. d a i PP MN PQ LMAns. a OP N 3rQ LMAns. r OP N aQ LMAns. r OP N 3p Q 3 2 2

2

(i) r =

a θ

2

(iii)

r2

=

a2

cos 2θ

3

(v) r2 cos 2θ = a2

2

4

(vii) r = a sec 2θ

2

LMAns. a OP N 2Q

(ii) r = a cos θ

3

LM Ans. a OP (iv) = sin nθ MN bn + 1g r PQ LMAns. 2 ar OP a (vi) r = b1 − cos θg 3 PQ 2 MN LM Ans. na OP (viii) r = a sin nθ N 2Q n

rn

an

n −1

3. If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode 2 2 r = a (1 + cos θ) which passes through the pole. Prove that ρ1 + ρ2

=

16a 2 · 9

1.3 SOME FUNDAMENTAL THEOREM 1.3.1 Rolle’s Theorem If a function f (x) is 1. continuous in a closed interval [a, b], 2. differentiable in the open interval (a, b) and 3. f (a) = f (b). Then there exists at least one value c of x in (a, b) such that f ′ (c) = 0 (No proof).

1.3.2 Lagrange’s Mean Value Theorem Suppose a function f (x) satisfies the following two conditions. 1. f (x) is continuous in the closed interval [a, b]. 2. f (x) is differentiable in the open interval (a, b). Then there exists at least one value c of x in the open interval (a, b), such that

bg

f (b) – f a b–a

= f ′ (c)

28


Proof. Let us define a new function φ(x) = f (x) – k·x

...(1)

where k is a constant. Since f (x), kx and φ (x) is continuous in [a, b], differentiable in (a, b). From (1) we have,

φ (a) = f (a) – k·a φ (b) = f (b) – k·b

∴

φ (a) = φ (b) holds good if f (a) – k·a = f (b) = k·b

i.e.,

k (b – a) = f (b) – f (a)

or

k =

bg bg

f b − f a b−a

...(2)

Hence, if k is chosen as given by (2), then φ (x) satisfy all the conditions of Rolle’s theorem. Therefore, by Rolle’s theorem there exists at least one point c in (a, b) such that φ′(c) = 0. Differentiating (1) w.r.t. x we have, φ′(x) = f ′(x) – k φ′(c) = 0 gives f ′(c) – k = 0

and

k = f ′(c)

i.e.,

...(3)

Equating the R.H.S. of (2) and (3) we have

bg bg

f b − f a b−a

= f ′(c)

...(4)

This proves Lagrange’s mean value theorem.

1.3.3 Cauchy’s Mean Value Theorem If two functions f (x) and g (x) are such that 1. f (x) and g (x) are continuous in the closed interval [a, b]. 2. f (x) and g (x) are differentiable in the open interval (a, b). 3. g′ (x) ≠ 0 for all x ∈ (a, b). Then there exists at least one value c ∈ (a, b) such that

bg g bbg – g b a g

f (b) – f a

=

b g· g ′ b cg f′ c

Proof: Let us define a new function φ (x) = f (x) – kg (x)

...(1)

where k is a constant. From the given conditions it is evident that φ (x) is also continuous in [a, b], differentiable in (a, b). Further (1), we have φ (a) = f (a) – k g (a); φ (b) = f (b) – k g (b)

29


∴

φ (a) = φ (b) holds good if f (a) – k g (a) = f (b) – k g (b)

i.e.,

k [g (b) – g (a)] = f (b) – f (a)

or

k =

bg bg gbbg − gb a g f b − f a

...(2)

Here, g (b) ≠ g (a). Because if g (b) = g (a) then g (x) would satisfy all the conditions Rolle’s theorem at least one point c in (a, b) such that g′(c) = 0. This contradicts the data that g′(x) ≠ 0 for all x in (a, b). Hence if k is chosen as given by (2) then φ (x) satisfy all the conditions of Rolle’s theorem. Therefore by Rolle’s theorem there exists at least one value c in (a, b) such that φ′(c) = 0. Differentiating (1) w.r.t. x we have, φ′(x) = f ′(x) – kg′(x) and φ′(c) = 0 f ′(c) – kg′(c) = 0

gives

f ′(c) = kg′(c)

i.e., Thus,

k =

bg bg

f′c

...(3)

g′ c

Equating the R.H.S. of (2) and (3) we have

bg bg bg bg

f b − f a gb −ga

=

bg g ′b c g f′c

This proves mean value theorem.

1.3.4 Taylor’s Theorem Taylor’s Theorem for a function of a single variable and Maclaurin’s series function: Suppose a function f (x) satisfies the following conditions: (1) f (x), f ′ (x), f ′′ (x), ..... f (n–1) (x) are continuous in the closed interval [a, b]. (2) f (n–1) (x) is differentiable i.e., f (n) (x) exists in the open interval (a, b). Then there exists a point c ∈ (a, b) such that f (b) = f (a) + (b – a) f ′ (a) +

where Rn =

bb – ag n!

bb – ag 2!

2

bb – ag bn – 1g!

n –1

f ′′ (a) + .......... +

f (n–1) (a) + Rn

n

f n (c)

...(1)

Taylor’s theorem is more usually written in the following forms. Substitute b = x in (1)

b x – ag f (x) = f (a) + (x – a) f ′(a) + 2!

2

b x − ag f ′′(a) + 3!

3

x n−1 + .... + f (n–1) (a) + Rn (x) + .... n −1 !

b g

...(2)

30


where Rn (x) =

b x – ag n!

n

fb

n

g b xg .

Thus f (x) can be expressed as the sum of an infinite series. This series is called the Taylor’s series for the function f (x) about the point a. If we a substitute a = 0, in Eqn. (2), we get f (x) = f (0) + x f ′(0) +

x2 x3 xn n f ′′(0) + f ′′′(0) + ...... + f (0) + ..... 2! 3! n!

...(3)

This is called the Maclaurin’s series for the function f (x). If f (x) = y and f ′ (x), f ′′ (x), .......... are denoted by y1, y2, ...... the Maclaurin’s series can also be written in the form: y = y (0) + xy1 (0) +

x2 xn y2 (0) + ...... y (0) + ...... 2! n! n

WORKED OUT EXAMPLES 1. Verify Rolle’s theorem for the function f (x) = x2 – 4x + 8 in the interval [1, 3]. Solution f (x) = x2 – 4x + 8 is continuous in [1, 3] and f ′ (x) = 2x – 4 exist for all values in (1, 3) ∴

f (1) = 1 – 4 + 8 = 5; f (3) = 32 – 4 (3) + 8 = 5

∴

f (1) = f (3)

Hence all the three conditions of the theorem are satisfied. Now consider

f ′ (c) = 0 2c – 4 = 0 ⇒ 2c = 4

i.e.,

4 = 2 ∈ (1, 3) 2 and hence Rolle’s theorem is verified. c =

2. Verify Rolle’s theorem for the function f (x) = x (x + 3)e–x/2 in the interval [–3, 0]. Solution

f (x) = x (x + 3)e–x/2 is continuous in [–3, 0] f ′(x) = (x2 + 3x)

and

FG – 1 IJ e H 2K

–x/2

1 2 (x – x – 6)e–x/2 2 Therefore f ′(x) exists (i.e., finite) for all x = –

Also

f (–3) = 0, f (0) = 0

+ (2x + 3)e–x/2

...(4)

31


So that

f (–3) = f (0)

Hence all the three conditions of the theorem are satisfied. f ′ (c) = 0

Now consider

–1 2 (c – c – 6)e–c/2 = 0 2

i.e.,

c2 – c – 6 = 0 (c + 2) (c – 3) = 0 c = 3 or –2 Hence there exists – 2 ∈ (–3, 0) such that f ′ (–2) = 0 and hence Rolle’s theorem is verified. 3. Verify Rolle’s theorem for the function f (x) = (x – a)m (x – b)n in [a, b] where m > 1 and n > 1. f (x) = (x – a)m (x – b)n is continuous in [a, b]

Solution

f ′ (x) = (x – a)m · n (x – b)n–1 + m (x – a)m–1 (x – b)n = (x – a)m–1 (x – b)n–1 [n (x – a) + m (x – b)] f ′ (x) = (x – a)m–1 (x – b)n–1 [nx – na + mx – mb] = (x – a)m–1 (x – b)n–1 [(m + n) x – (na + mb)]

...(1)

f ′ (x) exists in (a, b) Also

f (a) = 0 = f (b)

Hence all the conditions of the theorem are satisfied. f ′ (c) = 0

Now consider m–1

From (1) (c – a)

(c – b)n–1 [(m + n) c – (na + mb)] = 0

⇒

c – a = 0, c – b = 0, (m + n) c – (na + mb) = 0

i.e.,

c = a, c = b, c =

na + mb m+n

a, b are the end points. c=

na + mb is the x-coordinate of the point which divides the line joining [a, f (a)], [b, f (b)] m+n

internally in the ratio m : n. ∴

c =

na + mb ∈ (a, b) m+n

Thus the Rolle’s theorem is verified.

32


4. Verify Rolle’s theorem for the following functions: (i) sin x in [– π, π]

(ii) ex sin x in [0, π]

sin x in [0, π] ex

(iii)

(iv) ex (sin x – cos x) in

Solution f (x) = sin x is continuous in [– π, π]

(i)

f ′ (x) = cos x exists in [– π, π] f (– π) = sin (– π) = 0

and also

f (π) = sin π = 0 f (– π) = f (π)

so that

Thus f (x) satisfies all the conditions of the Rolle’s theorem satisfied. Now consider

f ′ (c) = 0

∴

cos c = 0 so that c = ±

π Both these values 2

lie in (– π, π). π ∈ (– π, π) 2 Hence Rolle’s theorem is verified.

∴

c = ±

f (x) = ex sin x is continuous in [0, π]

(ii)

f ′ (x) = ex cos x + sin x · ex = ex (cos x + sin x) f ′ (x) exists in (0, π) f (0) = e0 sin (0) = 0

And also

f (π) = eπ sin (π) = 0 ∴

f (0) = f (π) = 0

Therefore f (x) satisfies all the conditions of Rolle’s theorem. Now consider

f ′ (c) = 0

c

e (cos c + sin c) = 0 cos c + sin c = 0 sin c = – cos c as ec ≠ 0 or

tan c = –1 c =

–π 3π or c = 4 4

3π ∈ (0, π) 4 Hence Rolle’s theorem is verified. Hence there exists c =

LM π , 5π OP N4 4 Q

33


(iii)

f (x) = f ′ (x) = f ′ (x) =

sin x is continuous in (0, π) ex

b

g

e x cos x – sin xe x e2x

cos x – sin x ex

...(1)

f ′ (x) exists in (0, π) Also

f (0) =

sin 0 =0 e0

sin π =0 eπ f (0) ≠ f (π) = 0

f (π) =


Consider From (1),

cos c – sin c = 0 ec cos c – sin c = 0 sin c = cos c tan c = 1 c =

π 4

π ∈ (0, π) 4 ∴ Rolle’s theorem is verified.

Hence this exists

c =

(iv) Let f (x) = ex (sin x – cos x) is continuous in

LM π , 5π OP N4 4 Q

f ′ (x) = ex (cos x + sin x) + ex (sin x – cos x) f ′ (x) = 2ex sin x f (x) is differentiable in

f

FG π IJ H 4K

FG π , 5π IJ H4 4 K π /4 = e

f

FG 5π IJ H4K

FG π – cos π IJ H 4 4K FG 1 – 1 IJ = 0 H 2 2K FG sin 5π – cos 5π IJ H 4 4K

π /4 sin = e

5π / 4 = e

...(1)

34


FG H

5π/ 4 − = e

∴

f

FG π IJ H 4K

= f

FG 5π IJ H 4K

1 2

IJ 2K

1

+

= 0

= 0


Now consider

From (1), we have, 2 e sin c = 0 ( ec ≠ 0) c

∴

sin c = 0 = sin nπ c = nπ where n = 0, 1, 2, ..... c = π∈

But

FG π , 5π IJ H4 4 K

Thus Rolle’s theorem is satisfied.

LM x + ab OP MN ba + bg x PQ 2

5. Show that the constant c of Rolle’s theorem for the function f (x) = log a ≤ x ≤ b where 0 < a < b is the geometric mean of a and b. Solution. The given f (x) is continuous in [a, b] since 0 < a < b

in

LM x + ab OP MN ba + bg x PQ 2

f (x) = log

The given function can be written in the form f (x) = log (x2 + ab) – log (a + b) – log x f ′ (x) =

2x 1 –0– x x + ab 2

d

2 x 2 – x 2 + ab =

dx

2

i

+ ab x

i=

x 2 – ab

dx

2

f ′ (x) exists in (a, b)

L a + ab OP = log 1 = 0 f (a) = log M MN ba + bg a PQ L b + ab OP = log 1 = 0 f (b) = log M MN ba + bg b PQ 2

Also

2

∴

f (a) = f (b) = 0

Hence all the conditions of the theorem are satisfied. Now consider

i

+ ab x

f ′ (c) = 0

...(1)

35


c 2 – ab

dc

From (1), we have

i

+ ab c

2

= 0

c2 – ab = 0 i.e., ∴ that

c = ±

ab

c = +

ab ∈ (a, b) and we know

ab is the geometric mean of a and b.

6. Verify Lagrange’s mean value theorem for the following functions: (i) f (x) = (x – 1) (x – 2) (x – 3) in [0, 4]

(iv) f (x) = sin–1 x in [ 0, 1].

(iii) f (x) = log x in [1, e] Solution

LM π OP N 2Q

(ii) f (x) = sin2 x in 0,

bg bg

f b −f a = f ′ (c) b−a f (x) = (x – 1) (x – 2) (x – 3) is continuous in [0, 4] (i) We have the theorem

a f (b) f (a) We have f (x) ∴ f ′ (x) The theorem becomes,

∴

bg bg

f 4 – f 0 4–0

= = = = =

0, b = 4 by data f (4) = 3·2·1 = 6 and f (0) = (–1) (–2) (–3) = – 6 x3 – 6x2 + 11x – 6 in the simplified form. 3x2 – 12x + 11 exists in (0, 4)

= 3c2 – 11c + 11

b g

6 – –6 = 3c2 – 12c + 11 4 3c2 – 12c + 8 = 0

or ∴

c = c =

i.e.,

12 ± 144 – 96 6 12 ± 48 6

12 + 48 12 – 48 , 6 6 c = 3.15 and 0.85 both belongs to (0, 4) ∴ c = 3.15 and 0.85 ∈ (0, 4) Thus the theorem is verified.

c =

36


bg bg

f b – f a

(ii) We have the theorem

b–a

= f ′ (c)

LM π OP N 2Q

f (x) = sin2 x is continuous in 0,

Here,

f ′ (x) = 2 sin x cos x = sin 2x ∴ f (x) is differentiable in

FG 0, π IJ H 2K

π 2 f (a) = f (0) = sin2 (0) = 0

with

a = 0, b =

f (b) = f

FG π IJ H 2K

= sin 2

FG π IJ H 2K

= 1

The theorem becomes

f

FG π IJ – f b0g H 2K π –0 2

= sin 2c

1– 0 = sin 2c π 2 2 = sin 2c π

i.e.,

FG 2 IJ H πK 1 F 2I sin G J H πK 2

2c = sin–1

–1

c =

= 0.36 which lies between 0 and Here,

c = 0.36 ∈

Thus the theorem is verified. (iii) We have the theorem

FG 0, π IJ H 2K

bg bg

f b – f a b–a

π 2

= f ′ (c)

f (x) = log x is continuous in [1, e] f ′ (x) =

1 f (x) is differentiable in (1, e) x

37


with

a = 1, b = e f (a) = f (1) = log 1 = 0 f (b) = f (e) = log e = 1 The theorem becomes,

bg bg

f e – f 1 e –1

1 c

=

1 1– 0 = c e –1

1 1 = c e–1 c = e–1 2 < e < 3 since 1 < e – 1 < 2 < e c = e – 1 ∈ (1, e)

So that

Thus the theorem is verified. (iv) we have the theorem

b g b g = f ′ bcg b–a

f b – f a

f (x) = sin–1 x is continuous in [0, 1] f ′ (x) =

1 1 – x2

which exists for all x ≠ 1

Therefore f (x) is differentiable in (0, 1) By Lagrange’s theorem

bg b g

f 1 – f 0 1– 0

= f ′ (c)

sin–1 1 – sin–1 0 =

i.e.,

π –0 = 2 π = 2

i.e.,

1 1 – c2 1 1 – c2 1 1– c

c2 = 1 – c2 =

2

4 π2

π2 – 4 π2

⇒

4 = 1 – c2 π2

38


π2 – 4 = ± 0.7712 π

c = ±

or

Hence c = 0.7712 lies between 0 and 1. Therefore Lagrange’s theorem is verified for f (x) = sin–1 x in [0, 1]. 7. Prove that

b–a 1 – a2

< sin–1 b – sin–1 a <

b–a 1 – b2

where a < b < 1.

Solution f (x) = sin–1 x

Let ∴

1

f ′ (x) =

1 – x2

f (x) is continuous in [a, b] and f (x) is differentiable in (a, b) Applying Lagrange’s mean value theorem for f (x) in [a, b], we get a < c < b

sin –1 b – sin –1 a = b–a

1

...(1)

1 – c2

Now a < c ⇒ a2 < c2 ⇒ – a2 > – c2 ⇒ 1 – a2 > 1 – c2

1

Hence

1 – a2

1

<

...(2)

1 – c2

Also, c < b on similar lines, 1 1 – c2

1

<

...(3)

1 – b2

Combining (2) and (3), we get

1 1– a

2

1

or

1 – a2

1

<

1– c

<

2

<

1 1 – b2

sin –1 b – sin –1 a < b–a

1 1 – b2

On multiplying by (b – a) which is positive, we have

b–a 1– a

2

< sin–1 b – sin–1 a <

b–a 1 – b2

by using (1)

39


8. Show that b–a b–a π 3 4 π 1 –1 –1 , if 0 < a < b and deduce that + < tan –1 < + · 2 < tan b – tan a < 1 + a2 1–b 4 25 3 4 6

Solution Let f (x) = tan–1 x is continuous in [a, b], a > 0. 1 , f (x) is differentiable in (a, b). 1 + x2

Hence f ′ (x) =

Applying Lagrange’s theorem, we get

bg bg

f b – f a b –a

= f ′ (c) for some c : a < c < b

1 tan –1 b – tan –1 a = 1 + c2 b–a

Now c > a

∴

⇒

c 2 > a2

1 + c2 > 1 + a2

1 1 2 < 1+ c 1 + a2

∴ and c < b

⇒

⇒

c2 < b2

⇒

1 1 2 > 1+ c 1 + b2

1 1 1 < 2 < 2 1+b 1+ c 1 + a2 1 tan –1 b – tan –1 a 1 < < using (1) 1 + b2 b–a 1 + a2 On multiplying (b – a), we get b–a b–a –1 –1 2 < tan b – tan a < 1+b 1 + a2

In particular if a = 1, b =

...(2)

1 + c2 < 1 + b2

From (2) and (3), we obtain

i.e.,

...(1)

4 then 3

4 4 –1 –1 –1 4 –1 3 – tan 1 < 3 < tan 16 1+ 1 3 1+ 9 π 3 4 π 1 + < tan –1 < + 4 35 3 4 6

Hence proved.

...(3)

40


9. Verify Cauchy’s mean value theorem for the following pairs of functions. (i) f (x) = x2 + 3, g (x) = x3 + 1 in [1, 3].

LM π OP . N 2Q

(ii) f (x) = sin x, g (x) = cos x in 0, (iii) f (x) = ex, g (x) = e–x in [a, b], Solution (i) We have Cauchy’s mean value theorem

bg bg g bbg – g ba g f b – f a

Here,

∴

=

bg g ′ b cg f′ c

a = 1, b = 3 f (x) = x2 + 3 g (x) = x3 + 1 f ′ (x) = 2x

g′ (x) = 3x2 f (x) and g (x) are continuous in [1, 3], differentiable in (1, 3) – x ∈ (1, 3) g′ (x) ≠ 0 V Hence the theorems becomes

bg bg g b3g – g b1g f 3 – f 1

=

2c 3c 2

2 8 2 12 – 4 ⇒ = = 3c 26 3c 28 – 2 2 1 = 13 3c

or

c =

13 1 = 2 ∈ (1, 3) 6 6

Thus the theorem is verified. (ii) We have Cauchy’s mean value theorem

bg bg bg bg

f b – f a g b –g a Here,

=

bg bg

f′ c g′ c

f (x) = sin x g (x) = cos x f ′ (x) = cos x

∴

g′ (x) = – sin x g′ (x) ≠ 0

41


LM N

Clearly both f (x) and g (x) are continuous in 0, from Cauchy’s mean value theorem

FG π IJ – f b0g H 2K F πI g G J – g b 0g H 2K f

=

π 2

OP and differentiable in FG 0, π IJ . Therefore H 2K Q

bg bg

f′ c π for some c : 0 < c < 2 g′ c

1– 0 cos c = 0–1 – sin c

i.e.,

–1 = – cot c or cot c = 1 ∴

c =

π 4

Clearly

c =

π π ∈ 0, 4 2

FG H

IJ K

Thus Cauchy’s theorem is verified. (iii) We have Cauchy’s mean value theorem


=

bg bg

f′ c g′ c

f (x) = ex

Here

g (x) = e–x

and

f ′ (x) = ex g′ (x) = – e–x ∴ f (x) and g (x) are continuous in [a, b] and differentiable in (a, b) g′ (x) ≠ 0

and also

∴ From Cauchy’s mean value theorem


bg bg

f′ c g′ c

eb – e a ec = e –b – e – a – e– c

i.e.,

i.e.,

d

eb – ea = – e2c 1 1 – eb ea

e a eb e b – e a i.e.,

=

d

a

e –e

b

i

i

= – e2c

ea + b = e2c

42


i.e.,

a + b = 2c

or

c =

a+b 2

a+b ∈ (a, b) 2 Hence Cauchy’s theorem holds good for the given functions.

∴

c =

10. Expand ex in ascending powers of x – 1 by using Taylor’s theorem. Solution The Taylor’s theorem for the function f (x) is ascending powers of x – a is

b x – ag

f (x) = f (a) + (x – a) f ′ (a) + Here

2

f ′′ (a) + ......

2!

f (x) = ex and a = 1 f ′ (x) = ex f ′′ (x) = ex f ′ (1) = e f ′′ (1) = e1

and so on. ∴

x

e = e + (x – 1) e +

b x – 1g 2!

R| S| b g b x –21g T

2

e + .....

U| V| W

2

= e 1+ x –1 +

+ ..........

11. Obtain the Taylor’s expansion of loge x about x = 1 up to the term containing fourth degree and hence obtain loge (1.1). Solution We have Taylor’s expansion about x = a given by y (x) = y (a) + (x – a) y1 (a) +

b x – ag

2

y2 (a) +

2!

y (x) = loge x at a = 1 y (1) = loge 1 = 0 Differentiating y (x) successively, we get y1 (x) =

1 x

⇒

y1 (1) = 1

y2 (x) =

–1 x2

⇒

y2 (1) = –1

y3 (x) =

2 x3

⇒

y3 (1) = 2

b x – ag 3!

3

y3 (a) +

b x – ag 4!

4

y4 (a) + .......

43


–6 ⇒ y4 (1) = – 6 x4 Taylor’s series up to fourth degree term with a = 1 is given by

y4 (x) =

y (x) = y (1) + (x – 1) y1 (1) +

Hence,

b x – 1g

2

y2 (1) +

2!

loge x = 0 + (x – 1) (1) +

b x – 1g

3

y3 (1) +

3!

b x – 1g

2

(–1) +

2

b g b x –21g + b x –31g – b x –41g 2

3

b x – 1g

y4 (1)

4!

b x – 1g 6

4

3

(2) +

b x – 1g 24

4

(–6)

4

loge x = x – 1 – Now putting x = 1.1, we have

b g b01.2g + b01.3g – b01.4g 2

3

4

. – loge (1.1) = 01

= 0.0953 12. Expand tan–1 x in powers of (x – 1) up to the term containing fourth degree. Solution Taylor’s expansion in powers of (x – 1) is given by y (x) = y (1) + (x – 1) y1

b x – 1g (1) +

2

y2 (1) +

2!

b x – 1g 3!

3

y3 (1) +

b x – 1g 4!

4

y4 (1) + ......

y (x) = tan–1 x ⇒

y (1) = tan–1 (1) =

⇒

y1 (x) =

1 1 + x2

y (1) =

1 2

We have

y1 =

π 4

1 1 + x2

(1 + x2) y1 = 1

...(1)

Differentiate successively to obtain expressions Hence we have to differentiate (1) (1 + x2) y2 + 2 xy1 = 0 Putting 2y2 (1) + (2) (1)

x = 1 1 = 0 2

...(2)

44


–1 2 Differentiating (2) w.r.t. x, we get

∴

y2 (1) =

(1 + x2) y3 + 4x y2 + 2y1 = 0

...(3)

x = 1, 2y3 (1) – 2 + 1 = 0,

Putting

1 2 Differentiating (3) w.r.t. x, we get y3 (1) =

(1 + x2) y4 + 6xy3 + 6y2 = 0 Putting

...(4)

x = 1 2y4 (1) + 3 – 3 = 0 y4 (1) = 0

Substituting these values in the expansion, we get tan–1 x =

π 1 + 4 2

R| U S|b x – 1g – b x –21g + b x –61g |V| T W 2

3

13. By using Maclaurin’s theorem expand log sec x up to the term containing x6. Solution Let

y (x) = log sec x, y (0) = log sec 0 = 0

We have

y1 =

sec x tan x = tan x sec x

y1 (0) = 0 y2 = sec2 x y2 (0) = 1 To find the higher order derivatives Consider

y2 = 1 + tan2 x y2 = 1 + y12

This gives, Hence,

y3 = 2y1 y2 y3 (0) = 2y1 (0) y2 (0) = 0 y4 = 2 [ y1 y3 + y22] so that

and

y4 (0) = 2 [ y1 (0) y3 (0) + y22 (0)] = 2 [0·1 + 12] = 2 This yields

y5 = 2 [ y1 y4 + y2 y3 + 2y2 y3] = 2 [ y1 y4 + 3y2 y3] y5 (0) = 2 [ y1 (0) y4 (0) + 3y2 (0) y3 (0)] = 2 [0·2 + 3·1·0] = 0 y6 = 2 [ y1 y5 + y2 y4 + 3 {y2 y4 + y32}]

45


= 2 [ y1 y5 + 4y2 y4 + 3y32] This yields

y6 (0) = 2 [ y1 (0) y5 (0) + 4y2 (0) y4 (0) + 3y32 (0)] = 2 [0·0 + 4·1·2 + 3·0] = 16

Therefore by Maclaurin’s expansion, we have y = y (0) + xy1 (0) + log sec x = 0 + x · 0 +

=

x2 y (0) + ........ 2! 2

x2 x5 x3 x4 x6 ⋅2 + ⋅ 16 + ....... (1) + (0) + (0) + 2! 5! 3! 4! 6!

x 2 x4 x 6 + + + ........ 2 12 45

14. Expand log (1 + sin x) up to the term containing x4 by using Maclaurin’s theorem. Solution Let

y = log (1 + sin x) y (0) = 0

We have,

cos x 1 + sin x y1 (0) = 1 y1 =

FG π – xIJ H2 K = Fπ I 1 + cos G – x J H2 K 1Fπ I 1 Fπ I 2 sin G – xJ cos G – xJ K 2 H2 K 2H2 = 1 Fπ 2 cos G – xIJK 2H2 F π xI = tan G – J H 4 2K 1 F π xI = – sec G – J H 4 2K 2 1L F π xI O = – M1 + tan G – J P H 2 2K Q 2N sin

Now,

y1

2

∴

y1

∴

y2

2

2

y2 = y2 (0) =

–1 (1 + y12) 2 –1 (1 + 1) = – 1 2

46


y3 = – This gives,

1 2y1 y2 = – y1 y2 2

y3 (0) = –1 (–1) = 1 y4 = – [ y1 y3 + y22] y4 (0) = – [1 × 1 + (–1)2] = –2 y5 = – [ y1 y4 + y2 y3 + 2y2 y3] = – (y1 y4 + 3y2 y3)

which gives,

y5 (0) = – [1 (–2) + 3 (–1)] = 5

Therefore by Maclaurin’s theorem, we have y = y (0) + xy1 (0) + log (1 + sin x) = 0 + x · 1 + = x–

x2 y (0) + ....... 2! 2

b g

b g

bg

x2 x3 x4 x5 –1 + –2 + 5 + ......... + 2! 3! 4 ! 5!

x 2 x 3 2 x 4 5x 5 + + + ........ – 2 ! 3! 4! 5!

15. Expand log (1 + ex) in ascending powers of x up to the term containing x4. Solution Let

y = log (1 + ex), y (0) = log 2

Now

y1 = y1 (0) = y2 =

y2 (0) = ∴ ∴

ex 1 + ex 1 2

ex

d1 + e i 1F 1I 1– J G 2H 2K x 2

b

=

ex 1 . = y1 1 – y1 x 1 + e 1 + ex

=

1 4

y3 = y1 (–y2) + (1 – y1) y2 = y2 – 2y1 y2 y3 (0) =

1 1 1 – 2⋅ ⋅ = 0 4 2 4

y4 = y3 – 2 ( y1 y3 + y22) y4 (0) = 0 – 2

FG 1 ⋅ 0 + 1 IJ H 2 16 K

=

–1 8

g

47


Therefore from Maclaurin’s theorem, we get y = y (0) + xy1 (0) + log (1 + ex) = log 2 + x .

bg

FG IJ H K

1 x2 1 x3 x4 1 . + 0 + – + + ........ 2 2 ! 4 3! 4! 8

x x2 x4 + – + ....... . 2 8 192

log (1 + ex) = log 2 +

16. Expand by Maclaurin’s theorem

x2 y (0) + ....... 2! 2

ex up to the term containing x3. 1 + ex

Solution Let

y =

ex 1 + ex

y (0) = 1 Now

y = 1–

y1 =

1 1 + ex

ex

d1 + e i

x 2

=

ex 1 ⋅ = y (1 – y) = y – y2 1 + ex 1 + ex

1 1 1 – = 2 4 4 y2 = y1 – 2y y1

∴

y1 (0) =

∴

y2 (0) =

∴

y3 (0) = 0 – 2

1 1 1 – 2⋅ ⋅ = 0 4 2 4 y3 = y2 – 2 ( yy2 + y12)

FG 1 ⋅ 0 + 1 IJ H 2 16 K

=

–1 8

Therefore from Maclaurin’s theorem, y = y (0) + xy1 (0) +

bg

x3 x2 y2 (0) + y (0) + ..... 3! 3 2!

FG IJ H K

1 1 x2 1 x3 ex 0 + x ⋅ + + – + ....... = x 2 4 2! 3! 8 1+ e =

1 x x3 + – + ....... 2 4 48

48


17. Expand tan–1 (1 + x) as far as the term containing x3 using Maclaurin’s series. Solution y = tan–1 (1 + x)

Let ∴

y1 =

∴ Now

π 4

y (0) = tan–1 (1) =

1

b g

1+ 1+ x

y1 (0) =

2

=

1

d2 + 2 x + x i 2

1 2

(x2 + 2x + 2) y1 = 0

∴ (x2 + 2x + 2) y2 + (2x + 2) y1 = 0 2y2 (0) + 2y1 (0) = 0 ∴

y2 (0) = –

1 2

(x2 + 2x + 2) y3 + (2x + 2) y2 + (2x + 2) y2 + 2y1 = 0 (x2 + 2x + 2) y3 + 4 (x + 1) y2 + 2y1 = 0

i.e.,

2y3 (0) + 4

FG – 1 IJ H 2K

∴

+2

FG 1 IJ H 2K

= 0

y3 (0) =

1 2

Hence by Maclaurin’s theorem, we get y = y (0) + xy1 (0) + tan–1 (1 + x) =

=

bg

bg

x2 x3 y2 0 + y3 0 + ....... 2! 3!

π 1 1 x2 1 x3 + x– + + ....... 4 2 2 2 ! 3 3! π x x2 x3 + – + − ......... 4 2 4 18

18. Expand esin x up to the term containing x4 by Maclaurin’s theorem. Solution Let

y = esin x y (0) = 1 y1 = esin x · cos x y1 (0) = 1

i.e.,

y1 = cos x · y

49


y2 = cos x · y1 – sin x · y ∴

y2 (0) = 1 – 0 = 1 y3 = cos x · y2 – (2 sin x + cos x) y1

∴

y3 (0) = 1 – (0 + 1) 1 = 0 y4 = cos x · y3 – sin x · y2 – (2 sin x + cos x) y2 – (2 cos x – sin x) y1

∴

y4 (0) = 0 – (3·0 + 1) 1 – (2 – 0) 1 = –3

Therefore from Maclaurin’s expansion, we obtain esin x = 1 + x ⋅ 1 + = 1+ x + 19. Expand log

FH x +

x2 + 1

IK

y = log

∴

x2 x4 + ........ . –3 2! 4!

FH x +

x2 + 1

y (0) = 0

y1 =

=

1 x + x2 + 1

1 x + x2 + 1 1

y1 = ∴

x +1 2

y1 (0) = 1 y1

x2 + 1 = 1

y12 (x2 + 1) = 1

or

Differentiating, we get y12 · 2x + (x2 + 1) · 2y1 y2 = 0 (x2 + 1) y2 + xy1 = 0

i.e., ∴

y2 (0) = 0 2

b g

by using Maclaurin’s theorem up to the term containing x3.

Solution Let

bg

x2 x3 x4 0 + ⋅1+ – 3 + ........ 2! 3! 4!

(x + 1) y3 + 2xy2 + xy2 + y1 = 0

IK

OP LM 2x MM1 + 2 F x + 1I PP KQ N H LM x + 1 + x OP MN x + 1 PQ 2

2

2

50


i.e.,

(x2 + 1) y3 + 3xy2 + y1 = 0

∴

y3 (0) = –1

Therefore by Maclaurin’s theorem, we get log

FH x +

x2 + 1

IK

= x–

x3 + .......... 3!

EXERCISE 1.3 1. Verify Rolle’s theorem for the following functions in the given intervals: (i) x2 – 6x + 8 in [2, 4]

Ans. c = 3 Ans. c = − 1 2

(ii) 2x2 + 2x – 5 in [–3, 2]

Ans. c = − 1

(iii) x3 – 3x2 – 9x + 4 in [–3, 3]

LMAns. c = ± 1 OP 3Q N LMAns. c = 3 – 21 OP 3 PQ MN LMAns. c = 2 OP 3Q N LMAns. c = 4a + 3b OP 7 Q N

(iv) 2x3 – 2x + 5 in [–1, 1]

(v) x3 – 3x2 – 4x + 5 in [–2, 2]

(vi) x3 + 2x2 – 4x + 5 in [–2, 2]

(vii) (x – a)3 (x – b)4 in [a, b]

2. Find whether Rolle’s theorem is applicable to the following functions. Justify your answer.

bg Ans. f b x g is not differentiable at x = 1 Ans. tan x is discontinuous at x = π 2 ∈ b0, π g LMAns. sec x is discontinuous at x = π 2 and x = 3π 2OP both lie in b0, 2 πgQ N Ans. f b x g is not differentiable at x = 0

(i) f (x) = 2 + (x – 1)2/3 in [0, 2] (ii) f (x) = x – 1 in [0, 2] (iii) f (x) = tan x in [0, π] (iv) f (x) = sec x in [0, 2π]

(v) f (x) = x2/3 in [–8, 8]

Ans. f x is not differentiable at x = 0

3. Verify Lagrange’s mean value theorem for the following functions: (i) f (x) = 2x2 – 7x + 10 in [2, 5] (ii) f (x) = x (x – 1) (x – 2) in [0, 1/2]

Ans. c = 35 .

LMAns. c = 6 – 21 OP 6 PQ MN

51


LMAns. c = FG 4 – 1IJ OP H π K PQ MN Ans. log b e − 1g 1/ 2

–1

(iii) f (x) = tan x in [0, 1] (iv) f (x) = ex in [0, 1]

Ans. c = 1 + 3

(v) f (x) = (x2 – 4) (x – 3) in [1, 4]

LMAns. c = 1 sin 2 N

(vi) f (x) = cos2 x in [0, π/2]

OP Q

2 = 0.345 π

Ans. c = 6

x 2 – 4 in [2, 4]

(vii) f (x) =

–1

4. Verify Cauchy’s mean value theorem for the following pairs of functions: (i) f (x) =

1 in [a, b] x

x , g (x) =

Ans. c = ab

LMAns. c = FG 15IJ OP H 4 K PQ MN LM o 3 b12 – πg – 12t OP Ans. c = MM PP π 3 N Q LMAns. c = a + b OP 2 Q N LMAns. c = e OP e − 1Q N 2/3

2

(ii) f (x) = x , g (x) =

(iii) f (x) = tan–1

x in [1, 4]

L 1 , 1OP x, g (x) = x in M N3 Q

1/ 2

(iv) f (x) = sin x, g (x) = cos x in [a, b] (v) f (x) = log x, g (x) =

1 in [1, e] x

5. Expand sin x in ascending powers of x – π/2 using Taylor’s series.

LMAns. sin x = 1 – 1 FG x – π IJ 2! H 2K MN

2

+

FG H

π 1 x– 4! 2

IJ K

OP PQ

2

..........

6. Using Taylor’s Theorem expand log x in powers of x – 1 up to the term containing (x – 1)4.

LMAns. log x = b x − 1g − 1 b x − 1g 2 N

2

+

7. Using Maclaurin’s theorem prove the following: (i) sec x = 1 +

x 2 5x 4 + + ........... 2! 4!

b

g b

(ii) ax = 1 + log a x + log a (iii) sin–1 x = x +

g

2

b

x2 + log a 2!

x 3 3x 5 + + ......... 6 40

g

3

x3 + ......... 3!

b g

b g

OP Q

1 1 3 4 x −1 − x − 1 + ....... 3 4

52


(iv) log (1 + cos x) = log 2 – (v) log (1 + tan x) = x –

x2 x4 + ......... – 4 96

x2 4x3 + + ......... 2! 3!

(vi) log (sec x + tan x) = x + 5 (vii) log (1 + sin2 x) = x –

x 3 x5 + + ......... 6 24

5 4 x + .......... 6

x3 + ......... 3 (ix) Obtain the Maclaurin’s expansion for the function e

(viii) ex cos x = 1 + x –

LM MNAns. e

a sin −1 x

= 1+ a x +

d

a sin–1 x

.

i

d

OP PQ

i

2 2 a 2 22 + a 2 4 a2 2 a 1 + a x + x3 + x + ......... 2! 3! 4!

(x) Expand sin (m sin–1 x) by Maclaurin’s theorem up to the term containing x5.

LM m d1 – m i m d1 Ans. sin dm sin x i = mx + + MN 3! 2

2

−1

2

– m2

id3

2

5!

– m2

ix

5

OP PQ

+ .........

ADDITIONAL PROBLEMS (from Previous Years VTU Exams.) 1. Find the radius of curvature of the curve x3 + y3 = 3axy at the point

FG 3a , 3a IJ · H 2 2K

Solution. Refer page no. 7. Example 7. 2. State and prove Lagrange’s mean value theorem. Solution. Refer page no. 27. 3. State Taylor’s theorem for a function of single variable. Using Maclaurin’s series, obtain the series of esin x as for as the term containing x4. Solution. Refer page no. 29 for first part and page no. 48 for second part. Example 18. 4. State Rolle’s theorem. Verify Rolle’s theorem for f (x) = (x – a)m (x – b)n in [a, b] given m and n are positive integers. Solution. Refer page no. 27 for first part and page no. 29 for second part. Example 3. 5. Expand log (1 + cos x) by Maclaurin’s series up to the term containing x4. Solution. Let y = log (1 + cos x), then y (0) = log 2 We find that sin x ⇒ y1 (0) = 0 y1 = – 1 + cos x

53


y2 = –

R| bcos xgb1 + cos xg + sin S| b1 + cos xg T

2

2

x

U| = − R cos x + y U V| ST1 + cos x VW W 2 1

⇒ y2 (0) = – y3 = –

=

R| b– sin xgb1 + cos xg + cos x sin x + 2 y y U| S| V| b1 + cos xg T W

= ⇒ Thus,

1 2

2

sin x cos x sin x − − 2 y1 y2 1 + cos x 1 + cos x 2

b

g

RS cos x − 2 y UV ⇒ y (0) = 0 T 1 + cos x W R cos x − 2 y UV + y R|S – sin x b1 + cos xg + cos x sin x − 2 y U|V y S –1 + T 1 + cos x W |T b1 + cos xg |W R cos x − 2 y UV + y RS y – cos x − 2 y UV y S –1 + T 1 + cos x W T 1 + cos x W

= y1 –1 +

y4 =

1 2

2

3

2

2

1

2

2

1

2

1

3

3

1 4 for y = log (1 + cos x), we have

y4 (0) = –

y (0) = log 2, y1 (0) = 0, y2 (0) = –

1 , y (0) = 0 2 3

1, .... 4 According, the Maclaurin’s series expansion gives

y4 (0) = –

log (1 + cos x) = log 2 –

F I F I GH JK GH JK

1 x2 1 x4 − + ⋅⋅ ⋅⋅⋅⋅⋅ 2 2! 4 4!

x2 x4 + ⋅ ⋅⋅⋅⋅⋅ ⋅ – 4 96 6. Obtain the formula for the radius of curvature in polar form. = log 2 –

Solution. Refer page no. 19. 7. If ρ1 , ρ2 be the radii of curvature at the extremities of any focal chord of the cardiode 2 2 r = a (1 + cos θ) show that ρ1 + ρ2 =

16a 2 · 9

Solution. Let ρ1 be the radius of curvature at (r, θ) then ρ2 is the radius of curvature at (r, θ + π) on the curve r = a (1 + cos θ). Here, r = a (1 + cos θ)

54


⇒

r1 = – a sin θ r2 = – a cos θ ρ1 =

dr

+

2

r + 2

and

i

3 2 2 r1

2r12

− rr2

{a b1 + cos θg + a sin θ} a b1 + cos θg + 2a sin θ + a b1 + cos θg cos θ 2

2

=

=

2

2

2

2

o

gt o b a m3b1 + cos θgr RSa × 2 × 2 cos θ UV 2W T

=

2

3 2

FG where 1 + cos θ = 2cos θ IJ H 2K 2

θ a 2 × 3 × 2 cos2 2

FG IJ H K 4 F θ πI a cos G + J H 2 2K 3 4 θ+π a cos 3 2

4 θ a sin 3 2

FG 4 a cos θ IJ + FG – 4 a sin θ IJ H 3 2K H 3 2K 16 F θ + sin θ IJ a G cos H 2 9 2K 2

ρ12 + ρ22 = = =

∴

ρ12 + ρ22 =

t

2

ρ2 = – Now,

3 2

3 2

θ 2 = 4 a cos θ = θ 3 2 6a 2 cos2 2 ρ2 is obtained from ρ1 by replacing θ by θ + π. ρ2 =

i}

a 2 1 + 2 cos θ + cos2 θ + 2 sin 2 θ + cos θ + cos2 θ

8a 3 cos3

∴

3 2

2

{ d

2

=

2

a 2 1 + 2 cos θ + cos2 θ + sin 2 θ

a 2 × 2 1 + cos θ

=

2

2

16 2 a 9 16 2 a · 9

2

2

2

55


8. State and prove Cauchy’s mean value theorem. Solution. Refer page no. 28, Section 1.3.3. 9. Show that for the curve r2 sec 2θ = a2, ρ =

a2 · 3r

Solution. For the given curve, we have r2 = a2 cos 2θ Differentiating this w.r.t. θ, we get

2r ⋅

dr = – 2a2 sin 2θ dθ dθ r = – 2 dr a sin 2θ

or Therefore,

tan φ = r ⋅

dθ r2 = – 2 dr a sin 2θ cos 2θ sin 2θ = – cot 2θ = –

FG π + 2θIJ H2 K

= tan So that, Hence

Using the given equation

π + 2θ 2 p = r sin φ φ =

= r sin

FG π + 2θIJ H2 K

= r cos 2θ = r⋅

F where cos 2θ = r I GH a JK 2

r2 a2

2

r3 a2 This is the pedal equation of the given curve. From this, we get =

dp 3r 2 = dr a2 Therefore,

ρ = r⋅

or

dr a 2 = dp 3r

i.e., ρ is inversely proportional to r.

dr a2 = dp 3r 2

56


10. Obtain Taylor’s series expansion of loge x about the point x = 1 up to the fourth degree term and hence obtain loge 1.1. Solution. Refer page no. 42. Example 11. −1 11. For the curve θ = cos

FG r IJ − H kK

k 2 − r2 ds prove that r = constant. r dr

Solution. Refer page no. 24. Example 4, point (iii).

LM x + ab OP MN x ba + bg PQ 2

12. State Rolle’s theorem and verify the same for f (x) = log

in [a, b].

Solution. Refer page no. 27 and page no. 34. Example 5. 13. Find the first-four non-zero terms in the expansion of f (x) =

x e

x −1

using Maclaurin’s series.

Solution. We have Maclaurin’s series

bg bg x y b xg = e

bg

By data

f (x) =

x −1

bg

x2 x3 y2 0 + y3 0 + ⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅ 2! 3!

y (x) = y 0 + xy1 0 +

⇒ y (0) = 0

x x =e ⋅ x –1 e ⋅e e exy = ex

i.e.,

y =

x

or We differentiating this equation successively four times and evaluate at x = 0 as follows. exy1 + exy = e ⇒ y1 (0) = e exy2 + 2exy1 + exy = 0 ⇒ y2 (0) = – 2e exy3 + 3exy2 + 3exy1 + exy = 0 ⇒ y3 (0) = 3e Thus by substituting these values in the expansion, we get x e

x −1

F GH

2 = e x−x +

I JK

x3 ⋅⋅⋅⋅ ⋅⋅⋅⋅⋅⋅ · 2

b−a b−a < tan –1 b – tan –1a < · 1 + b2 1 + a2 Solution. Refer page no. 27 and page no. 39. Example 8. 15. Show that 14. State Lagrange’s mean value theorem. Prove that 0 < a < b

1 + sin 2x = 1 + x −

bg

x2 x 3 x 4 − + ⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ · 2 6 24

Solution. We have y (x) = y ( 0) + xy1 0 + Let

y =

1 + sin 2 x

bg

x2 y2 0 + ⋅⋅ ⋅⋅⋅⋅⋅ ⋅ 2!

57


Thus,

=

cos2 x + sin 2 x + 2 sin x cos x

=

bcos x + sin xg

y = y1 = y2 = y3 = y4 = Substituting these values

2

= cos x + sin x

cos x + sin x ⇒ y (0) = 1 – sin x + cos x ⇒ y1 (0) = 1 – cos x – sin x = – y ⇒ y2 (0) = – 1 – y1 ⇒ y3 (0) = – 1 – y2 ⇒ y4 (0) = 1 in the expansion of y (x), we get

1 + sin 2 x = 1 + x −

x2 x3 x4 − + ⋅⋅ ⋅⋅⋅⋅ ⋅⋅ ⋅ 2 6 24

16. For the curve r = a (1 – cos θ), prove that

e2 = constant. r

Solution. Refer page no. 22. Example 2. 17. State and prove Cauchy’s mean value theorem. Solution. Refer page no. 28.

OBJECTIVE QUESTIONS 1. The rate at which the curve is called (a) Radius of curvature (c) Circle of curvature

(b) Curvature (d ) Evolute.

Ans. b

2. The radius of curvature of r = a cos θ at (r, θ) is (a) a (b) 2a

1 a (d ) a2. 2 2 3. The radius of curvature of y = e–x at (0, 1) is (a) 1 (b) 2 (c)

1 2 4. The radius of the circle of curvature is (a) 1 (c)

(c)

1 ρ

(d )

1 · 3

Ans. d

Ans. c

(b) ρ (d ) ρ2.

Ans. b

58


5. The radius of curvature of the curve Pa2 = r3 is

a 2b 2 (b) p3

a2 (a) 3r (c)

a3 3

(d ) None of these.

6. The ellipse in the pedal form

Ans. a

1 1 1 r2 + − = the radius of curvature at the point p2 a 2 b 2 a 2b 2

( p, r) is (a)

a 2b p

(b)

a 2b 2 (c) p3

a 2b 2 b2

(d ) None of these.

Ans. c

7. Lagrange’s mean value theorem is a special case of (a) Rolle’s theorem (b) Cauchy’s mean value theorem (c) Taylor’s theorem

(d ) Taylor’s series.

Ans. b

8. The result “If f ′ (x) = 0 ∀ x in [a, b] then f (x) is a constant in [a, b]” can be obtained from (a) Rolle’s theorem (b) Lagrange’s mean value theorem (c) Cauchy’s mean value theorem

(d ) Taylor’s theorem.

Ans. b

x

9. The first-three non-zero terms in the expansion of e tan x is 2 (a) x + x +

1 3 x 3

(b) x +

x3 2 5 + x 3 5

x3 1 5 + x. Ans. c 3 6 10. In the expansion of tan x and tan–1 x, considering first-three non-zero terms (a) The first-three non-zero terms are same (b) The first-two non-zero terms are same 2 (c) x + x +

5 3 x 6

(c) All coefficients are same

(d ) x +

(d ) First-two coefficients are same.

Ans. c

11. The derivative f ′(x) of a function f (x) is positive or zero in (a, b) without being zero always. Then in (a, b) (a) f (b) < f (a) (b) f (b) > f (a) (c) f (b) – f (a) = f ′ (c), c ∈ (a, b)

Ans. b

(d ) f (b) = f (a). x

12. The Lagrange’s mean value theorem for the function f (x) = e in the interval [0, 1] is (a) C = 0.5413 (b) C = 2 . 3 (c) C = 0.3

(d ) None of these.

Ans. a

59


13. The Maclaurin’s series expension of ex is (a) 1 + x +

x2 x3 + + ⋅⋅⋅⋅ ⋅⋅⋅⋅ 2! 3!

(b) 1 + x –

x2 x3 + + ⋅⋅ ⋅⋅⋅⋅ ⋅⋅ 2! 3!

x2 x3 + + ⋅ ⋅⋅⋅⋅⋅ ⋅⋅ (d ) None of these. 2! 3! 14. The Maclaurin’s series expansion of log (1 + x) is (c) x –

(a) x –

x 3 x5 + + ⋅ ⋅⋅⋅ ⋅⋅⋅⋅ 3 5

(b) x +

Ans. a

x2 x 3 x 4 + + + ⋅⋅⋅⋅ ⋅⋅⋅⋅ 2! 3! 4!

x2 x3 x4 x2 x3 + + ⋅⋅⋅ ⋅⋅⋅⋅ ⋅ – 1 + x – + + ⋅⋅ ⋅⋅⋅⋅ ⋅⋅ · (d ) 2! 3! 4! 2! 3! 15. The Maclaurin’s series expansion of cos x is (c) x –

(a) 1 –

x2 x4 x6 + – + ⋅⋅ ⋅⋅⋅⋅ ⋅⋅ 2! 4! 6!

(b) x –

(c) 1 +

x2 x3 x4 + + + ⋅⋅⋅ ⋅⋅⋅⋅ ⋅ 2! 3! 4!

(d ) None of these.

x 3 x5 + + ⋅ ⋅⋅⋅ ⋅⋅⋅⋅ 3! 5!

16. The radius of curvature at a point (x, 4) of y = a cosh (a)

y2 a

(b)

a2 (c) y

FG x IJ H aK

Ans. a

is

x2 a Ans. a

(d ) None of these.

17. The radius of curvature of the curve y = 4 sin x – sin 2x at x = (a)

5 4

(b)

π is 2

5 4

5 5 (d ) None of these. 4 18. The radius of curvature at a point t on the curve x = at2 and y = 2at is

(c)

d i (c) 2a d1 + t i

(a) 2a 1 + t 2

3 2

Ans. c

Ans. c

(b) a (1 + t2)3

d

3 5 2

(d ) 2a 2 1 + t 2

i

3 2

Ans. a

.

19. The length of the perpendicular from the origin on to the line ax + by = c is (a) p =

(c) p =

c

d

i i

(d ) None of these.

b

da

2

a

(b) p =

1 2 2 2 a +b

1 + b2 2

da

2

+ b2

i

2

Ans. a

60


20. The radii of curvature of the curve 2ap2 = r3 is (a)

2 2ar 5

(b)

2 2ar 3

4 2ar (d ) None of these. 5 21. The radii of curvature of the curve r = aeθcot α is (a) r cosec α (b) r cot α (c)

(c) θ cot α 22. The radii of curvature of the curve (a) (c)

2r 3 2 a ar 3 2 2

(d ) None of these.

Ans. b

Ans. a

2a = 1 – cos θ is r (b)

4r 3 2 a

(d ) None of these.

Ans. a GGG

UNIT

11

Differential Calculus–II 2.1 INDETERMINATE FORMS If f (x) and g (x) are two functions, then we know that

bg bg

f x = g x

lim

x→a

bg lim g b x g lim f x

x→a

x→a

If lim f (x) = 0 and lim g (x) = 0, then the expression x→a

form

x→a

0 , at x = 0. 0

If lim f (x) = ∞, lim g (x) = ∞, then x→a

x→a

bg bg

f x g x

bg bg

f x is said to have the indeterminate g x

is said to have the indeterminate form

∞ . ∞

The other indeterminate forms are ∞ – ∞, 0 × ∞, 00, 1∞, ∞0.

2.1.1 Indeterminate Form

0 0

Here, we shall give a method called L’ Hospital’s rule to evaluate the limits of the expressions which ∞ 0 . and take the indeterminate forms ∞ 0 L’ Hospital’s Theorem Let f (x) and g (x) be two functions such that (1)

lim f (x) = 0 and lim g (x)= 0

x→a

x→a

(2) f ′ (a) and g′ (a) exist and g′ (a) ≠ 0 then, xlim →a

bg bg

f x g x

= lim

x→a

bg bg

f′ x . g′ x 61

62


Proof. Suppose f (x) and g (x) satisfy the conditions of Cauchy’s mean value theorem in the interval [a, x]. Then, we have

bg bg bg bg

bg bg

bg bg

bg bg

f x − f a f′ c = g x −ga g′ c

where c lies between a and x i.e., a < c < x

Since, f (a) = 0, g(a) = 0 and as x → a, c → a, we get f′c f x = g′ c g x

lim

Hence,

x→a

bg= gb x g f x

lim

c→a

bg bg

bg bg

f′ c f′ x = lim x a → g′ c g′ x

(replacing c by x)

f ′(a) = 0 = g′(a), then this theorem can be extended as follows:

If

lim

c→a

bg gb x g f x

bg bg f ′′′b x g lim , if g ′′′b x g

= xlim →a =

f ′′ x

g ′′ x

x→a

f ′′( a) = 0 = g′′(a)

and so on.

WORKED OUT EXAMPLES 1. Evaluate: lim

x →0

x − sin x ⋅ x3 L = lim

Solution

x →0

x − sin x x3

By L’ Hospital rule 1 − cos x x→0 3x 2

L = lim

= lim

sin x 6x

= xlim →0

1 cos x = · 6 6

x→0

2. Evaluate: lim

x→0

Solution

tan x − sin x ⋅ sin 3 x

L = lim

x→0

tan x − sin x sin 3 x

sin x − sin x = lim cos x 3 x →0 sin x

LM 0 OP N0Q LM 0 OP N0Q LM 0 OP N0Q

LM 0 OP N0Q

form

form form

form

63

DIFFERENTIAL CALCULUS–II

LM 0 OP N0Q

form

a x − bx sin x

LM 0 OP N0Q

form

a x log a − b x log b cos x

LM 0 OP N0Q

form

LM 0 OP N0Q

form

LM 0 OP N0Q

form

LM 0 OP N0Q

form

= lim

1 − cos x sin 2 x cos x

= lim0

sin x – sin 3 x + 2 sin x cos 2 x

= lim

1 1 1 = = ⋅ 2 – sin x + 2 cos x – 0 + 2 2

x →0

By L’ Hospital rule x→

and

x→0

3. Evaluate: lim

x→0

2

ax − bx . sin x L = lim

Solution

x→0

By L’ Hospital rule = lim

x →0

=

log a − log b 1

= log 4. Evaluate: lim

x →0

x sin x

de − 1i x

2

FG a IJ . H bK

.

L = lim

Solution

x→0

d

x sin x

i

e −1 x

2

By L’ Hospital rule L = lim

x→0

= lim

x cos x + sin x

d

2 2e 2 x – e x

2 = 1. 2 2 −1

b g x cos x – log b1 + x g lim · =

Solution

x

– x sin x + cos x + cos x

x →0

5. Evaluate:

i

2 e −1 ⋅e x

x →0

x2

L = lim

x→0

LM x cos x − log b1 + xg OP x N Q 2

64


By L’ Hospital rule

=

= = 6. Evaluate: lim

x →0

LM cos x − x sin x − 1 OP 1+ x PQ lim M 2x N LM – sin x − x cos x – sin x + 1 b1 + xg lim M MN 2

x→0

x →0

1 . 2

L = lim

cos hx − cos x x sin x

LM 0 OP N0Q

form

= lim

sin hx + sin x x cos x + sin x

LM 0 OP N0Q

form

= lim

cos hx + cos x – x sin x + cos x + cos x

LM 0 OP N0Q

form

LM 0 OP N0Q

form

x→0

By L’ Hospital rule x →0

x →0

=

x→0

2 = 1. 2

ax – bx . x L = lim

Solution

x→0

a x − bx x


x →0

a x log a – b x log b 1

= log a − log b = log

8. Evaluate: lim x→

Solution

π 4

OP PP Q

form

cos hx – cos x . x sin x

Solution

7. Evaluate: lim

2

LM 0 OP N0Q

FG a IJ . H bK

sec 2 x – 2 tan x . 1 + cos 4x L = limπ x→

4

sec 2 x − 2 tan x 1 + cos 4 x

65


By L’ Hospital rule = lim x→

π 4

2 sec 2 x tan x − 2 sec 2 x – 4 sin 4 x

b

g

LM 0 OP N0Q

form

sin hx − sin x x sin 2 x

LM 0 OP N0Q

form

= lim

cos hx − cos x sin x + 2 x sin x cos x

LM 0 OP N0Q

form

= lim

sin hx + sin x 2 sin x cos x + x ⋅ 2 cos 2 x + sin 2 x

= lim

sin hx + sin x 2 sin 2 x + 2 x cos 2 x

LM 0 OP N0Q

form

= lim

cos hx + cos x 4 cos 2 x + 2 x – 2 sin 2 x + 2 cos 2 x

= limπ x→

– sec 2 x tan x – 1 2 sin 4 x

4

= limπ − x→

4

|RS sec |T

2

b

g

x ⋅ sec 2 x + tan x – 1 ⋅ 2 sec 2 x tan x 8 cos 4 x

|UV |W

π +0 4 8 cos π

– sec 4 =

–4 1 = . –8 2

= 9. Evaluate: lim

x →0

sin hx − sin x . xsin 2 x L = lim

Solution

x →0

By L’ Hospital rule x →0

x→0

x→0

x→0

x→0

Solution

b

g

1+1 2 1 = = . 4+0+2 6 3

=

10. Evaluate: lim

2

b g. x log b1 + x g

e 2x − 1 + x

2

L = lim

x→0

b g x log b1 + x g

e2 x − 1 + x

2

LM 0 OP form N0Q

66



x →0

= lim

x→0

11. Evaluate: If lim

b

g

b g b g

4e 2 x − 2 4−2 = =1. 1 1 1 1 + + 2 1+ x 1+ x

b g

x 1 − a cos x + b sin x

x→0

x

3

LM 0 OP form N0Q

2e 2 x − 2 1 + x x + log 1 + x 1+ x

Solution

1 , find a and b. 3

=

b

g

LM 0 OP form N0Q

x 1 − a cos x + b sin x 1 = lim x→0 3 x3

Given By L’ Hospital rule

1 − a cos x + x a sin x + b cos x 3x 2 x = 0, the numerator = 1 – a + b.

= lim

x →0

At

In order that the limit should exist, we must have 1–a+b = 0 Applying L’ Hospital rule with this assumption, we get

b

...(1)

g

1 a sin x + a x cos x + sin x – b sin x = lim x→0 3 6x = lim

x→0

=

Solving (1) and (2), we get a = 12. Evaluate: lim

x →0

Since,

g

a cos x + a – x sin x + cos x + cos x – b cos x 6

a + 2a − b 6

1 3a − b = 3 6 3a – b = 2

Hence,

Solution

b

LM 0 OP form N0Q

...(2)

–1 1 ⋅ , b = 2 2

log sin x . cot x L = lim

x→0

log sin x cot x

log 0 = – ∞, cot 0 = ∞

–

LM ∞ OP form N∞ Q

67


By L’ Hospital rule L = lim

x →0

= lim

x →0

13. Evaluate: lim

x →0

cot x − cosec 2 x

LM ∞ OP N∞ Q

form

LM− ∞ OP N ∞Q

form

LM 0 OP N0Q

form

– sin x · cos x = 0.

log x . cosec x L = lim

Solution

x →0

log x cosec x


1 x L = lim x → 0 – cosec x cot x = lim

– sin 2 x x cos x

= lim

– 2 sin x cos x 0 = = 0. cos x − x sin x 1 – 0

x→0

x→0

14. Evaluate: lim x→

π 2

log cos x ⋅ tan x

L = lim

Solution

π x→ 2

log cos x tan x

LM– ∞OP form N ∞Q

By L’ Hospital rule = limπ x→

– tan x sec 2 x

2

= lim – sin x · cos x = 0. x→

15. Evaluate: lim

x →0

π 2

log sinax ⋅ log sinbx

Solution

L = lim

x→0

log sin ax log sin bx


x→0

a cos ax / sin ax b cosbx / sin bx

LM – ∞OP form N – ∞Q

68


= lim

a cot ax b cot bx

= lim

a tan bx b tan ax

x →0

x →0

LM ∞ OP N∞Q LM 0 OP N0Q

form form

ab sec 2 bx 1 = = 1. x → 0 ba sec 2 ax 1

= lim 16. Evaluate: lim

x→a

b

log x − a

d

x

log e − e

g

a

i

.

x→a

d

g

log e x − e a


i

b g / de – e i

LM ∞ OP N∞Q

form

LM 0 OP N0Q

form

1/ x − a

= lim

x→a

= lim

x→a

= lim

x→a

= lim

b

log x − a

L = lim

Solution

x→a

e

x

x

a

b

e x − ea x − a ex

b

ex x − a ex + ex

b

1 − x a +1

g

g g

= 1.

EXERCISE 2.1 1. Evaluate: (i) lim

x →0

x – sin x tan 3 x

1 − cos x (iii) xlim → 0 x log 1 + x

b g

(v) xlim →1

1 + log x − x

b1 − xg

2

(vii) lim

ex – e– x – 2x x 2 sin x

(ix) lim

xx − x x − 1 − log x

x→0

x →1

LMAns. 1 OP N 6Q LMAns. 1 OP N 2Q Ans. 1

LMAns. 1 OP N 3Q Ans. 2

(ii) lim

tan x – x x 2 tan x

LMAns. 1 OP N 3Q

(iv) xlim →0

e x − e sin x x – sin x

Ans. 1

(vi) lim

cos hx – cos x x sin x

Ans. 1

x→0

x→0

(viii) xlim →0 (x) lim

x →0

b g

x − log 1 + x 1 − cos x

sin x ⋅ sin –1 x x2

Ans. 1 Ans. 1

69


(xi) lim

x →0

(xiii) xlim →0 2. If lim

x →0

3. If lim

x 2 + 2 cos x − 2 x sin 3 x e x + sin x − 1 log 1 + x

b g

LMAns. 1 OP N 12 Q

(xii) lim

tan 2 x − x 2 x 2 tan 2 x

Ans. 2

(xiv) lim

e 3 x + e –3 x − 2 · 5x 2

x →0

x →0

sin 2 x + a sin x is finite, find a. x3

b

g

x 1 + a cos x − b sin x

x →0

x

3

log x cosec x

LMAns. a = − 5 , b = − 3 OP 2 2Q N

= 1, find a and b.

(ii) xlim →0

Ans. 0

log sin 2 x log sin x

FG H

log x −

log tan 2 x (iii) lim x → 0 log tan x log 1 − x (v) lim x →1 cot πx

b g

OP Q 9O 5 PQ 2 3

Ans. a = − 2

4. Evaluate: (i) xlim →0

LMAns. N LMAns. N

lim

Ans. 1

(iv)

Ans. 0

(vi) lim

x→π 2

x→∞

tan x x2

ex

2

·

Ans. 1 π 2

IJ K

Ans. 0 Ans. 0

2.1.2 Indeterminate Forms ∞ – ∞ and 0 × ∞ L’ Hospital rule can be applied to limits which take the indeterminate forms ∞ – ∞ and 0 × ∞. First 0 ∞ or and then by use L’Hospital rule to evaluate the we transform the given limit in the form 0 ∞ limit.

WORKED OUT EXAMPLES

1. Evaluate: lim

x→0

LM 1 − log b1 + xg OP . Nx x Q 2

LM 1 − log b1 + xg OP Nx x Q x − log b1 + x g lim

Solution

L = lim

Hence, required limit

L =

2

x →0

x →0

x

1−


= lim

x →0

2

1 1+ x 2x

x 1+ x = lim x →0 2x

∞ – ∞ form

LM 0 OP N 0Q

form

70


= lim

x →0

= 2. Evaluate: lim

x→0

RS 1 − cot x UV . Tx W

Solution Hence, we have

1 2 1+ x

b g

1 . 2

RS RTS T

UV WUV W

1 − cot x L = lim x →0 1 x cos x − L = lim x →0 x sin x

= lim

sin x − x cos x x sin x

= lim

cos x − cos x + x sin x x cos x + sin x

= lim

x sin x x cos x + sin x

= lim

x cos x + sin x cos x − x sin x + cos x

x→0


x →0

x→0

x →0

=

∞ – ∞ form

LM 0 OP N0Q

form

LM 0 OP N0Q

form

0 0 = = 0. 2−0 2

3. Evaluate: lim x log tan x. x→0

L = lim

Solution

= lim

Applying L’ Hospital rule

sec 2 x = lim tan x –1 x→0 x2 = lim – x→0

x 2 sec 2 x tan x

= lim –

LM 2 x sec N

= lim

– [2x + 2x2 tan x] = 0.

x→0

x →0

4. Evaluate: lim tan x log x. x→0

LM– ∞ OP form N ∞Q

log tan x 1 x

Hence, we have

x →0

[0 × (– ∞)] form

x log tan x

x →0

2

x + x 2 ⋅ 2 sec 2 x tan x sec 2 x

OP Q

LM 0 OP N 0Q

form

71


L = lim

Solution

= lim

x →0

Solution

LM 1 Nx

2

– sin 2 x x

= lim

– 2 sin x cos x = 0. 1

x→0

−

LM 0 OP N0Q

= lim

x→0

x→0

LM– ∞ OP form N ∞Q

log x cot x

1 x = lim x → 0 – cosec 2 x


5. Evaluate: lim

[0 × (– ∞)] form

tan x log x

x →0

OP Q

1 . sin 2 x

L = lim

x→0

LM 1 − 1 OP N x sin x Q 2

2

= lim

sin 2 x − x 2 x 2 sin 2 x

= lim

sin 2 x − x 2 x2 × sin 2 x x4

= lim

sin 2 x − x 2 x2 × lim x → 0 sin 2 x x4

x →0

x →0

x →0

FG lim sin x = 1 ∴ lim x = 1IJ H x K sin x x→0

= lim

x→0

x→ 0

sin 2 x − x 2 x4

By applying L’ Hospital rule = lim

2 sin x cos x − 2 x 4x3

= lim

sin 2 x − 2 x 4x3

x →0

x→0

= lim

x →0

= lim x →0 = lim

x →0

2 cos 2 x − 2 12 x 2 cos 2 x − 1

6x 2 – 2 sin 2 x 6x 2

form

[∞ – ∞] form

LM 0 OP form N0Q

LM 0 OP form N0Q LM 0 OP form N0Q (3 1 – cos 2x = 2 sin2 x)

72


FG IJ H K –1 F sin x IJ lim G H xK 3

= lim – x →0

=

6. Evaluate: lim

x→0

LM 1 Nx

2

2

x →0

=

–1 ×1 3

=

–1 . 3

OP Q

− cot 2 x .

FG 1 − cot xIJ Hx K F 1 − cos x I lim G H x sin x JK

L = lim

Solution

2

1 sin x 3 x

x→0

2

[∞ – ∞] form

2

2

=

x→0

2

2

LM 0 OP form N 0Q

= lim

sin 2 x − x 2 cos2 x x 2 sin 2 x

= lim

sin 2 x − x 2 cos2 x x2 × sin 2 x x4

= lim

sin 2 x − x 2 cos2 x x2 × lim x → 0 sin 2 x x4

x →0

x →0

x→0

FG lim sin x = 1IJ H x K x→0

sin x − x cos x x4 sin2 x – x2 cos2 x = sin2 x – x2 (1 – sin2 x) = (1 + x2) sin2 x – x2 2

2

2

= lim

x →0

∴

d1 + x i sin lim 2

=

x − x2

x4

x→0


2

d1 + x i ⋅ 2 sin x cos x + 2 x sin lim 2

=

d1 + x i sin 2 x + 2 x sin lim

x→0

Again by L’ Hospital rule

4x

2

x →0

LM 0 OP N0Q

form

x − 2x

x − 2x

3

d1 + x i ⋅ 2 cos 2 x + 2 x sin 2 x + 4 x sin x cos x + 2 sin 2

= lim

form

4x3

x →0

2

=

2

LM 0 OP N0Q

12 x

2

2

x−2

73


= lim x→

d

id

i

2 1 + x 2 1 − 2 sin 2 x + 4 x sin 2 x + 2 sin 2 x − 2 12 x 2

0

= lim

x →0

2 x 2 − 2 sin 2 x − 4 x 2 sin 2 x + 4 x sin 2 x 12 x 2 2−2×

= lim

FG sin x IJ H xK

2

− 4 sin 2 x + 4 ×

x→0

2−2× = lim

FG sin x IJ H xK

FG sin 2 x IJ H x K

12 2

− 4 sin 2 x + 4 × 2 ×

x→0

FG sin 2 x IJ H 2x K

12

2−2−0+8 8 2 = = . 12 12 3 7. Evaluate: lim (1 – tan x) sec 2x. =

x→

π 4

L = lim (1 – tan x) sec 2x

Solution

x→

π 4

= limπ x→

[0 × ∞] form

LM 0 OP form N 0Q

1 − tan x cos 2 x

4

By L’ Hospital rule = limπ x→

FG H

IJ b K

– sec 2 x –2 = = 1· – 2 sin 2 x – 2

4

g

8. Evaluate: lim log 2 –

x cot x – a . a

Solution

L = lim log 2 –

x→a

=


FG x IJ cotb x – ag H aK F xI log G 2 − J H aK lim tan b x − a g 1 F –1I ×G J FG 2 − x IJ H a K H aK lim sec b x − a g FG – 1 IJ H a K –1

x →a

=

=

x →a

2

x→a

1

=

a

.

[0 × ∞] form

LM 0 OP form N 0Q

74


EXERCISES 2.2 Evaluate the following:

LMAns. 1 OP N 2Q

FG x – 1 IJ · H x – 1 log x K F a xI lim G – cos J · H x aK

1. lim

x →1

3. 5. 7.

x →0

lim (sec x – tan x).

x→π

x →0

2

9. lim sec x →1

FG π xIJ H2 K

π sec x). 2

× log x.

FG 1 – 1 IJ · H sin x x K

Ans. 0

4. lim (cosec x – cot x).

Ans. 0

6. lim

x→0

x →1

2

lim (x tan x –

x→π

2. lim

Ans. − 1

FG 1 – x IJ · H log x log x K

8. lim x3 log x. x→0

LMAns. −2 OP N πQ

Ans. 0 Ans. 0 Ans. − 1 Ans. 0

2.1.3 Indeterminate Forms 00, 1∞, ∞0, 0∞

m b gr b g .

f x Let L = xlim →a

g x

If L takes one of the indeterminate forms 00, 1∞, ∞0, 00 then taking

logarithm on both sides, we get log L = lim g (x) log f (x) x→a

Then, log L will take the indeterminate form 0 × ∞ and which can be evaluated by using the method employed in preceding section.

WORKED OUT EXAMPLES 1. Evaluate: lim (1 + sin x)cot x. x →0

[1∞] form

(1 + sin x)cot x L = xlim →0

Solution Taking logarithm on both sides

log L = lim cot x log (1 + sin x) x→0

= lim

b

log 1 + sin x

x→0

[∞ × 0] form

g

LM 0 OP N0Q

tan x


x →0

log L = lim

x →0

∴

1

cos x / 1 + sin x sec 2 x

cos x 1 = = 1 1 1 +0 sec x 1 + sin x

L = e = e.

2

b

g

b g

form

75


2. Evaluate: lim (tan x)tan 2x. x→

π 4

L = lim (tan x)tan

Solution

(1∞) form

2x

π x→ 4

Taking logarithm on both sides log L = lim tan 2x log (tan x) x→

[∞ × 0] form

π 4

= lim

log tan x cot 2 x

= limπ

sec 2 x tan x – 2 cosec 2 2 x

π x→ 4

LM 0 OP form N 0Q

By L’ Hospital rule x→

4

= limπ x→

– =

− sec 2 x 2 tan x cosec 2 2 x

4

e 2j

2

2 ⋅ 1 ⋅ 12

log L = – 1 L = e–1 =

3. Evaluate: lim

x→0

FG tan x IJ H x K

Solution

1 · e

1 x2

.

L = lim

x →0

FG tan x IJ H x K

1 x2

[1∞] form

tan x = 1 as x → 0) x→0 x Taking logarithm on both sides

(Since, lim

1 tan x log 2 x →0 x x tan x log x = lim 2 x→0 x

log L = lim


LM N

1 x sec 2 x – tan x . tan x x2 x = lim x→0 2x

[∞ × 0] form

OP Q

LM 0 OP N0Q

form

76


= lim

x→0

= lim

x sec 2 x – tan x 2 x 2 tan x

x→0

form

LM 0 OP N0Q

form

sec 2 x + 2 x sec 2 x tan x – sec 2 x 2 x 2 sec 2 x + 2 x tan x

x→0

= lim

LM 0 OP N0Q

sec 2 x tan x x sec 2 x + 2 tan x

Again By L’ Hospital Rule log L = lim

x→0

∴

sec 2 x sec 2 x + tan x ⋅ 2 sec 2 x tan x

d

i

sec x + x 2 sec x tan x + 2 sec x 2

2

2

=

1 3

L = e1/3. –x

4. Evaluate: lim (1 + x2)e . x→∞

F∴ e = 0, I GH 1 + ∞ = ∞JK −∞

–x

L = lim (1 + x2)e

Solution

x→∞

[∞0] form

Taking log on both sides log L = lim e–x log (1 + x2) x→∞

= lim

d

log 1 + x 2

x→∞

e

i

x


d

2x 1 + x2

x→∞

e

2x

d

= lim

d1 + x i e

x→∞

LM ∞ OP N∞ Q

form

LM ∞ OP N∞ Q

form

i

x

= lim

x→∞

[0 × ∞] form

i

1 + x2 e x 2

2

x

+ 2 x ex

log L = 0 L = e0 = 1. 5. Evaluate: lim (sec x)cot x. x→

Solution

π 2

L = lim (sec x)cot x x→

π 2

log L = lim cot x log sec x x→

π 2

[∞0] form [0 × ∞] form

77


= limπ x→

LM ∞ OP N∞ Q

log sec x tan x

2

form


sec x ⋅ tan x sec x = lim π sec 2 x x→ 2

= limπ x→

tan x sec 2 x

2

log L = lim sin x cos x = 0 x→

∴

π 2

L = e0 = 1

EXERCISE 2.3 Evaluate:

b g F 1I lim G J H xK F sin x IJ lim G H x K

1. lim cos x x→0

1 x2

Ans. e −1 2

2. lim

π x→ 2

bsin xg

tan x

Ans. 1

tan x

3.

5.

x→0

1 x

x→0

1

7. lim

x →1

x b1− x g

d

i

d

i

2 9. lim 1 – x x →1

11. lim cos2 x x →0

1 log 1− x

b g

1 x2

Ans. 1

4. lim x x

Ans. 1

6. lim cos ax

Ans. 1

x →0

x→0

b

LMAns. 1 OP N eQ

8. lim x sin x

Ans. e

10. lim 2 – x

g

b x2

Ans. e − a

b g F sin hx IJ lim G H x K F 1I lim G J H xK

tan

x →1

12.

Ans. e

14.

b2

Ans. 1

x→0

LMAns. 1 OP N eQ

2

πx 2

Ans. e 2 π

LMAns. 1 OP N 6Q

1 x2

x →0

2 sin x

13. lim x log x x→0

15. lim

x →0

LM a N

x

+ bx 2

OP Q

1 x

Ans.

ab

Ans. 1

x →0

16. lim

x→∞

LM π – tan xOP N2 Q –1

1 x

Ans. 1

78


2.2 TAYLOR’S THEOREM FOR FUNCTIONS OF TWO VARIABLES Statement: If f (x, y) has continuous partial derivatives up to nth order in a neighbourhood of a point (a, b), then

LMh ∂ + k ∂ OP f (a, b) + 1 Lh ∂ + k ∂ O f ba, bg M P 2 ! N ∂x ∂y Q N ∂x ∂y Q LMh ∂ + k ∂ OP f ba, bg + R ∂O 1 L ∂ 1 + Mh + k P f ba , bg + .......... + 3 ! N ∂x ∂y Q bn – 1g ! N ∂x ∂y Q 1 L ∂ ∂O h + k P f (a + θh, b + θk) for some θ : 0 < θ < 1. M n ! N ∂x ∂y Q 2

f (a + h, b + k) = f (a, b) +

n –1

3

n

...(1)

n

where Rn =

Here, Rn is called the remainder after n times. By the Taylor’s theorem given in f (x, y) = f (a, b) +

+

1 2!

b x – ag

2

b

g b g b

g b g

1 x – a f x a, b + y – b f y a, b 1!

b g b

f xx a , b + 2 x – a

gb y – bg f ba, bg + b y – bg

2

xy

b g

f yy a, b + ............ ...(2)

This is called the Taylor’s expansion of f (x, y) about the point (a, b). If a = 0 and b = 0, we get the Maclaurin’s form of Taylor’s theorem. i.e.,

b g

f (x, y) = f 0, 0 +

+

b g

b g

1 x f x 0, 0 + yf y 0, 0 1!

b g

b g

b g

1 x 2 f xx 0, 0 + 2 xy f xy 0, 0 + y 2 f yy 0, 0 + ............ 2!

...(3)

Further, if the Taylor’s series of f (x, y) is approximated to some terms up to a particular degree the resulting expression of f (x, y) is called as the Taylor’s polynomial.

WORKED OUT EXAMPLES 1. Expand ex sin y by using Maclaurin’s theorem up to the third degree terms. Solution Let Now

f (x, y) = ex sin y fx = ex sin y

fy = ex cos y

fxx = ex sin y

fyy = – ex sin y

fxy = ex cos y

fxyy = – ex sin y

fxxx = ex sin y

fyyy = – ex cos y

fxxy = ex cos y

79


At (0, 0) f (0, 0) = 0

fx (0, 0) = 0

fy (0, 0) = 1

fxx (0, 0) = 0

fxy (0, 0) = 1

fyy (0, 0) = 0

fxxx (0, 0) = 0

fxxy (0, 0) = 1

fxyy (0, 0) = 0

fyyy(0, 0) = – 1 Hence, by Maclaurin’s theorem

b g

b g b g b0, 0g + 2 xy f b0, 0g + y f b0, 0g b0, 0g + 3x y f b0, 0g + 3xy f b0, 0g + y

f (x, y) = f 0, 0 + x f x 0, 0 + y f y 0, 0

1 2 2 x f xx xy yy 2! 1 3 2 2 x f xxx + xxy xyy 3! Substituting all the values in the expansion of f (x, y), we get 1 2 x ⋅ 0 + 2 xy ⋅ 1 + y 2 − 0 ex sin y = 0 + x ⋅ 0 + y ⋅ 1 + 2! 1 3 + x ⋅ 0 + 3x 2 y + 3xy 2 ⋅ 0 + y 3 –1 + ⋅⋅ ⋅⋅⋅⋅ ⋅ 3! +

3

b g

f yyy 0, 0 + ⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅

b g

ex sin y = y + xy +

x2 y y3 − + .......... . 2 6

2. Expand ex log (1 + y) by Maclaurin’s theorem up to the third degree term. Solution Let f (x, y) = ex log (1 + y) The function and it’s partial derivatives evaluated at (0, 0) is as follows: f (x, y) = ex log (1 + y)

→

0

x

→

0

x

→

1

x

→

0

→

1

→

–1

→

0

→

1

→

–1

→

2

fx = e log (1 + y) fy = e · 1/ (1 + y) fxx = e log (1 + y) x

fxy = e /1 + y x

2

fyy = – e /(1 + y) x

fxxx = e log (1 + y) x

fxxy = e /(1 + y) x

2

fxyy = – e /(1 + y) x

fyyy = 2e /(1 + y)

3

∴ By Maclaurin’s theorem

b g

b g

b g

f (x, y) = f 0, 0 + x f x 0, 0 + y f y 0, 0 +

+

b g

b g

b g

b g

b g

1 2 x f xx 0, 0 + 2 xy f xy 0, 0 + y 2 f yy 0, 0 2!

b g

b g

1 3 x f xxx 0, 0 + 3x 2 y f xxy 0, 0 + 3xy 2 f xyy 0, 0 + y 3 f yyy 0, 0 + ⋅⋅⋅⋅ ⋅⋅⋅⋅ 3!

80


Substituting all the values in the expansion of f (x, y), we get ex log (1 + y) = y +

d

i d

i

1 1 2 xy − y 2 + 3x 2 y – 3xy 2 + 2 y 3 . 2 6

3. Expand eax + by by using Maclaurin’s theorem up to the third term. Solution. Since the expansions required in powers of x, y the point (a, b) associated is (0, 0) and the expansion of f (x, y) about (0, 0) is given by

b g

b g

b g

b g

1 2 x f xx 0, 0 2!

f (x, y) = f 0, 0 + x f xx 0, 0 + y f yy 0, 0 +

b g b g 31! x f b0, 0g y f b0, 0g + 3xy f b0, 0g + y f b0, 0g + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

+ 2 xy f xy 0, 0 + y 2 f yy 0, 0 + + 3x 2

3

xxx

2

xxy

3

xyy

yyy

The functions and its partial derivatives evaluated at (0, 0) is as follows: + by

→

1

ax + by

→

a

ax + by

→

b

2 ax + by

→

a2

by

→

ab

2 ax + by

→

b2

f (x, y) = eax fx = ae fy = be

fxx = a e

fxy = abeax + fyy = b e

fyyy = b3eax

+ by

→

b3

fxxx = a3eax

+ by

→

a3

fxyy = ab2eax

+ by

→

ab2

fxxy = a2beax

+ by

→

a2 b

Substituting these values in the expansion of f (x, y), we get

g bax +2!byg + bax +3!byg

b

2

3

f (x, y) = 1 + ax + by +

+ ⋅⋅⋅ ⋅⋅⋅⋅⋅ .

4. Expand tan–1 y/x about the point (1, 1) using Taylor’s theorem up to the second degree terms. Solution. The expansion of f (x, y) about (1, 1) is given by

b g b g b g b g b g 1 + b x − 1g f b1, 1g + 2 b x − 1gb y − 1g f b1, 1g + b y − 1g 2!

f (x, y) = f 1, 1 + x − 1 f x 1, 1 + y – 1 f y 1, 1 2

xx

xy

2

b g

f yy 1, 1

b g f b1, 1g + 3 b x − 1g b y − 1g f b1, 1g + 3 b x – 1gb y – 1g + b y − 1g f b1g + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ +

1 x −1 3!

3

2

xxx

3

yyy

xxy

2

b g

f xyy 1, 1

81


Let

y x

−1 f (x, y) = tan

fx =

fy =

fxx =

fyy =

fxy =

–y

1 2

y x2 1

1+

1+

dx dx dx

2

2

x

y x2 2 xy

⋅

+y

2

=

–y x +y 2

2

x 1 = 2 x x + y2

i

2 2

– 2 xy 2 2

f (1, 1) = tan–1 (1) =

;

i + y i ⋅1 − x ⋅ 2x y –x = dx + y i dx + y i +y

2 2 2

2

2 2

2

2

;

fx (1, 1) =

–1 2

;

fy (1, 1) =

1 2

;

fxx (1, 1) =

1 2

;

fyy (1, 1) =

–1 2

;

fxy (1, 1) = 0

π 4

2

2 2

Substituting these values in the expansion of f (x, y), we get y π 1 1 1 1 2 2 tan −1 + – x −1 + y −1 + x −1 – y − 1 + ⋅⋅⋅ ⋅⋅⋅⋅⋅ ⋅⋅⋅ . = x 4 2 2! 2 2

LM b g b g OP N Q F πI 5. Expand e cos y by Taylors’s theorem about the point G 1, J up to the second degree terms. H 4K F πI Solution. The expansion of f (x, y) about G 1, J is given by H 4K F πI L F πI F πI F πI O f (x, y) = f G 1, J + Mb x − 1g f G 1, J + G y − J f G 1, J P H 4K N H 4K H 4K H 4KQ 1 L F πI F πI F πI + Mb x − 1g f G 1, J + 2 b x − 1g G y − J f G 1, J H K H 4K H 4K 2! N 4 F πI F πI + G y – J f G 1, J + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ H 4K H 4K F πI The function and its partial derivatives are evaluated at G 1, J . H 4K b g b g

x

x

y

2

xx

xy

2

yy

f (x, y) = ex cos y

→

fx = ex cos y

→

fy = – ex sin y

→

e 2 e 2 –e 2

82


fxx = ex cos y

→

fxy = – ex sin y

→

e 2 –e 2 –e

fyy = – ex cos y →

2 Substituting these values in the expansions of f (x, y), we get x

e cos y =

e 2

R|1 + L x − 1 − F y − π I O + 1 L x − 1 S| MNb g GH 4 JK PQ 2 MMb g N T

2

b g FGH y − π4 IJK − FGH y − π4 IJK

2

−2 x −1

OPU| PQV|W ·

EXERCISE 2.4 1. Expand x2y about the point (1, – 2) by Taylor’s theorem.

LMAns. N

b g b

g

b g

1 – 4 x −1 2! 2. Expand x2y + 3y – 4 about the point (1, – 2) by Taylor’s theorem. – 2– 4 x −1 + y +2 +

b g b

g b g

Ans. − 12 − 4 x − 1 + 4 y + 2 − 2 x − 1

2

2

b gb

b gb

+ 4 x −1 y + 2

g OPQ

g b g b y + 2g

+ 2 x −1 y + 2 + x −1

2

3. Expand exy about the point (1, 1) by using Taylor’s theorem up to second degree terms.

LMAns. e RS1 + b x − 1g+ b y − 1g + 1 b x − 1g 2! N T

4. Obtain the Taylor’s expansion of ex

b gb g b g + ⋅ ⋅⋅⋅⋅⋅ ⋅⋅UVWOPQ F πI sin y about the point G 0, J up to second degree terms. H 2K LMAns. 1 + x + 1 Lx − F y − π I O + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅OP M GH 2 JK PP 2 ! MN MN PQ Q 2

+ 4 x −1 y − 1 + y − 1

2

2

2

y about the point (1, 2) using Taylor’s theorem up to second degree terms: x 1 1 2 2 Ans. tan −1 2 + –2 x − 1 + y − 2 + 4 x − 1 − 6 x −1 y − 2 − 4 y − 2 5 50

−1 5. Expand tan

LM N

b g b

g

b g

b gb

g b

6. Obtain the Maclaurin’s expansion of the following functions: (i) exy up to second degree terms. (ii) log (1 – x – y) up to 3rd degree terms. (iii) eax sin by up to 3rd degree terms.

g OPQ

LMAns. 1 + xy + 1 x y + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅OP 2 N Q LMAns. b x − yg − 1 bx − yg + 1 bx − yg + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅OP 2 3 N Q LMAns. by + ab xy + 1 d3a b x y − b y i + ⋅ ⋅ ⋅ ⋅ ⋅OP 6 N Q 2

2

2

2

3

2

3

3

83


LM b x + yg + ⋅ ⋅ ⋅ ⋅ ⋅OP Ans. b x + y g − 3! PQ MN LM bax + byg + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅OP Ans. 1 − 2! PQ MN 3

(iv) sin (x + y) up to 3rd degree terms.

2

(v) cos (ax + by) up to 2nd degree terms.

2.3 MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES Maximum Value of a Function A function f (x, y) is said to have a maximum at a point (a, b), if these exists a neighbourhood N of (a, b) such that f (x, y) < f (a, b) for all (x, y) ∈ N. Minimum Value of a Function A function f (x, y) is said to have a minimum at (a, b), if there exists a neighbourhood N of (a, b) such that f (x, y) > f (a, b) for all (x, y) ∈ N.

2.3.1 Necessary and Sufficient Conditions for Maxima and Minima The necessary conditions for a function f (x, y) to have either a maximum or a minimum at a point (a, b) are fx (a, b) = 0 and fy (a, b) = 0. The points (x, y) where x and y satisfy fx (x, y) = 0 and fy (x, y) = 0 are called the stationary or the critical values of the function. Suppose (a, b) is a critical value of the function f (x, y). Then fx (a, b) = 0, fy (a, b) = 0. Now denote fxx (a, b) = A, fxy (a, b) = B,

fyy (a, b) = C

1. Then, the function f (x, y) has a maximum at (a, b) if AC – B2 > 0 and A < 0. 2. The function f (x, y) has a minimum at (a, b) if AC – B2 > 0 and A > 0. Maximum and minimum values of a function are called the “extreme values of the function”. Working rule to find the maximum and minimum value of a function f (x, y) 1. Find fx (x, y) and fy (x, y). 2. Solve the equations fx (x, y) = 0 and fy (x, y) = 0. Let (a, b) be a root of the above equations. Here (a, b) is called the critical point. 3. Then find fxx (x, y), fxy (x, y), fyy (x, y). 4. Then A = fxx (a, b), B = fxy (a, b), C = fyy (a, b). 5. If AC – B2 > 0 and A < 0, then f has a maximum at (a, b). 6. If AC – B2 > 0 and A > 0, then f has a minimum at (a, b).

84


7. If AC – B2 < 0, then f has neither a maximum nor a minimum at (a, b). The point (a, b) is called the ‘Saddle point’. 8. If AC – B2 = 0, further investigation is necessary.

WORKED OUT EXAMPLES 1. Find the extreme values of the function 2xy – 5x2 – 2y2 + 4x + 4y – 6. Solution f (x, y) = 2xy – 5x2 – 2y2 + 4x + 4y – 6

Let Now

fx = 2y – 10x + 4 fy = 2x – 4y + 4 fx = 0

and

2y – 10x + 4 = 0

and

2x – 4y + 4 = 0

5x – y – 2 = 0

and

x – 2y + 2 = 0

Now i.e., Solving, we get

x =

∴ The critical point of f is Now

fy = 0 implies

2 , 3

y =

4 3

FG 2 , 4 IJ H 3 3K

A = fxx = – 10,

B = fxy = 2

C = fyy = – 4 AC – B2 = (–10) (– 4) – (+2)2 = 36 > 0

and and

A = – 10 < 0

FG 2 , 4 IJ H 3 3K F 2 4 I 2 4 4 16 + 4 ⋅ 2 + 4 ⋅ 4 − 6 , Also maximum f (x, y) = f G , J = 2 ⋅ ⋅ – 5 ⋅ − 2 ⋅ H 3 3K 3 3 9 9 3 3 ∴ f attains its maximum at

= – 2. 2. Find the extreme values of the function x3y2 (1 – x – y). Solution Let

f (x, y) = x3y2 (1 – x – y)

i.e.,

f (x, y) = x3y2 – x4y2 – x3y3 fx = 3x2y2 – 4x3y2 – 3x2y3

Now

fy = 2x3y – 2x4y – 3x3y2 Then 2 2

3 2

fx 2 3

= 0

3x y – 4x y – 3x y = 0

and and

3

4

fy 3 2

= 0

2x y – 2x y – 3x y = 0

85


i.e., x2y2 (3 – 4x – 3y) = 0

and

∴

x = 0 or y = 0

or

4x + 3y = 3 and

x = 0 or y = 0

or

2x + 3y = 2

and

2x + 3y = 2

Solving

4x + 3y = 3

we get

x =

1 , 2

y =

FG 1 , 1IJ H 2 3K

Hence, the critical points are (0, 0) and Further,

x3y (2 – 2x – 3y) = 0

1 3

A = fxx = 6xy2 – 12x2y2 – 6xy3 = 6xy2 (1 – 2x – y) B = fxy = 6x2y – 8x3y – 9x2y2 = x2y (6 – 8x – 9y) C = fyy = 2x3 – 2x4 – 4x3y = 2x3 (1 – x – 3y)

(i) At the point (0, 0), A = 0, B = 0, C = 0, AC – B2 = 0 and further investigation is required.

FG 1 , 1IJ H 2 3K

(ii) At the point

Now

AC – B2 =

and

A =

A= –

–1 –1 1 ,B= ,C= 12 8 9

FG –1IJ FG −1IJ − FG – 1 IJ H 9 K H 8 K H 12 K

2

=

1 >0 144

–1 <0 9

FG 1 , 1IJ H 2 3K F 1 1 I 1 1 F 1 1I f G , J = ⋅ G1 – – J H 2 3K 8 9 H 2 3K

∴ f (x, y) attains it’s maximum at Maximum f (x, y) =

1 · 432 3. Find the extreme values of the function x3 + y3 – 3xy. Solution Let f (x, y) = x3 + y3 – 3xy =

fx = 3x2 – 3y

We have

fy = 3y2 – 3x fx = 0

Now ⇒ i.e.,

2

and

x – y = 0

and

2

and

x = y

fy = 0 implies 2

y – x = 0 x = y2

86


Eliminating y, we get x = (x2) 2

or

or ⇒ and

if x = 0, then if x = 1 then ∴The critical points are (0, 0) and (1, 1) Further,

x x – x4 x (1 – x3) x y y

= = = = = =

x4 0 0 0, x = 1 0 1

A = fxx = 6x, B = fxy = – 3 C = fyy = 6y

At (0, 0),

A = 0, B = – 3, C = 0

So that AC – B2 = 0 – 9 < 0 Hence, there is neither a maximum nor minimum at (0, 0) i.e., (0, 0) is a saddle point. At (1, 1), A = 6, B = – 3, C = 6 and AC – B2 = 6 · 6 – (– 3)2 = 36 – 9 = 27 > 0 and A = 6>0 Hence, f (x, y) attains its minimum value at (1, 1) Also, minimum,

f (x, y) = f (1, 1) = 13 + 13 – 3·1·1 = – 1.

4. Find the extreme values of the function x4 + 2x2y – x2 + 3y2. Solution f (x, y) = x4 + 2x2y – x2 + 3y2

Let

fx = 4x3 + 4xy – 2x

We have

fy = 2x2 + 6y

and Then

fx = 0 and fy = 0 implies 2

2x (2x + 2y – 1) = 0 and 2 (x2 + 3y) = 0 i.e., x = 0 or 2x2 + 2y – 1 = 0 and x2 + 3y = 0 which is same as {x = 0 and x2 + 3y = 0} or {2x2 + 2y – 1 = 0 and x2 + 3y = 0} i.e., x = 0 and y = 0. where x2 = – 3y ∴

2x2 + 2y – 1 = 0 2 (– 3y) + 2y – 1 = 0

which implies Hence, x2 =

y =

3 3 or x = ± 4 2

–1 4

87


Hence, the critical values are (0, 0), Further

F 3 , – 1I GH 2 4 JK

and

F − 3 , – 1I GH 2 4 JK

A = fxx = 12x2 + 4y – 2, B = fxy = 4x, C = fyy = 6 (i) At (0, 0), A = – 2, B = 0, C = 6 and AC – B2 = – 12 < 0 Hence, there is neither a maximum nor a minimum at (0, 0) (ii) At

F 3 , – 1I , GH 2 4 JK

FG IJ H K

A = 12 ⋅

–1 3 –2=6 +4 4 4

B = 4⋅

3 = 2 3, C = 6 2

e j

AC – B2 = 6 (6) – 2 3

Then,

2

= 24 > 0 and

A = 6>0 ∴ f (x, y) has a minimum at

F 3 , – 1I GH 2 4 JK

Hence, f (x, y) attains its minimum value at Also,

FG – 3IJ H8K

minimum f (x, y) =

(iii) Similarly, at

F − 3 , – 1I GH 2 4 JK

f (x, y) attains its minimum. Thus, f (x, y) attains minimum

–3 at 8

F± GH

F 3 , – 1I GH 2 4 JK

I JK

3 –1 . , 2 4

5. Determine the maxima/minima of the function sin x + sin y + sin (x + y). Solution Let We have

f (x, y) = sin x + sin y + sin (x + y) fx = cos x + cos (x + y) fy = cos y + cos (x + y)

Now

fx = 0 and fy = 0 implies cos (x + y) = – cos x and cos (x + y) = – cos y

i.e., or

– cos x = – cos y or cos x = cos y x = y

Then,

cos 2x = – cos x = cos (π – x)

88


2x = π – x or x =

or So that

y =

The critical point is Further,

FG π , π IJ H 3 3K

π 3

π 3

A = fxx = – sin x – sin (x + y) B = fxy = – sin (x + y)

At

FG π , π IJ , H 3 3K

C = fyy = – sin y – sin (x + y) A = – sin

π 2π 3 3 – sin = − − = – 3 3 3 2 2

B = – sin

2π 3 =– 3 2

C = – sin

π 2π 3 3 – sin =– – =– 3 3 3 2 2

2

and

AC – B =

e

je

A = – 3<0 π π , ∴ f (x, y) attains its maximum at 3 3 Also

and

FG H

maximum f (x, y) = f =

j FGH

– 3 – 3 – –

FG π , π IJ H 3 3K

3 2

I JK

IJ K

= sin

2

=

9 >0 4

FG IJ H K

π π 2π + sin + sin 3 3 3

3 3 3 3 3 + + = . 2 2 2 2

6. Examine the function f (x, y) = 1 + sin (x2 + y2) for extreme. Solution

f (x, y) = 1 + sin (x2 + y2) fx = 2x cos (x2 + y2) fy = 2y cos (x2 + y2) fx = 0 and fy = 0 implies

Now i.e.,

2x cos (x2 + y2) = 0 and 2y cos (x2 + y2) = 0

∴

x = 0, y = 0 and (0, 0) is the stationary point. A = fxx = – 4x2 sin (x2 + y2) + 2 cos (x2 + y2) B = fxy = – 4xy sin (x2 + y2) C = fyy = – 4y2 sin (x2 + y2) + 2 cos (x2 + y2)

At (0, 0);

A = 2, B = 0, C = 2

89


∴

AC – B2 = 4 > 0

Since,

AC – B2 > 0, A = 2 > 0, (0, 0) is minimum point and the minimum value of f (x, y) = f (0, 0) = 1.

EXERCISE 2.5 Find the extreme values of the following functions: 1. x2 – xy + y2 + 3x – 2y + 1 2. x2 + xy + y2 + 3x – 3y + 4 3. x2 + 2y2 – 4x + 4y + 6 4. x3 + 3xy2 – 15x2 – 15y2 + 72x 5. x3 + y3 – 63 (x + y) + 12xy

6. x2y2 (12 – 3x – 4y) 7. x3 – y3 – 3y2

LM Ans. Min. N

IJ OP KQ Ans. Min. = − 5 at b −3, 3g Ans. Min. = 0 at b2, − 1g Ans. Max. = 112 at b4, 0g LMAns. Max. = 784 at b−7, − 7gOP MN Min. = − 216 at b3, 3g PQ Ans. Max. = 8 at b2, 1g Ans. Max. = 0 at e3 4, − 2j LMAns. Max. = a at FG a , a IJ OP 27 H 3 3 K Q N LMAns. Min. = − 2 at FG 2, 1 IJ OP H 2K Q N OP LMAns. Min. = 0 at b0, 0g PP MM Max. = 125 at FG − 5 , 0IJ H 7 3 K MM and b−1, 2g, b−1, − 2g are saddle pointsPP Q N Ans. Min. = − 14 and 2 at b1, 1g and b −1, 1g LMAns. Min. = − 1 at FG − 1 , − 1IJ OP H 3 3K Q 3 N =

FG H

–4 –4 1 at , 3 3 3

3

8. xy (a – x – y), where a > 0 9. x2y (x + 2y – 4) 10. 2x3 + xy2 + 5x2 + y2

11. x4 + y4 – (x + y)4 12. x2 + xy + y2 + x + y

2.4 LAGRANGE’S METHOD OF UNDETERMINED MULTIPLIERS So far, we have considered the method of finding the extreme values of a function f (x, y, z), where these variables x, y, z are independent. Sometimes we may have to find the maximum or minimum

90


values of a function f (x, y, z), when x, y, z are connected by some relations say φ (x, y, z) = 0. In such cases we may eliminate z from the given conditions and express the function f (x, y, z) as a function of two variables x and y and obtain the extreme values of f as before. In such cases we have an alternative method for finding the critical points called “Lagrange’s method of undetermined multipliers”. Suppose we want to find the maximum or minimum values of the function µ = f (x, y, z)

...(1)

Subject to the condition φ (x, y, z) = 0

...(2)

d µ = fx dx + fy dy + fz dz

Then,

d φ = φx dx + φy dy + φz dz = 0

and

But the necessary conditions for the function f (x, y, z) to have a maximum or a minimum values are fx = 0, fy = 0, fz = 0. Hence, and

fx dx + fy dy + fz dz = 0

...(3)

φx dx + φy dy + φz dz = 0

...(4)

Multiplying (4) by λ and adding it to (3), we get ( fx + λ φx) dx + ( fy + λ φy) dy + ( f + λ φz) dz = 0 This is possible only if fx + λ φx = 0

...(5)

fy + λ φy = 0

...(6)

fz = λ φz = 0

...(7)

Solving the equations (2), (5), (6) and (7), we get the values of x, y, z and the undetermined multiplies λ. Thus, we obtain the critical points of the function f (x, y, z). But this method does not help us in identifying whether the critical points of the function gives the maximum or minimum.

Working Rules To find the extreme values of the function f (x, y, z) subject to the conditions φ (x, y, z) = 0 1. Form the auxiliary equation F (x, y, z, λ) = f (x, y, z) + λ φ (x, y, z) 2. Find the critical points of F as a function of four variables x, y, z, λ. i.e., solving the equations Fx = 0, Fy = 0, Fz = 0 and Fλ = φ = 0, we get λ = –

fy fx f =– =– z φx φy φz

The values of x, y, z thus obtained will give us the critical point (x, y, z).

91


WORKED OUT EXAMPLES 1. Find the minimum value of x2 + y2 + z2 subject to the condition ax + by + cz = p. Solution F = (x2 + y2 + z2) + λ (ax + by + cz)

Let

We form the equations Fx = 0, Fy = 0, Fz = 0 2x + λa = 0, 2y + λb = 0, 2z + λc = 0

i.e.,

λ =

or

– 2x , – 2y , – 2z λ= λ= a b c

– 2x – 2 y – 2z = = a b c

⇒

x y z = = k (say) = a b c x = ak, y = bk, z = ck

or ∴ But

ax + by + cz = p and hence, we have 2

a k + b2k + c2k = p ∴

k =

p a + b2 + c2 2

Hence, the required minimum value of x2 + y2 + z2 is a2k2 + b2k2 + c2k2 = k2 (a2 + b2 + c2)

d

p 2 a 2 + b 2 + c2 i.e.,

da

2

+ b2 + c2

i

i

2

=

thus, the required minimum value is

p2 a 2 + b2 + c2 p2 . a 2 + b2 + c2

2. The temperature T at any point (x, y, z) in space is T = 400 xyz2. Find the highest temperature at the surface of the unit sphere x2 + y2 + z2 = 1. Solution Let F = 400 xyz2 + λ (x2 + y2 + z2) We form the equations Fx = 0, Fy = 0, Fz = 0 i.e.,

– 200 yz 2 400 yz + λ · 2x = 0 or λ = x 2

– 200 xz 2 y 800 xyz + λ · 2z = 0 or λ = – 400 xy 400 xz2 + λ · 2y = 0 or λ =

Now,

−200 yz 2 x

=

−200 xz 2 = − 400xy y

92


Taking the equality pairs, we get

y x

=

x 2 , z = 2y2, z2 = 2x2 y

i.e., y 2 = x2 or y = x also z = 2 2 2 But x + y + z = 1 and hence, we have

2x

1 2

x2 + x2 + 2x2 = 1 i.e., 4x2 = 1 or x = ∴

x =

1 1 , y= ,z= 2 2

1 2

is a stationary point.

The maximum (height) temperature T = 400 xyz2 is = 400

FG 1IJ FG 1 IJ FG 1 IJ = 50. H 2K H 2K H 2K

1 1 1 + + = 1, show that the minimum value of the function a3x2 + b3y2 + c3z2 is x y z (a + b + c)3. Solution 3. If

F = (a3x2 + b3y2 + c3z2) + λ

Let we form the equations i.e.,

Fx = 0, Fy = 0, Fz = 0

FG – 1IJ Hx K F – 1I 2b y + λ G J Hy K F – 1I 2c z + λ G J Hz K 2a3x + λ

2

= 0 or λ = 2a3x3

2

= 0 or λ = 2b3 y3

2

= 0 or λ = 2c3z3

3

3

Now ⇒ ⇒ ∴ But

∴

2a3x3 = 2b3 y3 = 2c3z3 a3x3 = b3y3 = c3z3 ax = by = cz ax , z = ax b c 1 1 1 1 b c + + + + = 1 i.e., =1 x y z x ax ax

y =

a +b+c = 1 ax a +b+c x = a

FG 1 + 1 + 1IJ H x y zK

93


Also

y =

a +b+c , z = a +b+c b z

Required minimum value of the function a3x2 + b3y2 + c3z2 is given by 3 = a ⋅

FG a + b + c IJ H a K

2

+ b3

FG a + b + c IJ H b K

2

+ c3

FG a + b + c IJ H c K

2

= (a + b + c)2 (a + b + c) = (a + b + c)3 Thus, the required minimum value is (a + b + c)3. 4. Find the minimum value of x2 + y2 + z2, when x + y + z = 3a. Solution F = (x2 + y2 + z2) + λ (x + y + z)

Let

We form the equations Fx = 0, Fy = 0, Fz = 0 2x + λ = 0, 2y + λ = 0, 2z + λ = 0

i.e.,

λ = – 2x, λ = – 2y, λ = – 2z

or ⇒

– 2x = – 2y = – 2z or x = y = z

But

x + y + z = 3a

Substituting y = z = x, we get 3x = 3a x =a ∴

x = a, y = a, z = a

The required minimum value of x2 + y2 + z2 is a2 + a2 + a2 = 3a2. 5. Find the minimum value of x2 + y2 + z2 subject to the conditions xy + yz + zx = 3a2. [July 2003] Solution F = (x2 + y2 + z2) + λ (xy + yz + zx) = 0

Let

We form the equations Fx = 0, Fy = 0, Fz = 0 2x + λ (y + z) = 0, 2y + λ (x + z) = 0, 2z + λ (x + y) = 0

i.e.,

– 2x – 2y – 2z ,λ= ,λ= · y+z x+z x+y Equating the R.H.S. of these, we have ⇒

λ =

2y 2z 2x = = x+z x+y y+z Consider,

i.e., or

y x = x+z y+z 2 x + xz = y2 + yz or (x2 – y2) + z (x – y) = 0 (x – y) (x + y + z) = 0

⇒

x = y or x + y + z = 0

...(1)

94


we must have

x = y, since x + y + z cannot be zero.

Suppose

x + y + z = 0, then by squaring, we get

2

2

2

(x + y + z ) + 2 (xy + yz + zx) = 0 ⇒

x2 + y2 + z2 + 2 (3a2) = 0 x2 + y2 + z2 = – 6a2 < 0

or

which is not possible. Similarly by equating the other two pairs in (1), we get y = z, z = x thus x = y = z xy + yz + zx = 3a2, putting y = z = x, we get

But

3x2 = 3a2 ⇒ x = a Thus, x = a = y = z and the minimum value of x2 + y2 + z2 is a2 + a2 + a2 = 3a2.

EXERCISE 2.6 1. Find the minimum value of x2 + y2 + z2 subject to the condition xyz = a3. [Ans. 3a2] 2. Find the maximum and minimum values of x2 + y2 subject to the conditions 5x2 + 6xy + 5y2 = 8.

[Ans. 4 and 1] 2

2 2

2

2

2

3. Find the maximum value of x y z subject to the condition x + y + z = a2.

LM F a I OP MNAns. GH 3 JK PQ 2

3

4. Find the minimum value of x2 + y2 + z2 subject to the conditions (i) xy + yz + zx = 3a2

[Ans. 3a2]

(ii) xyz = a3

[Ans. 3a2]

LMAns. N a

(iii) ax + by + cz = P.

2

P2 + b 2 + c2

OP Q

ADDITIONAL PROBLEMS (From Previous Years VTU Exams.) 1. Evaluate:

Fa +b +c I lim G H 3 JK lim b2x tan x − π sec x g . x

(i)

x →0

(ii)

x→

π 2

x

x

1 x

95


Solution (i)

Fa lim G H

x

x→0

+ bx + cx 3

I JK

Let

1 x

[1∞] form

K =

Fa lim G H

x

x→0

+ bx + cx 3

Taking logarithm on both sides log K = lim

x →0

Apply L’ Hospital rule

I JK

1 x

F GH

1 a x + b x + cx log 3 x

I JK

R| FG 3 IJ F a log a + b G H a +b +c K H | lim S 1 || T 3 F log a + log b + log c I ×G JK 3 H 3 x

x

=

=

i.e.,

x →0

x

+ bx + cx 3

I JK

1 x

=

x

log b + c x log c 3

I U| JK | V| |W

b g

1 log abc 3

b g babcg

log K = log abc

Fa lim G H

x

x →0

log K =

K =

x

1 3

1 3

babcg

(ii) Let

1 3

b

.

K = limπ 2 x tan x − π sec x x→

2

= limπ x→

2

= lim

π x→ 2

Applying L’ Hospital rule K = lim

π x→ 2

g

LM2 x ⋅ sin x – π ⋅ 1 OP N cos x cos x Q LM 2 x sin x − π OP N cos x Q

LM 2 x cos x + 2 cos x OP = − 2 . N – sin x Q

[∞ – ∞] form

LM 0 OP N0Q

form

96


b

2. Evaluate: lim tan x x→

π 4

g

tan 2x

.

Solution. Refer page no. 75. Example 2. 3. Evaluate: (i) lim

x →0

LM 1 − 1 OP N x sin x Q 2

b g

(ii) lim sin x

2

π x→ 2

tan x

.

Solution. (i) Refer page 61. Example 5.

b g

(ii) lim sin x π x→ 2

[1∞] form

tan x

b g

K = limπ sin x

Let

x→

2

tan x

l

q

log K = limπ tan x log sin x x→

2

= lim

π x→ 2

Apply L’ Hospital rule

=

LM log sin x OP N cot x Q

LM 0 OP N0Q

form

LM 0 OP N0Q

form

LM 1 ⋅ cos x OP lim M sin x MN – cosec x PPQ L – cos x × sin x OP lim M N sin x Q

x→

2

π 2

2

=

x→

π 2

log K = 0 K = e0 = 1

i.e.,

b g

lim sin x

π x→ 2

tan x

= 1.

LM x b1 + a cos xg − b sin x OP x N Q L x b1 + a cos xg − b sin x OP lim M x N Q

4. Find the value of a and b such that lim

3

x →0

Solution. Consider

3

x →0

Applying L’ Hospital rule, we have = lim

x →0

b

g b

g

x – a sin x + 1 + a cos x − b cos x 3x

2

= 1.

97


1+ a − b and we must have 0 1+a–b = 0 To apply the L’ Hospital rule again. Hence, we have This is equal to

= lim

x →0

...(1)

LM x b– a cos xg − 2a sin x + b sin x OP 6x N Q

LM 0 OP N0Q

form

Applying the L’ Hospital rule again, we have = lim

x →0

b

g

x a sin x − 3a cos x + b cos x 6

– 3a + b which must be equal to 1 6 – 3a + b = 6

This is equal to

∴ Solving (1) and (2), we get a = 5. Evaluate: lim

b

log cos x

π x → 2

tan x

–5 2

and b =

g.

...(2)

–3 · 2

Solution. Refer page no. 67, Example 14. 6. Find the maximum and minimum distances of the function x4 + 2x2y – x2 + 3y2. Solution. Refer page no. 86, Example 4. 7. Expand eax + by in the neighbourhood of the origin up to the third degree term. Solution. Refer page no. 80, Example 3. 8. Expand log (1 + x – y) up to third degree terms about the origin. Solution. By Maclaurin’s series

b g b g b g 1 + ox f b0, 0g + 2 xy f b0, 0g + f b0, 0gt + ⋅⋅⋅⋅⋅ ⋅⋅⋅ 2!

f (x, y) = f 0, 0 + x f x 0, 0 + y f y 0, 0 2

xx

Here,

f (x, y) = log (1 + x – y)

xy

yy

⇒

f (0, 0) = log 1 = 0

fx =

1 1+ x − y

⇒

fx (0, 0) = 1

fy =

–1 1+ x − y

⇒

fy (0, 0) = – 1

⇒

fxx (0, 0) = – 1

fxx =

–1

b1 + x − yg

2

98


fxy = fyy = fxxx = fxxy = fxyy = fyyy =

1

b1 + x − yg

2

⇒

fxy (0, 0) = 1

2

⇒

fyy (0, 0) = – 1

3

⇒

fxxx (0, 0) = 2

3

⇒

fxxy (0, 0) = – 2

3

⇒

fxyy (0, 0) = 2

3

⇒

fyyy (0, 0) = – 2

–1

b1 + x − yg 2

b1 + x − yg –2

b1 + x − yg 2

b1 + x − yg –2

b1 + x − yg

and so on. Substituting these in (1)

b

g

t 1 + o2 x − 6 x y + 6 xy 3! b x − yg – 21 b x − yg + 13 b x − yg + ⋅⋅ ⋅⋅⋅

log (1 + x – y) = 0 + x − y +

o

1 – x 2 + 2 xy − y 2 2!

3

=

2

2

2

3

9. Divide the number 48 into three parts such that its product is maximum. Solution. Let x, y, z be the three parts of the number 48 ∴ Also, let

x + y + z = 48 u = xyz and F = xyz + λ (x + y + z)

We form the equations Fx = 0, Fy = 0, Fz = 0 i.e., or ⇒ and hence Since

yz + λ = 0; λ = – yz;

xz + λ = 0; λ = – xz

xy + λ = 0 and

λ = – xy

– yz = – xz = – xy x = y=z x + y + z = 48, we get x = y = z = 16

Thus, 16, 16, 16 are the three parts of 48 such that the product is maximum. 10. Find the minimum value of x2 + y2 + z2 when x + y + z = 3a. Solution. Refer page no. 93. Example 4.

t

− 2 y 3 + ⋅ ⋅⋅⋅ ⋅⋅⋅

99


11. Find the stationary value of x2 + y2 + z2 subject to the condition xy + yz + zx = 3a2. Solution. Refer page no. 93. Example 5. 12. If x, y, z are the angles of a triangle, show that the maximum value of cos x cos y cos z is

1 · 8

Solution. We need to find the maximum value of u = cos x cos y cos z subject to the condition x + y + z = π Let F = cos x cos y cos z + λ (x + y + z) We form the equations Fx = 0, Fy = 0, Fz = 0 i.e., – sin x cos y cos z + λ = 0 – cos x sin y cos z + λ = 0 – cos x cos y sin z + λ = 0 or λ = sin x cos y cos z λ = cos x sin y cos z λ = cos x cos y sin z Now, sin x cos y cos z = cos x sin y cos z = cos x cos y sin z From the first pair, we have sin x cos y = cos x sin y i.e., sin x cos y – cos x sin y = 0 i.e., sin (x – y) = 0 ⇒ x – y =0 or x =y similarly from the other pairs, we get y = z and z = x Combining these we have x = y = z But x+y+z =λ ∴ x+x+x =π or

x=

Hence

π 3

x =y=z=

π 3

∴ The maximum value of cos x cos y cos z = cos3x, where x = Thus, we have

cos3

FG π IJ = FG 1 IJ H 3 K H 2K

3

=

π 3

1 · 8

13. Find the point on the paraboloid z = x2 + y2 which is closest to the point (3, – 6, 4). Solution. Let A (3, –6, 4) and let P(x, y, z) be any point on the paraboloid x2 + y2 – z = 0 ∴ AP2 = (x – 3)2 + ( y + 6)2 + (z – 4)2 by distance formula

100


Let u (x, y, z) = (x – 3)2 + ( y + 6)2 + (z – 4)2 and we need to find the point P1 = (x1, y1, z1) Satisfying z = x2 + y2 such that AP12 is minimum. Now, let F = [(x – 3)2 + ( y + 6)2 + (z – 4)2] + λ (x2 + y2 – z) We form the equation Fx = 0, Fy = 0, Fz = 0

b g

− x −3

i.e., 2 (x – 3) + 2λx = 0

or

λ=

2 ( y + 6) + 2λy = 0

or

λ=

2 (z – 4) – λ = 0

or

λ = 2z – 8

∴

−1 +

3 6 = −1 − x y

Also

−1 +

3 = 2z – 8 ⇒ x

b

x

= −1+

g = −1− 6

− y+6 y

y

−6 3 = y x

⇒

3 x

or y = – 2x

3 = 2z or x

7+

z =

FG H

1 3 7+ x 2

IJ K

But we have x2 + y2 = z

7x + 3 7x + 3 or 5x2 = 2x 2x 3 or 10x – 7x – 3 = 0 x = 1 is a root by inspection and is the only real root.

∴

Also

x2 + 4x2 =

y = – 2x and

z =

FG H

1 3 7+ x 2

IJ K

gives

y = – 2, z = 5 Thus the required point is (1, – 2, 5). 14. Find the dimensions of the rectangular box, open at the top, of the maximum capacity whose surface area is 432 sq. cm. Solution. Let x, y, z respectively be the length, breadth and height of the rectangular box. Since it is open at the top, the surface area (S) is given by S = xy + 2xz + 2yz = 432 (using the data) Volume (V) = xyz We need to find x, y, z such that V is maximum subject to the condition that xy + 2xz + 2yz = 432 Let F = xyz + λ (xy + 2xz + 2yz) We form the equation Fx = 0, Fy = 0, Fz = 0 i.e., yz + λ ( y + 2z) = 0 or λ = – yz/( y + 2y) xz + λ (x + 2z) = 0 or λ = – xz/(x + 2z) xy + λ (2x + 2y) = 0 or λ = – xy/2(x + y) Now

− xz − xy − yz = = x + 2z 2 x+ y y + 2z

b

g

101


or

y x = y + 2z x + 2z

Also

z y = x + 2z 2x + 2 y

Hence, But

gives

x = y

gives

y = 2z

x = y = 2z

xy + 2xz + 2yz = 432 x2 + x2 + x2 = 432 3x2 = 432 or x2 = 144

or

we have x = 12 and hence y = 12, z = 6 Thus the required dimensions are 12, 12, 6.

OBJECTIVE QUESTIONS 1. The necessary conditions for f (x, y) = 0 to have extremum are (a) fxy = 0, fyx = 0

(b) fxx = 0, fyy = 0

(c) fx = 0, fy = 0

(d ) fx = 0, fy = 0 and fxx > 0, fyy > 0. [Ans. c]

2. The stationary points of f (x, y) = x2 + xy2 + y4 is (a) (1, 0)

(b) (0, 1) (d ) (1, 1).

(c) (0, 0) 3. Minimum value of

x2

+

y2

+ 6x + 14 is

(a) 5

(b) 2 (d ) – 2.

(c) 3 4. If p = q = 0, rt –

s2

(b) Saddle point

(c) Maximum

(d ) None of these.

5. The conditions for f (x, y) to be minimum or maximum is p = ......, q = ....., rt – (a) 0, 0, + ve

(b) 1, –1, + ve

(c) 0, 0, – ve

(d ) None of these.

x →−5

(a)

[Ans. c] s2

= ..... is [Ans. a]

x 2 + 3x − 10 is equal to x 3 + 125

7 25

(c) −

[Ans. a]

> 0, r < 0 then f (x, y) is

(a) Minimum

6. Lim

[Ans. d ]

7 75

(b)

7 125

(d ) None of these.

[Ans. c]

102


LM x − Nx − 3 x

7. Lim x→3

2

9 − 3x

OP is equal to Q

(a) 0

(b) 3

(c) 9

(d ) 2.

LM x − Nx − 3 x

8. Lim x→3

(a)

2

9 − 3x

OP is equal to Q

1 2

(b) 0 (d ) None of these.

(c) 1 x →0

1 2

(c)

x →0

(a)


2

10. Lim


(c) 2

x →∞

FG x IJ H x + 1K

is equal to (b) e

(c) 1

(d ) None of these.

π x→ 4

[Ans. a]

x

(a) 2

12. Lim

[Ans. c]

x2 + x is equal to x + sin x

1 2

11. Lim

[Ans. a]

x is equal to 1 − cos x

9. Lim

(a)

[Ans. d]

[Ans. c]

1 − tan x is equal to π −x 4

(a) 2

(b) – 2

(c) 1

(d ) –1.

[Ans. a]

13. Lim x cot x is equal to x→0

(a) 0

(b) 1

(c) doesn’t exist

(d ) None of these.

[Ans. b]

103


14. Lim x →0

sec x − 1 is equal to tan 2 x

(a) 2

(b) 1

(c) 0

(d )

1 · 2

(b)

1 15

(d )

3 · 5

[Ans. d]

15. Lim tan 5x cot 3x is equal to x→0

(a) 15

5 3

(c)

16. Lim h→ 0

b x + hg

2

− x2

h

[Ans. c]

is equal to

(a) 1

(b) x

(c) 0

(d ) 2x.

[Ans. d]

17. The function (x) is continuous for

18.

(a) all x

(b) x ≠ 0

(c) x > 0

(d ) x < 0.

F 1I Lim G 1 − J H xK

−x

x →∞

is equal to

(a) e–1 (c)

(b) e

e–x

19. If rt –

(d ) e x. s2

(b) Minimum value of f (x, y)

(c) Saddle point

(d ) None of these.

s2


(c) Saddle point

(d ) None of these.

21. If rt –

[Ans. a]

> 0, r > 0 then f (a, b) is

(a) Maximum value of f (x, y) s2

[Ans. b]

> 0, r < 0 then f (a, b) is

(a) Maximum value of f (x, y) 20. If rt –

[Ans. a]

[Ans. b]

< 0 then (a, b) is

(a) Maximum value of f (x, y)


(c) Saddle point f (x, y)

(d ) None of these.

[Ans. c]

22. If L (x, y, z, λ) = f (x, y, z) + λ φ (x, y, z) is called (a) Particular function

(b) Auxillary function

(c) General function

(d ) None of these.

[Ans. b]

104


23. To find maxima or minima of f (x, y, z) subject to (a) φ (x, y, z) = 0

(b) f (x, y, z) = 0

(c) φ (x, y, z) ≠ 0

(d ) f (x, y, z) ≠ 0.

[Ans. a]

24. A set of necessary conditions for f (x, y) to have a maximum or minimum is that (a)

∂f ∂f ≠ 0 = 0 and ∂y ∂x

(b)

∂f ∂f = 0 ≠ 0 and ∂y ∂x

(c)

∂f ∂f = 0 = 0 and ∂y ∂x

(d ) None of these.

[Ans. c]

25. The stationary points of f (x, y) = y2 + 4xy + 3x2 + x3 is

FG 2 , 1IJ H3 K F 2 4I (c) (0, 2) and G , J H 3 3K (a) (0, 1) and

(b) (0, 0) and

FG 2 , −4 IJ H3 3 K

(d ) None of these.

[Ans. b] GGG

106


Similarly, we can define another

z LMNz b g OPQ z LMMNz b g OPPQ z LMMNz d

b

c

a

f x , y dx dy

For continuous function f (x, y), we have

zz

b g

b

d

d

R

b

f x , y dy dx =

f x , y dx dy =

a

c

c

a

O b g PP Q

f x , y dx dy

If f (x, y) is continuous on a bounded region S and S is given by S = {(x, y)/a ≤ x ≤ b and φ1 (x) ≤ y ≤ φ2 (x)}, where φ1 and φ2 are two continuous functions on [a, b] then

zz

b g

LM MN

S

OP f b x , y g dy dx PQ bg

z z b

f x , y dx dy =

a

bg

φ2 y

y

y = B2 (x)

φ1 x

The iterated integral in the R.H.S. is also written in the form

z z

bg

φ2 x

b

dx

zz

y = B1(x)

S = {(x, y)/c ≤ y ≤ d and φ1 ( y) ≤ x ≤ φ2 ( y)}

Similarly, if

then

bg

b g

f x , y dy

φ1 x

a

S

b g

zz d

f x , y dx dy =

S

c

LM N

bg φ b yg φ2 y 1

a

b

x

Fig. 3.1

O f b x , y g dx P dy Q

If S cannot be written in neither of the above two forms we divide S into finite number of subregions such that each of the subregions can be represented in one of the above forms and we get the double integral over S by adding the integrals over these subregions.

1. Evaluate: I = Solution

zz

WORKED OUT EXAMPLES

1

2

0

0

xy 2 dy dx. I =

=

z LMN z z LMN OPQ z 1

2

0

0

1

xy 3 3

0

=

1 3

1 = 3

1

OP Q

xy 2 dy dx 2

dx 0

8 x dx

0

LM 8x OP N2Q

2 1

0

=

4 ⋅ 3

(Integrating w.r.t. y keeping x constant)

107

INTEGRAL CALCULUS

zz

2. Evaluate:

1

2

0

1

xy dy dx.

Solution. Let I be the given integral

z RST z z LMN OPQ 1

Then,

I =

y dy dx

0

1

y2 x⋅ 0 2 1

= 3. Evaluate the following:

zz zz 2

(i) (iii)

3

2

2

3y

y

dx. = 1

(ii)

( x + y ) dx dy

0

1

x

0

x2

0

z FGH z z z d 2

I =

ex

1

ex e y

1

2

= = = I = I = =

cos θ

0

dx

2

i

– e2

z

2

x

1

3

2

x 2

3

2

2

1

1

z z z z 1

x

0

x2

1

x x2

0

1

2

0

3 1

0

1 1 1 – = = 2 3 6

zz z z 2

(iii)

I =

1

2

=

1

2

=

r sin θ dr dθ

e y dy dx 3

1

RS b x + yg dxUV dy T W LM x + xyOP dy N2 Q 3y

y

2

x =3y

x=y

6 y 2 dy = 2 y 3

2 1

3 4·

dy dx

i e dx –e i e – e ide – e i · RS dyUV dx T W y dx = d x − x i dx LM x − x OP N2 3Q 3

2

=

x dx =

e x e 3 − e 2 dx

1

de de de

1

IJ K

3

2

2

=

(ii)

3 2

π

(iv)

Solution. Let I be the given integral. Then (i)

z zz zz

2

e x + y dy dx

1

1

UV W

2

x

= 14

108


(iv)

I =

=

z z

= cos θ = – sin θ dθ = sin θ dθ =

where ∴

=

z

sin θ

0

0

z

0

π

1 sin θ ⋅ cos2 θ dθ 2 0 t dt – dt 1 π 2 t ⋅ – dt 2 0

z

b g

LM t OP N3Q

3 π

–1 = 2

0

–1 π cos θ 0 = 6 –1 1 –1 − 1 = ⋅ = 6 3

b

zz 1

4. Evaluate:

0

0

1+x 2

g

dy dx ⋅ 1 + x2 + y2

z z 1

Solution

I =

0

R| S| T

1 + x2

0

U| V| W

dy dx 1 + x2 + y2 1 + x2

a2 = 1 + x2 or a =

where

=

R| S| T

zz z LMN z 1

0

1

=

0

1

=

= = Note :

cos θ

2

π

0

R| r dr U| dθ S| V| T W L r O dθ sin θ M P N2Q cos θ

π

LM RS N T

π log x + x 2 + 1 4

UVOP WQ

1

= 0

0

a

0

U| V| W yO a PQ

dy dx 2 a + y2 1 tan –1 a

a

dx 0

π 1 π ⋅ dx = a 4 4

z

dx

1

0

x2 + 1

LM { x + 1}OP N Q π log e 2 + 1j 4 π log x + 4

π sin h −1 x 4

1 0

=

1

2

0

bg

π sin h −1 1 4

109

INTEGRAL CALCULUS

z z zd c

5. Evaluate:

b

a

i

x 2 + y 2 + z 2 dz dy dx.

– c –b – a

z z zd c

Solution

I =

b

a

i

x 2 + y 2 + z 2 dz dy dx

x = – c y = –b z = – a

Integrating w.r.t. z, x and y – constant.

z z z z z z c

=

b

=

b

=

z =–a

3

b

3

z z z

2

LM2ax y + 2ay + 2a yOP dx 3 3 Q N LM2ax bb + bg + 2a db + b i + 2a bb + bgOP dx 3 3 N Q LM4ax b + 4ab + 4a b OP dx 3 3 Q N LM4ab F x I + 4ab b xg + 4a b b xgOP 3 MN GH 3 JK 3 PQ F 2c I + 4ab ⋅ b2cg + 4a b b2cg 4ab G 3 H 3 JK 3 c

=

3

b

3

2

y = –b

x=–c c

=

3

2

3

3

x=–c c

=

3

3

2

x=–c

3

=

3

c

3

–c

3

=

3

3

8abc 3 8ab 3c 8a 3bc + + 3 3 3 8abc 2 a + b 2 + c2 ⋅ I = 3 =

d

zz z a

6. Evaluate:

e x + y + z dz dy dx.

0

zzz a

Solution

i

x x+y

0 0

I =

3

2

2

x = – c y = –b

Integrating w.r.t. y, x – constant.

2

2

x = – c y = –b c

a

3

2

x = – c y = –b c

LM x z + y z + z OP dy dx 3Q N LMx ba + ag + y ba + ag + F a + a I OP dy dx GH 3 3 JK PQ MN F 2ax + 2ay + 2a I dy dx GH 3 JK

x

x+y

x = 0 y =0 z = 0

e x + y ⋅ e z dz dy dx

110


zz zz d i i z zd R| L O U| z S|T MN PQ V|W RS d i d iUV zT W I JK z FGH x

a

x+ y

e x+ y ⋅ ez

=

0

x = 0 y =0 a

dy dx

x

e x + y e x + y – 1 dy dx

=

x = 0 y =0 a

x

e 2 x ⋅ e 2 y – e x ⋅ e y dy dx

=

x = 0 y =0 a

e

=

e2 y 2

2x

x =0 a

=

x =0 a

=

x =0

LM e N8 Fe GH 8

4x

=

4a

= =

z zz log 2

7. Evaluate: Solution

0

x

0

I =

x

– ex ey

x

0

dx

0

e2 x 2 x e − 1 − e x e x − 1 dx 2

e4 x 3 2 x − e + e x dx 2 2

− −

OP Q I F1 3 I + e J – G − + 1J K H8 4 K

3e 2 x + ex 4 2a

3e 4

a

0

a

3 e 4 a 3e 2 a – + ea – 8 4 8

d

i

1 4a e – 6e 2 a + 8e a – 3 ⋅ 8

x + log y

ex + y + z dz dy dx.

0

I = = = =

z z z z z

But

elog y = y

∴

I =

log 2

x =0 log 2

x=0 log 2

x=0 log 2

x=0

log 2

x=0

z z z z z z d x

x + log y

y =0

z=0

x

y=0 x

y=0 x

y=0

x

y=0

ex + y ez

ex + y. ez dz dy dx x + log y 0

dy dx

e x + y e x + log y − 1 dy dx e x + y e x ⋅ e log y – 1 dy dx

i

e 2 x ⋅ y ⋅ e y – e x e y dy dx

INTEGRAL CALCULUS

= = = =

z z z z

111

d

log 2

x=0 log 2

e2x

x=0 log 2

x=0 log 2

x =0

{d x e

Thus, 8. Evaluate:

zz a

0

a2 – y2

dx e

3x

– e 3 x + e x dx 3x

e 3x + ex – 3

3x

OP Q

O +e P Q O 4 – 0P – de Q 9

log 2

0

log 2

x

0

i d

b g b g

8 log 2 28 – +1 3 9 8 log 2 19 – . 3 9

a 2 – x 2 – y 2 dx dy.

I =

zz zz LM zN z 0

a

0

|RS |T RS T

a2 – y2

0 b

da

=

0

a

=

0

UV W

OP dy Q da – y i dy

x b2 x b2 – x 2 + sin –1 b 2 2

b π π . dy = 2 2 4 2

LM N

|UV |W

i

– y 2 – x 2 dx dy

b 2 – x 2 dy

0

b2 = a2 – y2 a

2

π 2 y3 a y– = 4 3

OP Q

a

= 0

z

a

0

πa 3 6

2

i

– 1 + e log 2 – 1

3 log 2

=

0

i

i

8 log 2 4 – 8–1 + 2 –1 3 9

= where

d

– e x – 0 – 1 – e x e x – 1 dx

i

LM x . e – e N 3 9 LM x e – 4 e N 3 9 LM log 2 . e N 3

a

Solution

i b g}

x

=

I =

dx

– e 3 x + e 2 x – e 2 x + e x dx

3 log 2

=

y=0

3x

3x

=

x

dx e 3x

=

i

e2 x y e y – e y – e x e y

2

b

0

112


EXERCISE 3.1 I. Evaluate the following double integrals:

zz zz zz b zz zz z z 3

1.

0

5.

x2

0

1

1– x

0

0

9.

g

2

dy dx

x 2 y dy dx

0

4a

2 ay

0

y 2 /4a

π

10.

LM Ans. 1 e − 1OP N 2 Q LMAns. 1 OP N 4Q LMAns. a OP N 15 Q LM Ans. 16 a OP N 3 Q

θ=0

4.

ar

r =0

8.

a2 – r 2

zzz zzz zzz z zz b z z z zz z zz z 2

3. 5. 7.

0

1

1

1

9.

xy 2 z dz dy dx

1

y

0

0

0

1

1

2

0

x +y 2

0

1

z

x+z

–1

0

x–z

2

ea

0

0

0

1

1– x

1– x – y

0

0

0

a2 – x2

0

0

0

1

1– x

0

b

g

sin x + y dy dx

LM Ans. 1 OP N 24 Q LMAns. 1 OP N 2Q LMAns. 1 OP N 2Q

xy dy dx

0

y2

0

0

∞

∞

0

y

e x / y dx dy x e– x

2

/y

Ans. 2

dx dy

LMAns. a FG π − 1IJ OP H 2 KQ N 2

Ans. 26

LM Ans. 1 OP N 16 Q L 3O xyz dz dy dx M Ans. P N 8Q x + y + zg dy dx dz

a sin θ

2

a

10.

2

0

2

dr dθ

xyz dx dy dz

π

8.

3

π

0

1

6.

II. Evaluate the triple integrals: 1.

2

2

dx dy

a cos θ

2

2.

5

a2 – x2

a

7.

x+y

z z zz zz zz π

Ans. 24

ce y dy dx

0

0

g

xy x + y dy dx

1

1

3.

b

2

2

j

– r 2 /a

2.

z zzb zz z zz z 3

1

–3

0

1

4. 6.

1

2

0

y

4

2 z

0

0

2

1

g

x + y + z dx dy dz

1– x

0

x dz dx dy

4 z – x2

dy dx dz

Ans. 12

LM Ans. 4 OP N 35 Q Ans. 8π

0

Ans. 0

LMAns. 5πa OP 64 Q N LM Ans. 1 FG log 2 − 5IJ OP 8K Q N 2H LMAns. a OP N 48 Q 3

r dr dθ

dz dy dx

b1 + x + y + zg

3

6

a2 – x2 – y2

xyz dz dy dx

113

INTEGRAL CALCULUS

3.3.1 Evaluation of a Double Integral by Changing the Order of Integration In the evaluation of the double integrals sometimes we may have to change the order of integration so that evaluation is more convenient. If the limits of integration are variables then change in the order of integration changes the limits of integration. In such cases a rough idea of the region of integration is necessary.

3.3.2 Evaluation of a Double Integral by Change of Variables Sometimes the double integral can be evaluated easily by changing the variables. Suppose x and y are functions of two variables u and v. i.e.,

x = x (u, v) and y = y (u, v) and the Jacobian

J =

b g ∂ b u, v g

∂ x, y

∂x ∂u = ∂y ∂u

∂x ∂v ≠ 0 ∂y ∂v

Then the region A changes into the region R under the transformations

zz

Then

b g

zz

x = x (u, v) and y = y (u, v)

f x , y dx dy =

A

b g

f u, v J du dv

R

x = r cos θ, y = r sin θ

If

J =

zz

∴

b g

f x , y dx dy =

A

b g= ∂ b r , θg

∂ x, u

zz

b g

∂x ∂r ∂y ∂r

∂x ∂θ ∂y ∂θ

F r , θ r dr dθ.

=

cos θ – r sin θ sin θ r cos θ

= r

...(1)

R

3.3.3 Applications to Area and Volume 1.

zz zz zzz

dx dy = Area of the region R in the Cartesian form.

R

2.

r ⋅ dr dθ = Area of the region R in the polar form.

R

3.

dx dy dz = Volume of a solid.

V

4. Volume of a solid (in polars) obtained by the revolution of a curve enclosing an area A about the initial line is given by V=

zz

A

2πr 2 sin θ ⋅ dr dθ.

114


5. If z = f (x, y) be the equation of a surface S then the surface area is given by

zz

F ∂z I F ∂z I 1+ G J + G J H ∂x K H ∂y K 2

A

2

dx dy

Where A is the region representing the projection of S on the xy-plane.

WORKED OUT EXAMPLES Type 1. Evaluation over a given region 1. Evaluate

zz

xy dx dy where R is the triangular region bounded by the axes of coordinates

R

and the line

x y + = 1. a b

Solution. R is the region bounded by x = 0, y = 0 being the coordinates axes and

FG 0, b FG1 – x IJ IJ H H aKK F xI when x is held fixed and y varies from 0 to b GH 1 – JK a

x y + = 1 a b

being the straight line through (0, a) and

x y + = 1 a b y x = 1– b a

∴ ⇒ ⇒

∴

y =

zz

F xI b G1 – J H aK R| FGH IJK U| |S xy dy|V dx || || T W FG IJ L y O H K dx x⋅M P N2Q R|x ⋅ b FG1 – x IJ U| dx S| 2 H a K V| T W b F x x I x–2 + J dx G 2 H a a K 2x b Lx x O – + M P 2 N2 3a 4a Q

z z

b 1–

a

xy dx dy =

x=0

R

z z

y=0

a

=

2

a

2

2

0

z

2 a

=

x a

b 1–

0

0

=

x a

2

3

2

0

2

=

2

3

4

a

2

0

x y —+—=1 a b

O

(a, 0)

Fig. 3.2

115

INTEGRAL CALCULUS

2. Evaluate

zz

Solution

LM N

=

b2 a 2 2 2 1 2 – a + a 2 2 3 4

=

a 2b 2 24

OP Q

xy dx dy over the area in the first quadrant bounded by the circle x2 + y2 = a2.

zz

xy dx dy =

LM MM N

z z a

x =0

z z

2

z

2

2

2

2

Y

0

0

2

a2 – x2

2

a

2

2

L y O dx x⋅M P N2Q F a – x I dx xG H 2 JK 1 da x – x i dx 2 1L x x O a – P M 2N 2 4Q 1 La a O a – P = . M 2N2 4Q 8 0

=

2

y=0

a

=

R|3 x + y = a S|⇒ y = a – x |T y = a – x

OP xy dy P dx QP

a2 – x2

2

2

2

x +y =a

2

a

=

2

3

O

(a, 0)

X

0

2

=

4

a

2

Fig. 3.3

0

4

= 3. Evaluate

zz

4

4

x dx dy where R is the region bounded by

R

x2 y2 + = 1 and lying in the first a 2 b2

quadrant. Solution. From the ellipse

x2 y2 + a 2 b2

Y

= 1

2

y2 x2 1 – = b2 a2 b a2 − x2 y = a x changes from 0 to a and y changes from 0 to

zz

R

x dx dy =

R| S| |T

z z a

x=0

b a

O

b a

U| x dy V dx || W

a2 – x2

y=0

2

x y —2 + —2 = 1 a b

a2 – x2 Fig. 3.4

X (a, 0)

116


z z FGH a

=

b a 0

x y

a2 – x2

0

a

=

IJ K

b a

x

0

dx

a 2 – x 2 dx

Putting x = a sin θ, dx = a cos θ dθ

a 2 – x 2 = a cos θ ∴ θ varies from 0 to π/2

b = a

z

π2

a sin θ ⋅ a cos θ ⋅ a cos θ dθ

0

z

π2 2 = a b

sin θ cos2 θ dθ

0

4. Evaluate

zz

2 = a b×

1 a 2b = ⋅ 3 3

xy dx dy where R is the region in the first quadrant included between

R

x2 y2 x y + 2 = 1 and + = 1. 2 a b a b

Y

x y + = 1 a b

Solution ⇒

FG x IJ H aK b ba – xg a

y = b 1– =

x2 y2 + a 2 b2

zz

R

X

O (a, 0) 2

xy dx dy =

R| S| |T

U| xy dy V dx b g ||W

z z a

0

2

y x —+— =1 b a

y x —2 + —2 = 1 b a

= 1

y2 x2 1 – = b2 a2 b a2 – x2 y = a

⇒

∴

(a, b)

b a

a2 – x2

y=

b a–x a

Fig. 3.5

( y ≥ 0)

117

INTEGRAL CALCULUS

z

Ly O xM P dx N2Q b g Lb O 1 b x M da – x i – a – x g P dx b 2 a Na Q b d2ax – 2 x i dx 2a a

=

b a

2

b a–x a

0

z a

=

2

2

2

2

z a

2

2

2

3

0

LM N LM 2a N3

b2 x3 x4 2 – a = 3 2 2a 2

5. Evaluate

zz

b2 2a 2

=

2

2

2

0

=

a2 – x2

4

a4 2

–

OP Q

a

0

OP = Q

a 2 b2 ⋅ 12

xy 2 dx dy where R is the Triangular region bounded by y = 0, x = y and

R

x + y = 2. Solution. Given y = 0, x = y, x + y = 2 where

y = 0, y + y = 2

⇒

2y = 2

⇒

y = 1

Y

where x = y, x = 2 – y

(1, 1)

∴ y varies from 0 to 1 x varies from y to 2 – y

zz

xy 2 dx dy =

zz z LMN OPQ z {b g z b g

x=2–y

x=y

2– y

1

xy 2 dx dy

y=0 x=y

R

1

=

y

2

y=0

=

1 2

1 = 2

=

x2 2

2–y

O

dy

x= y

Fig. 3.6

1

y2

2– y

2

}

– y 2 dy

0

1

y 2 4 – 4 y dy

0

LM N

1 4 3 y – y4 2 3

OP Q

1

= 0

(2, 0)

1 . 6

X

118

6. Evaluate

zz


b

g

xy x + y dx dy over the region between y = x2 and y = x.

R

Solution. The bounded curves are y = x2 and y = x. The common points are given by solving the two equations.

Y

So, we have

y=x

2

2

x = x = x (x – 1) = 0 ⇒

y=x

x = 0 or 1

when x = 0, we have y = 0 and when x = 1, y = 1 (from y = x) ∴

zz

b

z z z LMN z |RS|T FGH z FGH 1

g

xy x + y dx dy =

x

x = 0 y = x2

R

1

=

0

OP Q

X O

x

dx x2

I F JK GH 5 x x I x – – dx 6 2 3 JK LM 5 x – x – x OP N 6 5 14 24 Q x x

0

1

=

g

xy 2 y 3 x + 2 3

1

=

b

xy x + y dy dx

x2 x4 x3 x6 – – + 2 2 3 3 6

Fig. 3.7

I |UV dx JK |W

7

4

0

5

=

7

8 1

0

7. Evaluate

zz

=

1 1 1 3 – – = . 6 14 24 56

xy dx dy where R is the region bounded by the x-axis, ordinate at x = 2a and

R

x2 = 4ay. Solution

Y

x = 2a and x2 = 4ay

When ∴

4a2 = 4ay

⇒

y = a

2

x = 4ay (2a, 0)

∴ The point of intersection of x = 2a and x2 = 4ay is (2a, a)

Now

zz

R

zz

2a

xy dx dy =

x2 4a

X (2a, 0)

xy dy dx

x =0 y=0

Fig. 3.8

119

INTEGRAL CALCULUS

z z

2a

=

0

2a

=

0

Ly O xM P N2Q 2

x2 4a

dx 0

x5 dx = 32a 2

LM x OP N 32a × 6 Q 6

2a

=

2

0

a4 3

Type 2. Evaluation of a double integral by changing the order of integration

zz

a 2 ax

1. Change the order of integration and hence evaluate

0

Solution

y = 2 ax

⇒

y2 = 4ax

x 2 dy dx.

0

when x = a on y2 = 4ax, y2 = 4a2 ⇒

y = ± 2a

So, on y = 2 ax , y = 2a when x = a The integral is over the shaded region. Y

Y 2

y x= — 4a

y = 2 ax (a, 2a)

X

X

O

Fig. 3.9

zz

Fig. 3.10

0

z z

2a

a 2 ax

x 2 dy dx =

a

y=0

0

z

2a

=

0

x=

2

y 4a

LM x OP N3Q 3

x 2 dx dy

a

dy y2 4a

(By changing the order)

120


z

F a – y I dy GH 3 192 a JK LM a y – y OP N 3 192a × 7 Q

2a

=

3

6

3

0

3

=

7

2a

3

4

7

0

4

2a 2 a – 3 192 × 7

=

4 = a

FG 2 – 2 IJ H 3 21K

=

4 4 a . 7


2 – x2

0

x

x x2 + y2

2 – x2

Solution

y=

⇒

y2 = 2 – x2

⇒

zz 1

dy dx .

Y (0, 2 )

x2 + y2 = 2

y=x (1, 1)

This circle and y = x meet if x2 + x2 = 2 ∴

2x2 = 2 ⇒ x = 1

2

Now

I=

zz zz 2

R where φ ( y) = S T

x x + y2 2

x

0

=

O

2 – x2

1

bg

φ y

x x + y2 2

y=0 x=0

dy dx

dx dy Fig. 3.11

for 0 ≤ y ≤ 1

y

2 – y 2 for 1 ≤ y ≤ 2

(Note that x = φ ( y) is the R.H.S. boundary of the shaded region) So, the required integral is

zz z ze y

1

I=

x +y 2

y=0 x=0 1

=

x +y 2

2

y 0

2

1

dy +

z ze 1

j

2 y – y dy +

0

dx dy + 2

0

=

z z

2 – y2

2

x

2

1

LM N

y =1

x +y 2

j

2

x=0

OP Q

2 – y dy

x x + y2 2

2 – y2

dy 0

2

x +y =2 X

So, (1, 1) is the meeting point.

dx dy

121

INTEGRAL CALCULUS

=

LMe N

=

j OPQ + LMN 2 y – y2 OPQ 2 –1 F 2 1I + 2 e 2 – 1j – G – J H 2 2K 2 y2 2 –1 2

1

2

2

0

1

1 . 2

= 1–

zz

∞ ∞


0 x

Solution. The region of integration is the portion of the first quadrant between y = x and the y-axis. So, by changing the order of integration.

zz ∞

∞

x =0 y =

e– y dy dx = y x

zz z z ∞

=

0

Y

y=x

y

y =0 x =0

∞

e–y dy dx. y

e– y dx dy y

e– y x y

y 0

dy O X

∞

=

e – y dy

0

∞

y = – e–

0

= 1.

Fig. 3.12

z zb 3


4– y

y = 0 x =1

Solution

⇒

x= 4 – y

x+ y= 4

g

x + y dx dy.

Y

Limits for x are from 1 to 4 – y when

x = 1 on x + y = 4

we have

z zb 3 4– y

So,

0

(1, 3)

1+y= 4 ⇒ y = 3

g

z zb 4

x + y dx dy =

4–x

x =1

1

g

x + y dy dx

x+y=4

0

by changing the order of integration.

z z 4

=

1

4

=

1

LMxy + y OP dx N 2Q LM b4 – x g x b4 – xg + 2 MN 2

4– x

(4, 0)

O

0

2

OP PQ dx

Fig. 3.13

X

122


z FGH 4

=

8–

1

IJ K

1 2 x dx 2

LM8x – x OP N 6Q

3 4

=

=

1

27 · 2

z zb 3


4– y

0

Solution

x =

⇒

g

x + y dx dy.

1

4– y

(0,4)

(1,3)

2

x = 4–y y = 4 – x2, a parabola.

Here, the limits 1 and

y=4–x

4 – y are for x, 0 and

3 are for y. O (1,0)

When x = 1, on y = 4 – x2, y = 3

z zb 3

Now,

y=0

4– y

g

x + y dx dy =

x =1

z zb

4 – x2

2

g

x + y dy dx

x =1 y = 0

Fig. 3.14

(By changing the order of integration)

z z FGH 2

=

xy + y 2 2

4 – x2 0

dx

1

2

=

4x – x3 + 8 – 4x2 +

1

=

LM2 x N

= 6–

2

–

I JK

x4 dx 2

x4 4 x5 + 8x – x 3 + 4 3 10

OP Q

2

1

15 28 31 241 +8– + = · 4 3 10 60

Type 3. Evaluation by changing into polars

zz

∞∞

1. Evaluate

e−

cx

2

+ y2

h dx dy

by changing to polar coordinates.

0 0

Solution. In polars we have x = r cos θ, y = r sin θ ∴

x2 + y2 = r2 and dx dy = r dr dθ

(2,0)

2

123

INTEGRAL CALCULUS

Since x, y varies from 0 to ∞

Y

r also varies from 0 to ∞ In the first quadrant ‘θ’

P (x, y)

varies from 0 to π/2

zz

π2

Thus

I =

r

∞ 2

e – r r dr dθ

G

θ=0 r=0

dt ∴ r dr = 2

2

Put

O

r = t

t also varies from 0 to ∞

zz z zb z

π2

I =

∞

e –t

θ=0 t =0

1 = 2

= +

=

z z a

2. Evaluate

0

1 2

dθ

g

0 – 1 dθ

0

π2

1 . dθ

0

+1 θ 2

π2 0

=

+1 π π ⋅ = ⋅ 2 2 4

a2 – y2

y x 2 + y 2 dx dy by changing into polars.

0

I =

z z

y =0

x=

0

θ=0

a

Solution

∞

– e –t

π2

Fig. 3.15

dt dθ 2

π2

–1 = 2

X

a2 – y2

y x 2 + y 2 dx dy

x=0

a 2 − y 2 or x2 + y2 = a2 is a circle with centre origin and radius a. Since, y varies from 0 to a

the region of integration is the first quadrant of the circle. In polars, we have x = r cos θ, y = r sinθ ∴

x2 + y2 = r2

i.e.,

r 2 = a2

⇒

r = a

Also x = 0, y = 0 will give r = 0 and hence we can say that r varies from 0 to a. In the first quadrant θ varies from 0 to π/2, we know that dx dy = r dr dθ

124


zz zz zb z b π2

a

∴

I =

r sin θ r r dr dθ

r =0 θ=0 π2

a

=

r 3 sin θ dr dθ

r =0 θ=0 a

=

r 3 – cos θ

r =0 a

=

–r

3

0

g

π2 0

dr

Lr O 0 – 1g dr = M P N4Q 4

a

= 0

a4 4

4

a . 4

I =

Type 4. Applications of double and triple integrals 1. Find the area of the circle x2 + y2 = a2 by using double integral. Solution Y Since, the circle is symmetric about the coordinates axes, area of the circle is 4 times the area OAB as shown in Figure. B For the region OAB, y varies from 0 to 2 a – x 2 and x varies from 0 to a.

z z

a2 – x2

a

∴ Area of the circle = 4

dy dx

y =0

0

O

z z a

= 4

y

a2 – x2 y=0

2

y= a –x

2

Q

p

y=0

A

X

dx

0

a

= 4

a 2 – x 2 dx

0

=

Lx 4M N2

a2 x a –x + sin –1 a 2 2

2

Fig. 3.16

OP Q

a

= πa2 sq. units 0

2. Find by double integration the area enclosed by the curve r = a (1 + cos θ) between θ = 0 and θ = π. Solution

Area =

zz

r dr dθ

where r varies from 0 to a (1 + cos θ) and θ varies from 0 to π

z

a2 − x2 =

x 2 a2 x a − x2 + sin −1 a 2 2

125

INTEGRAL CALCULUS

i.e.,

z z z LMN OPQ z b g z RST FGH IJK UVW z FGH IJK b

π

a 1 + cos θ

θ=0

r =0

A =

π

=

r dr dθ

r2 2

θ=0

g

b

a 1 + cos θ

g

dθ

r=0

π

1 a 2 1 + cos θ = 2

2

dθ

0

=

π

a2 2

2 cos2

0

π

2 = 2a

θ 2

2

dθ

θ dθ 2

cos4

0

θ/2 = φ, dθ = 2dφ

Put and φ varies from 0 to π/2

z z

π2

∴

A = 2a2

cos4 φ ⋅ 2dφ

0

π2

= 4a 2

cos4 φ ⋅ dφ

0

3 1 π ⋅ ⋅ (by the reduction formula) 4 2 2 A = 3πa2/4 sq. units. = 4a 2 ⋅

Area, 3. Find the value of Solution Let

zzz

z dx dy dz where V is the hemisphere x2 + y2 + z2 = a2, z ≥ 0.

I =

zzz z z z z

z dx dy dz

V

a

=

a2 – x2

z

a2 – x2 – y2

x = –a y = – a2 – x2 a

=

1 + cos θ = 2 cos2

θ 2

a2 – x2

x = –a y = – a – x 2

z=0

LM z OP N2Q

a2 – x2 – y 2

2

2

z dz dy dx

dy dx 0

126


1 = 2

1 = 2

z zd

a2 – x2

a

i

a 2 – x 2 – y 2 dy dx

x = –a y = − a2 – x2

z a

x = –a

1 4 . = 2 3 2 .2 = 3

LMda N

zd zd

2

–x

a

a2 – x2

–a

a

a2 – x2

0

i

y3 y– 3

i

dx

2

i

32

32

OP Q

a2 – x2

dx y=– a –x 2

2

dx

x = a sin θ

Put

dx = a cos θ dθ θ varies from 0 to π/2 4 = 3

zd z

π2

a 2 cos2 θ

θ=0

4a 4 = 3

Thus,

32

a cos θ dθ

π2

cos4 θ dθ

0

=

4a 4 3 1 π . . . 3 4 2 2

=

πa 4 4

I =

i

πa 4 · 4

4. Using multiple integrals find the volume of the ellipsoid Solution

zzz

The volume (V ) is 8 times in the first octant (V1) i.e., z varies from 0 to c

V = 8V1 = 8 1 –

(By applying reduction formula)

x2 y2 – a2 b2

dz dy dx

x2 y2 z2 + + = 1. a 2 b2 c2

127

INTEGRAL CALCULUS

y varies from 0 to (b/a)

a2 – x2

x varies from 0 to a x2 y2 – a 2 b2

z z z z z z z |RS|T a

a2 – x2

x=0

y =0

c 1–

V = 8V1 = 8

bb/ a g

a

= 8

a2 – x2

c 1–

x=0

y =0

bb / a g

a

= 8c

x=0

We shall use

z

a2 – x2

y=0

y α2 – y2

α 2 – y 2 dy =

2

+

dz dy dx

z=0

x2 y2 – dy dx a 2 b2

F GH

1 2 x2 b 1– 2 b a

I |UV – y JK |W

2

FG IJ H K

y α2 sin –1 2 α

where α2 = b2 {1 – x2/a2} = b2 (a2 – x2)/a2 ∴

8c V = b

8c = b 8c = b

=

8c b

zz LM z MN z z α

a

α 2 – y 2 dy dx

x =0 y=0 a

x=0 a

0+

x =0

FG H

y α 2 − y2 α 2 y + sin –1 α 2 2

bg

α2 sin –1 1 – sin –1 0 dx 2

a

d

i

π 1 b2 2 a – x 2 dx . 2 2 2 a x =0

2bcπ = a2

LMa x – x OP N 3Q 3

a

2

0

2bcπ 2a 3 4π abc . = = 3 3 a2 Thus the required volume (V ) =

bg

IJ OP K PQ

4 πabc cubic units. 3

α

dx 0

dy dx

128


1. Evaluate

zz zz

EXERCISE 3.2 xy 2 dx dy over the region bounded by y = x2, y = 0 and x = 1.

R

2. Evaluate

b

g

xy x + y dx dy taken over the region bounded by the parabolas y2 = x and

R

y = x2. 3. Evaluate

zz zz

x 2 y dx dy over the region bounded by the curves y = x2

R

4. Evaluate

LM Ans. 3 OP N 28 Q and y = x. L Ans. 1 O MN 35 PQ

xy dx dy where R is the region in the first quadrant bounded by the line

R

x + y = 1. Evaluate the following by changing the order of integration (5 to 9)

z zd zz z zb zz zz a

5.

x a

i

x a

7.

8.

2

x dx dy.

0

0

a

a2 − x2

–a

0

a

a

0

ax

3

g

a − x dy dx.

2

y 2 dy dx y4 – a2 x2

.

LMAns. 3a OP N 8Q

a 2a − x

9.

4

xy dy dx.

0

LMAns. a + a OP N 28 20 Q LMAns. 4a OP N 7Q LMAns. πa OP N 2Q LMAns. πa OP N 6Q 4

a 2 ax

6.

LM Ans. 1 OP N 6Q 3

x 2 + y 2 dy dx.

0

LM Ans. 1 OP N 24 Q

x2 a

z z

2a

10. Evaluate

0

2 ax − x 2 2

x dy dx by transforming into polar coordinates.

0

11. Find the area of the cardioid r = a (1 + cos θ) by double integration.

LMAns. 5πa N 8 LMAns. 3π a 2 N

4

2

OP Q OP Q

12. Find the volume of the region bounded by the cylinder x2 + y2 = 16 and the planes z = 0 Ans. 48π and z = 3.

129

INTEGRAL CALCULUS

3.4 BETA AND GAMMA FUNCTIONS In this topic we define two special functions of improper integrals known as Beta function and Gamma function. These functions play important role in applied mathematics.

3.4.1 Definitions 1. The Beta function denoted by B (m, n) or β (m, n) is defined by

z

b g

1

β (m, n)

=

x m −1 1 − x

n −1

b

g

dx , m, n > 0

...(1)

0

2. The Gamma function denoted by Γ (n) is defined by

z

∞

Γ (n)

=

x n −1 ⋅ e – x dx

...(2)

0

3.4.2 Properties of Beta and Gamma Functions 1.

β (m, n)

= β (n, m)

zb z z z z ∞

2.

β (m, n)

=

x m −1

1+ x

0

g

m+ n

zb

∞

dx =

0

x n −1

1+ x

g

m+ n

dx

...(3)

π2

3.

β (m, n)

=

2

sin 2 m−1 θ cos 2 n −1 θ dθ

...(4)

0

π2

=

2

sin 2 n −1 θ cos 2 m−1 θ dθ

0

4.

L p + 1 , q + 1OP βM N 2 2 Q

π2

=

2

sin p θ cosq θ dθ

0

π2

=

2

sin q θ cos p θ dθ

...(5)

0

5.

Γ (n + 1)

6. Γ (n + 1) Proof 1. We have

= n Γ (n) = n!, if n is a + ve real number.

z zb 1

β (m, n)

...(6)

=

b g

x m −1 1 − x

n −1

dx

0

1

=

1− x

0

g

m −1

b g

1− 1− x

n −1

dx

130


z a

Since

bg

f x dx

z b g zb g b z b g a

=

f a − x dx

0

0

1

=

m −1

1− x

1−1+ x

g

n −1

dx

0

1

=

x n −1 1 − x

m −1

dx

0

= β (n, m) Thus,

β (m, n)

= β (n, m)

Hence (1) is proved. (2) By definition of Beta function,

z

b g

1

β (m, n)

=

x m −1 1 − x

n −1

dx

0

Substituting x =

–1 1 then dx = 1+ t 1+ t

b g

Therefore,

z z zb zb zb 0

β (m, n)

=

∞ 0

=

∞

LM 1 OP N1 + t Q FG 1 IJ H1+ tK

∞

=

0

=

0

1+ t

=

0

Since,

LM1 – 1 OP R|S –1 dt U|V N 1 + t Q |T b1 + t g |W FG t IJ R|S – 1 dt U|V H 1 + t K |T b1 + t g |W n −1

2

m −1

n −1

2

g

m −1+ n −1+ 2

1+ t

g

m+n

g

dt =

m+n

= β (n, m), we get

β (m, n)

=

zb 0

g

1+ x

zb ∞

x n −1

1+ x

x n −1

m +1

dx =

β (m, n) =

z 0

m+ n

0

b g

x m −1 1 − x

n −1

x m−1

1+ x

(3) By definition of Beta functions 1

g

dx

dx

β (m, n)

∞

zb 0

x m−1

1+ x

dt ∞

t n −1

∞

Similarly, β (n, m)

dt when x = 0, t = ∞ and when x = 1, t = 0.

t n −1

∞

β (m, n)

m −1

2

dx

g

m+ n

dx

131

INTEGRAL CALCULUS

Substitute

x = sin2 θ then dx = 2 sin θ cos θ dθ

Also when

x = 0, θ = 0 x = 1, θ =

when

zd z z z

π 2

π2

∴

β (m, n) =

sin 2 θ

0

π2

= 2

sin

i ⋅ d1 − sin θi m −1

2 m− 2

0

2

d

θ cos2 θ

i

n −1

n −1

⋅ 2 sin θ cos θ dθ

⋅ sin θ cos θ dθ

π2

= 2

sin 2 m− 2 θ cos2 n − 2 θ sin θ cos θ dθ

0

π2

= 2

sin 2 m−1 θ cos2 n −1 θ dθ

0

β (m, n) = β (n, m), we have

Since,

z z

π2

β (m, n) = 2


0

π2

= 2

sin 2 n −1 θ cos2 m−1 θ dθ

0

(4) Substituting 2m –1 = p and 2n – 1 = q So that

m =

F p + 1 , q + 1IJ βG H 2 2K

p+1 q +1 , n= in the above result, we have 2 2

z z

π2

= 2


0

π2

= 2

sin q θ cos p θ dθ

0

(1) Substituting q = 0 in the above result, we get β

LM p + 1 , 1 OP N 2 2Q

z

π2

= 2

z

π2

sin θ dθ = 2 p

0

cos p θ dθ ⋅

0

(2) Substituting p = 0 and q = 0 in the above result

F 1 1I βG , J H 2 2K

z

π2

= 2

0

dθ = π

132


(5) Replacing n by (n + 1) in the definition of gamma function.

z

x n −1 ⋅ e − x dx

z

x n ⋅ e − x dx

∞

Γ (n) =

0

where n = (n + 1)

∞

Γ (n + 1) =

0

On integrating by parts, we get

Γ (n + 1) =

d

x n ⋅ − e− x

z

∞

= 0+n

i

∞

zd ∞

–

0

i

− e − x ⋅ n x n −1 dx

0

e − x x n −1 dx = n Γ (n).

0

LMsince lim x e N x →∞

Thus,

/ (n + 1) = n / (n),

for n > 0

n x

= 0, if n > 0

OP Q

This is called the recurrence formula, for the gamma function. (6) If n is a positive integer then by repeated application of the above formula, we get

Γ (n + 1) = n /(n) = n /(n – 1 + 1) = n (n – 1) / (n – 1) (using above result) = n (n – 1) (n – 2) / (n – 2) ..................................... ..................................... = n (n –1) (n – 2)........1/ (1) = n! / (1)

z

∞

But

/ (1) =

x 0 e − x dx

0

−x = −e

Hence

∞ 0

= – (0 – 1) = 1

/ (n + 1) = n!, if n is a positive integer.

For example / (2) = 1! = 1, / (3) = 2! = 2, / (4) = 3! = 6 If n is a positive fraction then using the recurrence formula Γ (n + 1) = n Γ (n) can be evaluated as follows.

133

INTEGRAL CALCULUS

FG 3IJ H 2K F 5I (2) / G J H 2K F 7I (3) / G J H 2K (1) /

FG 1 IJ H 2K F3 I / G + 1J H2 K F5 I / G + 1J H2 K

= / 1+

=

=

=

=

FG IJ HK 3 F 3I Γ G J H 2K 2 5 F 5I ΓG J 2 H 2K 1 1 / 2 2

=

FG IJ HK

=

5 3 1 1 ⋅ ⋅ ⋅Γ 2 2 2 2

=

15 1 Γ ⋅ 8 2

FG IJ HK

3.4.3 Relationship between Beta and Gamma functions The Beta and Gamma functions are related by β (m, n) =

b g bg b g

Γm Γn Γ m+n

z zd i z z ∞

Γ (n) =

Proof. We have

...(7)

x n −1 ⋅ e − x dx

0

2

Substituting x = t ,

dx = 2t dt, we get ∞

/ (n) =

t2

n −1

e − t ⋅ 2 t dt 2

0

∞

=

2

t 2 n −1 ⋅ e − t dt 2

0

∞

/ (n) =

x 2 n −1 e − x dx

2

2

...(i)

0

Replacing n by m, and ‘x’ by ‘y’, we have

z

∞

/ (m) =

2

y 2 m−1 e – y dy 2

...(ii)

0

Hence / (m) · / (n) =

R|2 S| T

z zz ∞

x 2 n −1 e – x dx

0

∞ ∞

=

4

2

0 0

U| R|2 V| S| WT

z

∞

y 2 m−1 e – y dy 2

0

x 2 n −1e − x y 2 m−1 ⋅ e − y dx dy 2

2

U| V| W

134


zz

∞ ∞

=

4

e

e

− x 2 + y2

j x 2n−1 y 2m−1 dx dy

...(iii)

0 0

We shall transform the double integral into polar coordinates. Substitute x = r cos θ, y = r sin θ then we have dx dy = r drdθ As x and y varies from 0 to ∞, the region of integration entire first quadrant. Hence, θ varies from 0 to

π and r varies from 0 to ∞ and also x2 + y2 = r2 2

Hence (iii) becomes,

b g bg

Γm Γn

zz z z ∞

=

4

π2

e−r

2

r = 0 θ =0 ∞

=

4

r 2b

br cos θg br sin θg

g

m + n −1

2 n −1

e − r dr ×

r =0

∞

r

b

g

2 m + n −1

e

−r 2

and from (iv),

z

sin 2 n −1 θ cos2 n −1 θ dθ

...(iv)

0

dr

∞

=

1 2

=

1 Γ m+n 2

r =0

π 2

z

⋅ rdθ dr

r2 = t, in the first integral. We get,

Substituting

z

2

π 2

2 m −1

t m+ n −1 e − t dt

0

b

sin 2 m−1 θ cos2 n −1 θ dθ =

0

g

b g

1 β m, n 2

b g bg b g b g Γ b mg Γ b ng Thus, β (m, n) = · Hence proved. Γ bm + ng F 1I Corollary. To show that Γ G J = π H 2K Therefore (iv) reduces to Γ m Γ n = Γ m + n β m, n

Putting m = n =

1 in this result, we get 2

L1 1O βM , P N2 2Q

=

LM 1 OP ⋅ Γ LM 1 OP N2Q N2Q Γ 1

/ (1) = 1

But ∴

Γ

β

LM 1 , 1 OP N2 2Q

=

RSΓ FG 1 IJ UV T H 2K W

2

...(8)

135

INTEGRAL CALCULUS

z π 2

Now consider β (m, n) = 2


0

Now we have from (8), L.H.S.

L1 1O βM , P N2 2Q

z π 2

=

2

sin 0 θ cos0 θ dθ = 2 θ

0

π =

R1U ΓS V T2W

2

∴ Γ

LM 1 OP N2 Q

=

π 2 0

=π

π.

WORKED OUT EXAMPLES 1. Evaluate the following: (i)

bg Γ b 5g Γ 7

(ii)

Solution

bg Γ b5g F 5I ΓG J H 2K F 3I ΓG J H 2K F 8I ΓG J H 3K F 2I ΓG J H 3K Γ 7

(i)

(ii)

(iii)

=

=

=

=

(i)

z 0

x e

−x

FG 3 + 1IJ 3 Γ FG 3 IJ H 2 K = 2 H 2K = 3 F 3I Γ FG 3 IJ 2 ΓG J H 2K H 2K F 5 I 5 Γ FG 5IJ Γ G + 1J H 3 K = 3 H 3K F 2 I Γ FG 2 IJ ΓG J H 3K H 3K 5 F 2 I 5 2 F 2I Γ G + 1J × ΓG J 3 H 3 K 3 3 H 3K = 2I F F 2I ΓG J ΓG J H 3K H 3K 10 . 9

z

(ii)

0

z

∞

∞

dx

b g Γ b2 3g

Γ 83

6! = 30 4!

2. Evaluate: 4

(iii)

Γ

=

∞

b g Γ b 3 2g Γ 52

6

x e

−3 x

dx

(iii)

0

x 2 e −2 x dx . 2

136


Solution

z z

∞

(i)

x 4 e − x dx =

0

∞

(ii)

z

∞

bg

x 5–1 e − x dx = Γ 5 = 4!

0

= 24 x 6 e −3 x dx

0

t dt then dx = 3 3

3x = t ⇒ x =

Substituting

z

z

∞

∞ 6

x e

−3 x

dx =

0

0

FG t IJ H 3K

z

6

⋅ e −t ⋅

∞

1 = 7 3

6

t e

dt 3

1 dt = 7 3

−t

0

bg

z

∞

t 7 −1 e − t dt

0

1 Γ 7 37 1 80 = 7 6! = 243 3

=

2x2 = t ⇒ x2 =

(iii) Substitute Then

2x dx = dx =

z

∞

x 2 e −2 x dx = 2

0

=

=

=

=

t 2

dt 2 2 dt dt = 4x 4 t

z

∞

0

=

t ⇒x= 2

2 t −t dt ⋅e ⋅ 2 4 t

2 8

2 8

z z ∞

1

t 2 e − t dt

0

∞

3

t2

−1

LM3 Γ FG 1 IJ N H 2K

e − t dt

0

FG IJ HK 2 F1 I Γ G + 1J H2 K 8 2 3 Γ 8 2

2 1 ⋅ 8 2

π=

= 2π 16

FG IJ HK

2 1 1 ⋅ Γ 8 2 2

=

π

OP Q

137

INTEGRAL CALCULUS

3. Evaluate the following:

z z

∞

(i)

z z

∞

x e − x dx 3

(ii)

0

∞

(iv)

2

(iii)

0

–3 x 2

0

0

∞

d1 − e i dx −x

(v)

3 −4x dx . 2

0

Solution (i)

x3 = t so that 3x2 dx = dt

Substitute

1 t3

dt

where

x =

when

3t 3 x = 0, t = 0 and when x = ∞, t = ∞

z

Hence,

z

∞

∞

xe

− x3

dx =

, dx =

1

1 = 3

=

=

1 3

z z

2

dt

t 6 e −t

2

0

0

3t 3

∞

t

−1 2

e − t dt

0

∞

0

1

t2

−1

e − t dt

FG IJ HK

1 1 1 Γ = π. 3 2 3 1

2

(ii) Substitute,

⇒

x = t2 =

2x dx = dt ⇒ dx =

So that

z

∞

Hence,

x = t

x 4 e − x dx = 2

0

z

∞

0

1 = 2 =

1 2

z z

3 2 t

t

dt 2 t

dt

t 2 e −t ∞

2 t

e − t dt

0

∞

0

5

t2

z

∞

x 4 e − x dx

−1

FG IJ HK

e −t dt

=

1 5 Γ 2 2

=

1 3 1 1 3 ⋅ ⋅ Γ = π· 2 2 2 2 8

FG IJ HK

1

x 4 e−

x

dx

138


x = t ⇒ x = t2, dx = 2t dt

(iii) Substitute,

z

∞

Hence,

z

∞

1

x 4 e−

x

dx =

0

0

t 1 2 e − t 2t dt

z z

∞

= 2

3

t 2 e − t dt

0

∞

= 2

0

= 2Γ

5

t2

–1

e − t dt

FG 5 IJ H 2K

FG IJ H K

3 1 1 3 ⋅ Γ = π· 2 2 2 2

= 2⋅

(iv) On integrating by parts, we get

z

∞

0

−3 x2

d1 − e i dx −x

=

LM 1 − e F – 2 x MNd i GH −x

z

∞

= 0+2

= 2

z

I OP JK P Q

∞

− 0

z

∞

0

e − x dx

0

∞

1 −1 x2

0

FG 1 IJ H 2K

= 2Γ

–1 2

x

–1 2

e − x dx

= 2 π⋅ a = elog a, a > 0, we have

(v) Since 3−4 x

z

2

=

e log 3

z

∞

∞

e

−4 x 2

dx =

e

b

−4 x 2

g

=e

− 4 log 3 x 2

b

g

– 4 log 3 x 2

dx

0

0

(4 log 3) = x2 = t we get,

Setting

x2 =

t t ⇒ x= 4 log 3 2 log 3

(4 log 3) 2x dx = dt (4 log 3) 2 ⋅

t dx = dt 2 log 3

F – 2x I e GH JK –1 2

−x

dx

139

INTEGRAL CALCULUS

z

∞

3

−4 x 2

=

=

=

z

1

e −t ⋅

t ⋅ 4 log 3

0

0

∞

z

∞

dx =

3−4 x dx = 2

0

z z ∞

1 4 log 3

t

4 log 3

zb z 1

0

1

(iii)

log x

0

g

4

0

1

t2

−1

e – t dt

FG IJ HK

π 4 log 3

(ii)

zb

x log x

0

g

dx

log x = – t so that x = e+ t

1 dx = – dt or dx = – x dt = – e–t dt x t = – log 0 = ∞ and

Also when x = 0,

zb 1

0

t = – log 1 = 0 (note that log 0 = – ∞)

log x

g

4

dx

zb g z z z z 0

=

–t

∞

4

⋅ – e – t dt

∞

=

t 4 e – t dt

0

∞

=

t 5−1 e – t dt

0

zb 1

(ii)

3

.

Solution Substitute

(i) Hence

e – t dt

1

F 1I log G J H xK

when x = 1,

dt

1 1 Γ 2 4 log 3

dx

dx

t ⋅ 4 log 3

0

∞

1

−1 2

4. Evaluate: (i)

1

⇒ dx =

t ⋅ 4 log 3 dx = dt

0

x log x

g

= /(5) = 4 ! = 24. 3

dx =

0

e −1 ( − t )

∞

= −

∞

0

3

d– e dt i

t 3 e – 4 t dt

–t

dt

140


Put

4t = u

z 1

∴

b x log xg

3

z

dx = –

0

(iii)

z

1

1

0

log

FG 1 IJ H xK

Solution Now,

0

2

z

∞

0

u 3 ⋅ e – u du

–1 256

=

–1 –3 3! . = Γ 4 =– 256 256 128

dx =

a –bx dx =

1 du 4

⋅ e –u ⋅

=

z z z

4

0

∞

u 4 −1 e – u du

0

bg

0

– e – t dt

∞

∞

t –1 2

t

e – t dt

0 ∞

1 −1 2

t

e – t dt

0

= Γ

z

∞

3

b4 g

=

∞

z z

–1

4dt = du 1 dt = du 4

=

=

5. Prove that

FG u IJ H 4K

⇒

FG 1IJ = H 2K

π.

π where a and b are positive constants. 2 b log a

2

a – bx dx = =

zo z ∞

e log a

0

∞

e –b

t

– bx 2

g

b log a x 2

since a = elog a

dx

0

Substitute

(b log a) x2 = t, dx =

So that,

dx =

z

∞

0

2

e – bx dx =

b

g

t b log a

x =

∴

dt b log a ⋅ 2 x

dt 2 t b log a

z

∞

0

e –t ⋅

dt 2 t b log a

141

INTEGRAL CALCULUS

=

=

=

=

z

6. Prove that

∞

0

z z

1 2 b log a

2 b log a

π

Γ

n

na

e – t dt

1

t2

–1

e – t dt

FG 1 IJ H 2K

⋅

2 b log a 1

∞

Γ

2 b log a

bm+1g

–1 2

0

1

n

t

0

1

x m e –ax dx =

∞

FG m + 1IJ , where m and n are positive constants. H n K

Solution Substitute

ax

Then

n

FtI = t so that x = G J H aK

dx =

1 na n

Therefore,

z

∞

m

x e

– ax

n

0

dx =

= =

7. Prove that

z

1

0

b

x m log x

Solution Substitute

g

1

1

z

∞

0

⋅t

n

–1

LMF t I MMGH a JK N

1 n

1

m + 1 /n na b g

1 n

dt

OP PP Q

z

m

1

e ⋅ –t

∞

0

tn

–1

1 na n

dt

b m + 1g

t

n−1

e – t dt

L m + 1OP. n Q

ΓM m+ n N nab 1g / 1

b – 1g ⋅ n! , where n is a positive integer and m > –1. bm + 1g n

n

dx =

n+1

log x = – t or x = e– t dx = – e–t dt

Then

when x = 0, t = ∞ and when x = 1, t = 0. Therefore,

z

1

0

b g

x m log x

n

dx =

zd 0

∞

e –t

i b – t g ⋅ d– e i dt m

n

–t

142

since setting

z g z RST

=

b– 1g

n

=

b– 1

n

(m + 1) t = u

∞

∞

0

g

u m+1

UV e W

– m +1 t

z z

dt

n

–u

⋅

du m+1

b–1g u e du bm + 1g b–1g ub ge du bm + 1g b –1g Γbn + 1g bm + 1g b–1g n! where Γ (n + 1) = n!. bm + 1g n+1

∞

n

–u

0

∞

n

=

b

tn e

0

n

=


n +1

n + 1−1

−u

0

n

=

=

n +1

n +1

8. Prove that (i)

(ii)

z z

∞

0

∞

0

x n −1 e –ax cos bx dx =

x n −1 e –ax sin bx dx =

Solution Consider

I = =

Substitute

da

2

+

bg

i

Γn

da

z z

2

∞

+

RS T

b a

RS T

b . a

cos n tan –1

n b2 2

sin n tan –1

i

n b2 2

I =

=

UV W

0 ∞

x n −1 e – b a −ib g x dx

0

dt a − ib

t a − ib

z

∞

0

=

UV W

x n −1 e – ax e ibx dx

(a – ib) x = t, so that dx = x =

Hence,

bg

Γn

b

RS t UV T a − ib W 1

a − ib

g z n

∞

n–1

e– t ⋅

t n −1 e – t dt

0

Γ n ba − ibg b g 1

n

dt a − ib

143

INTEGRAL CALCULUS

ba + ibg ⋅ Γbng da + b i n

I =

2

...(1)

n

2

a + ib = r (cos θ + i sin θ)

Since where

a 2 + b 2 and θ = tan–1

r =

Hence (1) reduces to, I = Apply De Moivre’s theorem =

bg b da

Γ n r cos θ + i sin θ 2

+ b2

=

n

n

bg b g da + b i Γ bng . d a + b i bcos n θ + i sin n θg da + b i Γ b ng (cos n θ + i sin n θ) da + b i Γ n r n cos n θ + i sin n θ 2

2

=

i

g

FG b IJ H aK

2

2

n

n2

2

2

2

n

n2

2

On equating the real and imaginary parts, we get

z z

bg

∞

(i)

x n – 1 e – ax cos bx dx =

0

da

Γn 2

bg

∞

(ii)

x n – 1 e – ax sin bx dx =

0

where θ = tan–1

d

+ b2 Γn

a + b2 2

i i

n2

cos n θ

n2

sin n θ

b · a

9. Evaluate (i) β (3, 5)

(ii) β (3/2, 2)

Solution Using the relation β (m, n) =

(i)

β (3, 5) =

b g bg b g Γ b3g Γ b5g = Γ b3 + 5g

(iii) β (1/3, 2/3).

Γm Γn Γ m+n

2! 4! 1 = 7! 105

144


β

(ii)

LM 3 , 2OP N2 Q

=

=

β

(iii)

FG 1 , 2 IJ H 3 3K

FG 3 IJ Γb2g ΓFG 3IJ Γb2g H 2 K = H 2K F3 I F 7I Γ G + 2J ΓG J H2 K H 2K 1 F 1I Γ G J . 1! 4 2 H 2K = 5 3 1 F 1I 15 . . ΓG J 2 2 2 H 2K F 1I F 2 I Γ G J . ΓG J H 3K H 3 K F 1 2I ΓG + J H 3 3K F 1 I F 1I ΓG J ΓG1 – J H 3K H 3K Γ

=

=

π sin n π

where Γ(n) Γ(1 – n) =

π

=

sin

π 3

2π . 3 10. Evaluate each of the following integrals =

z z zd 1

(i)

x

4

0

d1 – x i dx

(ii)

0

1

a

(iii)

y

4

2

2

a – y dy

(iv)

0

0

2

(v)

4 – x2

0

Solution

z

i

32

x4 1 – x

3

b g

dx = β 5, 4 =

0

(ii) Substitute Then ∴

bg bg Γ b5 + 4g

Γ5 Γ4

x = 2t

z 2

0

x2 2–x

dx

1– x dx x

dx

b g

1

(i)

z z 2

3

dx = 2dt x2 2–x

z 1

dx =

0

4t 2 2 – 2t

2 dt

=

4 ! 3! 1 = 8! 280

145

INTEGRAL CALCULUS

z

b g

1

= 4 2

t2 1 – t

–1 2

dt

0

LM 1 OP N 2Q F 1I Γb3g Γ G J H 2K = 4 2 L 1O Γ M3 + P N 2Q F 1I Γb3g Γ G J H 2K 64 2 = = 4 2 F 7I 15 ΓG J H 2K = at = 4 2 β 3,

y2

(iii) Substitute or

2

y = a t dy =

a

Given integral becomes,

ze 1

0

a t

j

4

a dt a –a t 2 t 2

2

dt

2 t

a6 = 2

z

b g

1

12

t3 2 1– t

dt

0

LM OP N Q F 5 I F 3I ΓG J . ΓG J H 2K H 2K F 5 3I ΓG + J H 2 2K

5 3 a6 β , = 2 2 2

=

z 1

(iv)

0

1– x dx = x

a6 2

z

or

x–1 2 1 – x

12

FG 1 , 3IJ H 2 2K

Γ

0

= β

(v) Substitute

b g

1

x2 = 4t x = 2 t

=

=

π a6 · 32

dx

FG 1 IJ ⋅ ΓFG 3 IJ H 2K H 2 K F 1 3I ΓG + J H 2 2K

=

π . 2

146


then

dx =

dt t

Given integral reduces to,

z 1

b4 – 4t g

32

z

–1 2 1– t = 8 t

t

0

b g dt L1 5O 8β M , P N2 2 Q F 1I F 5I ΓG J Γ G J H 2K H 2K 8 Γ b3g 3 1 F 1I 8 π . . ΓG J 2 2 H 2K 1

dt

32

0

=

=

=

2

= 3 π. 11. Evaluate each of the following integrals:

z z z

π2

(i)

(ii)

π2

sin θ cos θ dθ 4

5

(iv)

0

0

π2

(v)

tan θ dθ .

0

Solution From the relation

z

F p + 1 , q + 1IJ βG H 2 2K

= 2

0

LM N

OP Q

LM N 1 L7 1O β , 2 MN 2 2 PQ

OP Q

1 p +1 q +1 β , 2 2 2

p = 6

z

q = 0

π2

we get


0

sin q θ cosq θ dθ =

(i) Taking

z

π2

π2

∴

cos4 θ dθ

0

0

π2

(iii)

z z π

sin 6 θ dθ

sin 6 θ dθ =

0

=

1 6+1 0+1 β , 2 2 2

sin1 2 θ cos 3 2 θ dθ

147

INTEGRAL CALCULUS

=

z

=

cos4 θ dθ = 2 .

0

=

5π . 32

LM N F 1I F 5I ΓG J ΓG J L 1 5 O H 2K H 2 K = βM , P = Γ b3g N2 2Q

cos θ dθ = 2 4

0

(iii) Here

z

FG 7 IJ . ΓFG 1 IJ H 2K H 2K Γ b 4g

π2

π

(ii)

1 2

Γ

0+1 4+1 1 β , 2 2 2

OP Q

3π . 8

p = 4

z

q = 3 from the above relation

π2

LM OP N Q 1 L5 O β ,3 2 MN 2 PQ F 5I Γ G J Γb3g 8 1 H 2K = 2 F 11I 315 ΓG J H 2K

sin 4 θ cos5 θ dθ = 1 β 4 + 1 , 5 + 1 2 2 2 0 =

=

(iv) Here

z

p =

LM MM N 1 L 3 5O β , 2 MN 4 4 PQ F 3I F 1 I ΓG J ΓG J 1 H 4K H 4K

1 3 +1 +1 1 β 2 , 2 sin1 2 θ cos3 2 θ dθ = 2 2 2

π2

0

1 3 ,q= 2 2

=

=

2

FG H

1

=

1 1 Γ 1– 8 4

=

1 π 8 sin π 4

=

2π . 8

IJ ΓFG 1 IJ K H 4K

OP PP Q

FG where Γbng Γb1 − ng = H

π sin nπ

IJ K

148


z

π2

(v)

z

π2

tan θ dθ =

0

dθ =

cos θ

0

z

π2

sin θ

sin1 2 θ cos −1 2 θ dθ

0

LM MM N

OP PP Q

1 –1 +1 +1 1 1 3 1 β 2 , 2 = β , = 2 2 2 2 4 4

FG 3 IJ ΓFG 1 IJ 1 H 4K H 4K 2 Γ b1g 1 F 1I F 1I ΓG1 – J ΓG J 2 H 4K H 4K

LM OP N Q

Γ

=

=

=

z

∞

12. Evaluate: (i)

0

1 π π . = . 2 sin π 2 4

z

∞

x dx 1 + x6

(ii)

0

y 2 dy · 1 + y4

Solution x6 = t or x = t1/6

(i) Let

dx =

1 −5 6 t dt 6

The given integral becomes,

z

∞

0

1

t6

FG 1 IJ t H 6K

−5 6

1+ t

dt

z zb ∞

−2

1 t 3 dt = 6 1+ t 0

1 = 6

=

∞

0

1

1+ t

LM N

−1

t3

g

2 3+1 3

1 1 2 β , 6 3 3

OP Q

dt

LM x Using the relation, β bm, ng = MN b1 + xg

z

∞

0

m –1 m+n

zb ∞

dx =

0

xn –1

1+ x

g

m+n

dx

OP PQ

149

INTEGRAL CALCULUS

FG 1IJ Γ FG 2 IJ H 3K H 3 K Γ b1g 1 F 1I F 1I Γ G J . Γ G1 – J H K H 6 3 3K Γ

1 = 6

=

=

1 . 6

π

F πI sin G J H 3K

=

π 3 3

y4 = t, or y = t1/4

(ii) Substituting then

dy =

1 t 4

so that,

z

y 2 dy 1 + y4 0

z

∞

∞

=

−3 4

Ft GH

1 4

dt

I FG 1IJ t JK H 4 K 2

1 = 4

=

=

= =

=

z zb ∞

0

dt

1+ t

0

1 = 4

−3 4

−1

t4 dt 1+ t 3

∞

0

t4

1+ t

−1

g

3 1 + 4 4

dt

LM OP N Q F 3I F 1 I ΓG J ΓG J 1 H 4K H 4K 4 Γ b1g 1 F 1I F 1I Γ G1 – J Γ G J H 4 4K H 4K 1 3 1 β , 4 4 4

1 π 4 sin π 4 π 2 2

.

.

150


z

π2

13. Prove that

0

sin θ

Solution

z

π2

L.H.S.

−1 sin 2

z

π 2

dθ

z

×

π2

θ dθ ×

0

sin θ dθ = π.

0

1

sin 2 θ dθ

0

LM – 1 + 1 OP LM 1 + 1 OP MM 22 , 0 2+ 1PP × 12 β MM 2 2 , 0 2+ 1PP N Q N Q 1 L1 1O L3 1O βM , P×βM , P 4 N4 2 Q N4 2Q L ΓFG 1 IJ ΓFG 1 IJ O L Γ FG 3 IJ Γ FG 1 IJ O 1 M H 4K H 2K P M H 4K H 2K P M F 3 I PP MM F 5 I PP 4M MN Γ GH 4 JK PQ MN Γ GH 4 JK PQ RSΓFG 1 IJ UV ΓFG 1 IJ 1 T H 2K W H 4K FG since ΓFG 1 IJ = π IJ = π. 1 F 1I H 2K K 4 H ΓG J 4 H 4K

1 β = 2

=

=

2

=

z

∞

14. Prove that

0

e –x

x

z ∞

2

dx ×

4

x 2 e –x dx =

0

π 4 2

.

Solution Substituting we get

x2 = t or x = dx =

dt 2 t

zd z z ∞

I1 =

t in the first integral,

e –t

i

1 0 t1 2 2 ∞ −3 4

1 t = 2

.

1 2 t

e – t dt

0

=

=

1 2

∞ 1 −1 t4 0

FG IJ HK

1 1 Γ 2 4

e – t dt

. dt

151

INTEGRAL CALCULUS −3

x4 = u or x = u1/4 in the second integral, we obtain, dx =

Taking

z zd z z ∞

1 4 u du 4

4

x 2 e – x dx

I2 =

0

∞

x1 4

=

0

1 = 4

∴ The given integral = I1 × I2

0

1– x

0

∞

3

0

FG IJ HK

1 π π . = · 8 sin π 4 2 4

=

z 1

4

dx ×

−1

u 4 e – u du

FG IJ FG IJ HK HK 1 F 1I F 1I Γ G J Γ G1 – J 8 H 4K H 4K

=

z

1 4 u du 4

1 1 3 Γ Γ 8 4 4

=

15. Prove that

−3

. e–u .

1 3 Γ 4 4

=

x2

2

−1 1 u 4 e – u du 4

=

1

∞

i

dx 1+ x

0

4

=

π 4 2

.

Solution. In the first integral setting x2 = sin θ or x = π , when x = 0, θ = 0. 2 Therefore, 1 x2 I1 = dx = 4 0 1– x

sin θ we get dx =

x = 1, θ =

z

z

π2

0

1 = 2 =

1 2

sin θ

z z

cos θ dθ 1 – sin 2 θ 2 sin θ

π2

0

.

sin θ cos θ . dθ cos θ sin θ

π2

0

sin1 2 θ d θ

cos θ 2 sin θ

dθ when

152


LM MM N

1 +1 0+1 1 1 . β 2 , = 2 2 2 2

=

FG H

1 3 1 β , 4 4 2

IJ K

In the second integral substitute x2 = tan θ or x =

OP PP Q tan θ then dx =

when x = 0, θ = 0 when x = 1, θ = π/4

z

π4

Hence,

I2 =

0

1 = 2 1 = 2 1 = 2

=

where

sec 2 θ .d θ 1 + tan 2 θ 2 tan θ 1

z z z

π4

sec θ tan θ

0

π4

dθ

dθ sin θ ⋅ cos θ

0

π4

dθ

FG 1 IJ sin 2θ H 2K

0

1 2

.

z

π4

dθ sin 2θ

0

t = 2θ, dt = 2 · dθ, ⇒ dθ = θ = 0, t = 0, when θ = =

=

=

1 2

z

π2

1 2 2 1 2 2

0

1 dt ⋅ 2 sin t

z zb

π2

0

1 sin t

π2

0

sin t

g

dt

−1 2

dt

1 dt 2

π π , t= 4 2

sec 2 θ d θ 2 tan θ

.

153

INTEGRAL CALCULUS

1

1 . β = 2 2 2

=

1

β

4 2

∴ The given integral is I1 × I2 =

=

zb 1

16. Show that

1+ x

1 16 2

q –1

LM 1 , 1 OP N4 2 Q

OP PQ

LM 3 , 1 OP ⋅ β LM 1 , 1 OP N4 2 Q N4 2 Q F 3I F 1 I F 1 I F 1 I ΓG J ⋅ ΓG J ΓG J ⋅ ΓG J H 4K H 2K ⋅ H 4K H 2K F 5I F 3I ΓG J ΓG J H 4K H 4K

16 2

p–1

0+1 2

β

1

g b1 – xg

LM –1 + 1 MN 2 2 ,

=

π 4 2

⋅

b g

dx = 2 p + q – 1 β p , q .

–1

Solution Substitute:

1 + x = 2t, dx = 2dt

when

x = –1, t = 0, when x = 1, t = 1 Given Integral,

zb 1

=

1+ x

zb g

g b1 – xg p –1

q –1

(where x = 2t − 1)

dx

0

b

1

=

2t

p –1

z

0

g

1 – 2t – 1

b g

1

p+q –1 t p–1 1 – t = 2

q –1

q –1

2 dt

dt

0

zb

= 2p + q – 1 β ( p, q).

b

17. Show that

x–a

g bb – xg m–1

n –1

b g

dx = b – a

m+ n –1

b g

β m, n .

a

Solution Substitute

x – a = (b – a) t

so that

dx = (b – a) dt

when

x = a, t = 0 and

when

x = b, t = 1

z b

∴

a

b x – ag bb – xg m –1

zb 1

n–1

dx =

g

b–a t

0

m –1

b

g

b–a– b–a t

n–1

bb – ag dt

154


=

bb – ag

z

b g

1

m + n –1

tm–1 1 – t

n–1

dt

0

= (b – a)m + n – 1 β (m, n). Hence proved.

zb 1

18. Prove that

xm – 1 + xn –1

0

1+ x

g

m+n

b g

dx = β m, n .

Solution From the relation,

zb zb ∞

β (m, n) =

0

xm – 1

1+ x

1

=

0

g

dx

m+ n

1+ x

g

zb ∞

xm – 1

dx +

m+n

1

xm –1

1+ x

g

m+n

dx

In the second integral on R.H.S. of (1) Substitute x = 1/t so that dx = – dt/t2 when

x = 1, t = 1, and

when

x = ∞, t = 0

zb ∞

Hence,

1

xm – 1

1+ x

g

m+ n

z FG

FG 1IJ H tK

0

dx =

1

H

1+

z 0

= –

t

zb zb 1

0

0

Therefore from (1), we get

1+ t

0

1

=

0

g

2

tm+ n

b1 + t g

m+n

1+ x

g

m+ n

xm – 1

1+ x

g

m+n

zb 1

dx +

0

g

dt t2

dx

xm – 1 + xn – 1 1+ x

.

m+n

dt

xn – 1

zb zb 1

β (m, n) =

.

m–1

FG – dt IJ Ht K

m+n

tn–1

1

=

IJ K

1 t

1

1

=

m–1

m+ n

dx

xn – 1

1+ x

g

m+n

dx

Hence proved.

...(1)

155

INTEGRAL CALCULUS

zd

π2

19. Prove that

0

cos 2m – 1θ sin 2n – 1θ a cos 2 θ + b sin 2 θ

i

m+n

1

dθ =

2a mb n

b g

β m, n .

Solution. Let I be the given integral,

zd zd

π2

then

I =

cos2 θ

0

π2

=

cos2 m – 1 θ sin 2 n – 1 θ

0

i da + b tan θi m+n

2

tan 2 n – 1 θ sec 2 θ dθ a + b tan 2 θ

i

m+n

Substituting tan θ = t, we get sec2 θ dθ = dt when

θ = 0, t = 0 and

when

θ = π/2, t = ∞

zd ∞

Then

I =

0

t 2n + 1 a + bt 2

i

m+ n

a

bt2 = ay or t =

Now substitute

so that

dt

y

b

a dy b 2 y

dt =

Limits remain the same.

z

∞

Hence,

I =

0

=

=

zb ∞

where β (m, n) =

0

yn – 1

1+ y

g

m+ n

dy

e

a y

ba + byg

1 m m 2a b 1 2a b

m m

zb ∞

b

j

2 n −1

yn – 1

g β bm, n g 0

.

m+n

1+ y

Hence proved.

m+ n

a dy b2 y

dy

dθ

156


EXERCISE 3.3 1. Evaluate

(1)

bg 2 Γ b4g Γb 3g

(3) Γ

FG 1 IJ Γ FG 3IJ ΓFG 5 IJ H 2K H 2K H 2 K

b g FGH 52 IJK F 11I · ΓG J H 2K

Γ3 Γ (5)

2. Evaluate

Ans. 30

(2)

LMAns. 3π π OP 8 PQ MN

(4)

LMAns. N

LMAns. MN

FG –5IJ H 2K F –9 I (3) Γ G J H 2K

(1) Γ

LMAns. MN

16 315

z z z z zb z

OP PQ –32 π O P 945 PQ –8 π 15

x e

–x

dx

Ans. 120

∞

x 3 2 e – x dx

0

∞

x 6 e –2 x dx

0

∞

(7)

e

– x3

dx

0

1

(9)

x log x

g

4

0

∞

(11)

0

LMAns. 16 π OP MN 105 PQ LMAns. – 3 ΓFG 2 IJ OP H 3K Q N

FG –7 IJ H 2K –1 (4) Γ FG IJ · H 3K

z z z zb g z FGH IJK z FGH IJK

LMAns. π OP MN 2 PQ LMAns. 3 OP N 8Q LMAns. 105 π OP 8 PQ MN

∞

5

0

(5)

LMAns. 4 OP N 3Q

(2) Γ

∞

(3)

LMAns. 16 OP N 105Q

OP Q

3. Evaluate (1)

b g FGH 23IJK F 9I ΓG J H 2K F 7I ΓG J H 3K F 4I ΓG J H 3K Γ3 Γ

Γ7

2

2 –3x dx

dx

LMAns. 3 π OP 4 PQ MN LMAns. 45 OP N 8Q LMAns. 1 ΓFG 1IJ OP N 3 H 3K Q LMAns. N

LMAns. MN 2

OP Q π O P 3 log 2 PQ 94 625

(2)

x e – x dx

0

∞

(4)

x 3 e –2 x dx

0

∞

(6)

2

x 5 e – x dx

0

1

(8)

log x

3

dx

Ans. – 6

0

1

(10)

0

1 log x

1

(12)

x 2 log

0

32

1 x

dx 3

dx .

Ans. – 2 π

LMAns. 2 OP N 27 Q

157

INTEGRAL CALCULUS

z

∞

4. Show that

e – st t

0

π , s > 0. s

dt =

z 1

5. Prove that Γ (n) =

0

FG log 1 IJ H xK

z

n –1

dx , n > 0.

∞

6. Prove that Γ (n) = a

n

0

e – ax x n – 1 dx.

z ∞

n 7. Prove that Γ (n) = 2a

2

x 2 n – 1 e – ax dx.

0

8. Prove that

z z

∞

(1)

x n – 1 cos ax dx =

0

∞

(2)

x n – 1 sin ax dx =

0

b g FGH IJK 1 F nπ I Γ bn g sin G J ⋅ H2K a 1 nπ Γ n cos n 2 a

n

[Hint. choose a = 0, b = a, in solved example 8.] 9. Evaluate:

LMAns. 1 OP N 60 Q LMAns. 5π OP N 16 Q

(1) β (4, 3)

F 7 1I (3) β G , J H 2 2K F 5 1I (5) β G , J . H 6 6K (1)

z z

b g

x3 2 1 – x

0

2

(3)

0

z 3

(5)

0

x2 2–x

dx

dx 3x – x 2

2

Ans. 2π

10. Evaluate the following integrals: 1

LMAns. π OP N 6Q LMAns. RΓF 1 I U 2 πOP MN ST GH 4 JK VW PQ

FG 3 , 5 IJ H 2 2K F 1 1I (4) β G , J H 4 2K (2) β

12

dx

LMAns. π OP N 16 Q LMAns. 64 2 OP 15 PQ MN

z z 1

(2)

0

4

(4)

Ans. π

b

u3 2 4 – u

0

z 1

(6)

0

LMAns. π OP N 2Q

x dx 1– x

dx 1 – x3

g

52

Ans. 12π

du

LM MMAns. MN

FG 1IJ OP H 3K P F 5I P ΓG J P H 6K Q

πΓ

158


z 1

(7)

e

0

x dx

zd 2

(9)

LMAns. 1 OP N 21Q

j

x3 1 –

x 8 – x3

0

i

13

LMAns. N

dx

16π 9 3

OP Q

z z

LMAns. 8 OP N 15Q LMAns. 2 OP N 105Q

(1)

sin θ dθ 5

0

π2

(3)

sin 3 θ cos4 θ dθ

0

z

π2

(5)

13

sin

θ cos

−1 3

θ dθ

0

z z

π2

(7)

cot θ dθ

0

z

(10)

z z

(2)

0

(4)

0

LMAns. π OP 3Q N

(6)

LMAns. π OP 2Q N

(8)

z zd

OP PQ LMAns. 5 2 π OP MN 128 PQ LMAns. MN

∞

(3)

0

x 2 dx

1 + x4

i

3

·

π 2 4

OP PQ F 1I O π ΓG J P H 4K P F 3I P ΓG J P H 4K Q πa 6 32

LMAns. 16 OP N 35 Q LMAns. 5 2 π OP sin θ cos θ dθ 64 PQ MN LM F 1I O ΓG J 1 H 4K P MMAns. 2 F 3 I PP dθ ΓG J P cos θ MN H 4K Q LM F 5I F 1 I O ΓG J ΓG J P 60 H 6 K H 4 K M PP Ans. sin θ dθ M 13 π cos θ MN PQ LMAns. π sec FG pπ IJ OP N 2 H 2 KQ cos7 θ dθ

π2

z

12

72

π2

0

z

π2 3

0

12. Evaluate each of the following integrals: dx 1 + x4 0

LM MMAns. MN

LMAns. MN

π2

tan q θ d θ , 0 < p < 1.

∞

1 – x 4 dx

0

0

(1)

a 2 – x 2 dx

1

π2

(9)

x4

0

11. Evaluate each of the following integrals: π2

z a

(8)

z

∞

(2)

0

8

dx

b g

x 1+ x

Ans. π

13. Evaluate:

zb zb 3

(1)

1

7

(2)

4

0

dx

gb

x –1 3– x

g

Ans. π

LM R F 1I U O 2 SΓ G J V P MMAns. T H 4 K W PP 3 π Q N 2

gb

g

7 – x x – 3 dx

159

INTEGRAL CALCULUS

14. Show that

z z z z ∞

(1)

z

∞

2

x e – x dx ×

0

0

∞

(2)

0

4

(4)

dx = π π

4

x 2 e – x dx × e – x dx =

z

16 2

2

x

∞

0

∞

e– x

0

∞

(3)

z z ∞

2

x e – x dx ×

π

4

x 2 e – x dx =

0

8 2

∞

π . 2 P +1

sin θ dθ × sin p + 1 θ dθ = p

0

zb ∞

15. Prove that

0

b

0

xn – 1 x+a

g

dx =

m+n

1 an

g

b g

β m, n .

ADDITIONAL PROBLEMS (From Previous Years VTU Exams.)

zz 1

1. Evaluate

0

1− y 2

x 3 y dx dy.

0

Solution. We have

z z z LMN OPQ zd zd zd 1− y 2

1

I =

i.e.,

I =

x 3 y dx dy

y=0

x=0

1

x4 y 4

y=0

=

=

=

=

1 4 1 4 1 4 1 4

1− y 2

dy x=0

1

y 1 − y2

y =0

i

2

dy

i

1

y 1 − 2 y 2 + y 4 dy

y =0 1

i

y − 2 y 3 + y 5 dy

y=0

LM y N2

2

−

2 y4 y6 + 4 6

OP Q

1

y=0

160


LM 1 − 1 + 1 OP N2 2 6Q

1 4

=

1 · 24

I =

zz 1


0

Solution Let

I =

dx dy .

y

Y

z z 1

1

(0, 1)

1

(1, 1)

dx dy

y =0 x=

y

On changing the order of integration, x =

zz z

dy dx

X (0, 0)

x =0 y =0 1

=

y

x2 0

dx =

z

Lx O dx = M P N3Q

3 1

1

x =0

I =

x=1

x =y

x2

1

I =

⇒

y

x= y

2

x

x =0

2

=

0

1 · 3

3. Change the order of integration and evaluate

1 3

(1, 0)

Fig. 3.1

z zb 3

4− y

0

0

g

x + y dx dy.

Solution. Refer page no. 122. Example 5. 4. Change the order of integration and hence evaluate

zz

4a 2 ax

0

Y

xy dy dx .

y = 2 ax

2

x 4a

(0, 4a)

Solution. We have

We have

z z

2

y = x /4a

4a

2 ax

x =0

x2 y= 4a

I =

x2 = 2 ax 4a

i.e., x (x3 – 64x3) = 0

(4a, 4a)

xy dy dx

(4a, 0) X (0, 0)

or

x4 = 64a3x

⇒

x = 0 and x = 4a Fig. 3.2

x = 4a

161

INTEGRAL CALCULUS

x2 From y = , we get y = 0 and y = 4a. 4a x2 and y = 2 ax are (0, 0) and 4a (4a, 4a) on changing the order of integration we have y varying from 0 to 4a and x varying Thus the points of intersection of the parabola y =

y2 to 2 ay . from 4a Thus I =

=

z z z LMN OPQ z LMN z FGH

4a

2 ay

y =0

y2 x= 4a

4a

x2 y⋅ 2

xy dx dy

y =0

2 ay

dy x=

y2 4a

OP Q 1 y I 4ay − J dy 2 16a K 1 L 4ay 1 y O − ⋅ P M 2N 3 6Q 16a O 1 L F 64a I 1 4a G 4096 a iP − M d J 2 MN H 3 K 96a PQ 1 L 256a 128a O − M P 2N 3 3 Q

1 = 2

4a

y 4ay −

y=0 4a

=

y4 dy 16a 2 5

2

2

y=0

a 6 4

3

=

2

y=0

3

=

6

2

3

=

= I =

zz

4

64a 4 3 64a 4 · 3

b

g

xy x + y dx dy taken over the region enclosed by the curves y = x and 5. Find the value of 2 y = x. Solution. Refer page no. 118. Example 6.

zz 1


0

Solution. Refer page no. 120. Example 2.

2− x2

x

x x + y2 2

dy dx .

162


z zb 3


4− y

g

x + y dx dy .

y=0 x=1

Solution. Refer page no. 121. Example 4. 8. With usual notation show that β(m, n) =

b g b g· Γ b m + ng

Γ m ⋅Γ n

Solution. Refer page no. 133.

zb 1

9. Show that

1+ x

−1

g b1 − xg p −1

q −1

dx = 2p + q – 1 β (p, q) .


z

π2

10. Using Beta and Gamma functions evaluate

z

π2

sin θ dθ ×

0

0

dθ sin θ

·


OBJECTIVE QUESTIONS 1. The area bounded by the curves y2 = x – 1 and y = x – 3 is (a) 3

(b)

7 2

9 7 (d ) · 2 3 2. The volume of the tetrahedron bounded by the coordinate planes and the plane

(c)

Ans. c

x y z + + = 1 is a b c (a)

abc 2

(b)

abc 3

(c)

abc 6

(d )

24 · abc

zz zz zz

∞ ∞

3. For

b g

f x , y dx dy , the change or order is

0 x

∞ ∞

(a)

b g

f x , y dx dy

∞ y

0 0

zz zz

∞ ∞

(b)

x 0

(c)

Ans. c

0

b g

f x , y dx dy

∞ x

(d )

b g

f x , y dx dy

y

0 0

b g

f x , y dx dy .

Ans. c

163

INTEGRAL CALCULUS

z 2

4. The value of the integral

–2

dx is x2

(a) 0

(b) 0.25

(c) 1

(d ) ∞.

zz

Ans. d

1 2− x

5.

xy dx dy is equal to

x2

0

(a)

3 4

(b)

3 8

(c)

3 5

(d )

3 · 7

zzb

g

2 x

6.

Ans. b

x + y dx dy = ..........

0 0

(a) 4

(b) 3

(c) 5

(d ) None of these.

zz

Ans. a

1 1− x

7.

0

dx dy represents.....

0

Ans. Area of the triangle having vertices (0, 0), (0, 1), (1, 0)

z

∞

8.

2

e – x dx = ......

0

π 2

(a) (c) 9. β

(b)

π 4

FG 1 , 1 IJ H 2 2K

π 2

(d ) None of these. = .....

(a) 3.1416

(b) 5.678

(c) 2

(d ) None of these.

b g

Ans. a

Ans. a

. = .... 10. Γ 35 (a)

15 8

(b)

15 π 8

(c)

10 7

(d ) None of these.

Ans. b

164


11. The surface area of the sphere x2 + y2 + z2 + 2x – 4y + 8z – 2 = 0 is (a) 59 π (b) 60 π (c) 92 π

zzz

(d ) None of these.

Ans. c

2 3 2

12.

xy 2 z dz dy dx = .....

0 1 1

13.

(a) 26

(b) 28

(c) 30

(d ) 50.

zz

Ans. a

dx dy is

R

(a) Area of the region R in the Cartesian form (b) Area of the region R in the Polar form (c) Volume of a solid

Ans. a

(d ) None of these. 14. Γ

FG 1 IJ H 2K

is

π

(a)

(b)

(c) 2 15. Γ

FG – 7 IJ H2K

(a)

(d ) None of these.

Ans. a

is

16 15

(b)

16 18 16. Γ (n + 1) is (a) n (c)

(c) (n + 1) ! 17. β

π 2

FG 7 , – 1IJ H2 2 K

16 315

(d ) None of these.

Ans. b

(b) n + 1 (d ) n !.

Ans. d

is

(a)

– 15π 8

(b)

15 8

(c)

π 8

(d )

15π · 8

Ans. a

165

INTEGRAL CALCULUS

z

∞

18.

3

x 2 e – x dx is

0

(a) (c)

z

3 4 π 4

(b)

3 π 4

(d ) None of these.

Ans. b

π2

19.

sin 6 θ d θ is

0

5π 32 π (c) 32

(a)

z

(b)

5 32

(d ) None of these.

Ans. a

π2

20.

sin 4 θ cos3 θ d θ is

0

(a)

5π 32

(b)

16 35

(c)

2 35

(d )

6 · 35

Ans. c GGG

UNIT

18

Vector Integration and Orthogonal Curvilinear Coordinates 4.1 INTRODUCTION In the chapter we shall define line integrals, surface integrals and volume integrals which play very important role in Physical and Engineering problems. We shall show that a line integral is a natural generalization of a definite integral and surface integral is a generalization of a double integral. Line integrals can be transformed into double integrals or into surface integrals and conversely. Triple integrals can be transformed into surface integrals. The corresponding integral theorems of Gauss, Green and Stokes are discussed. The concept of Gradient, Divergence, Curl and Laplacian already discussed in the known Cartesian system. These will be discussed in a general prospective in the topic orthogonal curvilinear coordinates.

4.2 VECTOR INTEGRATION 4.2.1 Vector Line Integral If F is a force acting on a particle at a point P whose positive vector is r on a curve C then represents physically the total work done in moving the particle along C. Thus, total work done is

z

z

F ⋅ dr

C

F ⋅ dr = 0

C

WORKED OUT EXAMPLES 1. If F = (5xy – 6x2) i + (2y – 4x) j . Evaluate

F ⋅ dr along the curve y = x3 in the x-y

C

plane from (1, 1) to (2, 8). Solution. We have

z

F

= (5xy – 6x2) i + (2y – 4x) j and r = xi + yj will give

dr

= dxi + dyj 166


167

∴

F ⋅ dr = (5xy – 6x2) dx + (2y – 4x) dy Since y = x3 we have dy = 3x2 dx and varies from 1 to 2

z

F ⋅ dr

= (5x . x3 – 6x2) dx + (2 . x3 – 4x) . 3x2 dx

C

zd 2

=

i

5 x 4 – 6 x 2 + 6 x 5 – 12 x 3 dx

1

=

2

x 5 − 2 x 3 + x 6 − 3x 4

H 2. If F = (3x2 + 6y) i – 14yz j + 20 xz2 k . Evaluate

= 35

z

1

F ⋅ dr from (0, 0, 0) to (1, 1, 1) along

C

the path x = t, y =

t2,

z=

Solution

t3.

H F = (3x2 + 6y) i – 14yz j + 20xz2 k .

dr = dxi + dyj + dzk F ⋅ dr = (3x2 + 6y) dx – 14yz dy + 20xz2dz x = t, y = t2, z = t3

∴ Since

dx = dt, dy = 2t dt, dz = 3t2 dt

we obtain

F ⋅ dr = (3t2 + 6t2) dt – (14t5) 2tdt + (20t7) 3t 2 dt

∴

F ⋅ dr = (9t2 – 28t 6 + 60t9) dt ; 0 ≤ t ≤ 1 i.e., ∴ t varies from 0 to 1

z

zd 1

F ⋅ dr =

t =0

C

LM 9t N3

3

=

3. Evaluate:

z

i

9t 2 – 28t 6 + 60t 9 dt 28t 7 60t 10 − + 7 10

= 3–4+6 = 5.

OP Q

1

0

F ⋅ dr where F = yz i + zx j + xyk and C is given by r = t i + t 2 j + t 3 k ; 0 ≤ t ≤ 1.

C

Solution

F = yz i + zx j + xyk r = ti + t 2 j + t 3 k

∴ ∴ where ∴

2 dr = dti + 2tdt j + 3t dt k

F ⋅ dr = yz dt + zx × 2tdt + xy3t2.dt r = ti + t 2 j + t 3 k = xi + y j + zk x = t, y = t2, z = t3

168


∴ ∴ ∴

F ⋅ dr = t5 dt + 2t5 dt + 3t5 dt = (t5 + 2t5 + 3t5) dt F ⋅ dr = 6t5 dt t varies from 0 to 1

z

z 1

F ⋅ dr =

C

6t 5 dt

0

LM6 t OP N 6Q

6 1

= =

4. Evaluate z = t, 0 ≤ t ≤ π.

z

t

6 1 0

0

=1–0

= 1. 2 2 2 F ⋅ dr where F = x i + y j + z k and C is given by x = cos t, y = sin t,

C

F x y z

Solution Here,

z

= = = =

x2 i + y2 j + z2 k cos t ⇒ sin t ⇒ t ⇒

dx = – sin t dt dy = cos t dt dz = dt

dr = dxi + dy j + dzk , 0 ≤ t ≤ π

zd zo π

F ⋅ dr =

x 2 dx + y 2 dy + z 2 dz

t =0

C

i

π

=

– cos 2 t sin t dt + sin 2 t cos t dt + t 2 dt

0

LM cos t + sin t + t OP N 3 3 3Q 3

=

= − 5. If F = x 2 i + xy j . Evaluate

z

Solution

x.

3

π

0

π π −2 1 1 − +0+ = 3 3 3 3 2

2

F ⋅ dr from (0, 0) to (1, 1) along (i) the line y = x (ii) the

C

parabola y =

3

t

2 F = x i + xy j

dr = dxi + dy j

F ⋅ dr = x2dx + xy dy

169


(i) Along ∴

z

y = x; we have 0 ≤ x ≤ 1 and dy = dx.

zd z 1

F ⋅ dr =

i

x 2 + x 2 dx

x =0

C

1

2 x 2 dx =

=

x =0

(ii) Along

z

y =

y2

x,

zd

=

0

2 3

i

LM y + y OP N3 4Q

4 1

6

=

=

2 y 5 + y 3 dy

y =0

C

3 1

= x ⇒ 2y dy = dx, 0 ≤ y ≤ 1

1

F ⋅ dr =

LM 2 x OP N3Q

0

1 1 7 + = · 3 4 12

b

g

b

g

6. Use the line integral, compute work done by a force F = 2y + 3 i + xz j + yz − x k when it moves a particle from the point (0, 0, 0) to the point (2, 1, 1) along the curve x = 2t2, y = t, z = t3. Solution

Work done =

z

F ⋅ dr

C

where, Here,

t varies from 0 to 1

F x y z ( y

b

g

= 2 y + 3 i + xz j + = 2t2 ⇒ = t ⇒ 3 = t ⇒ = t)

b yz − xg k dx = 4t dt dy = dt dz = 3t2 dt

dr = dx i + dy j + dzk

z

2 = 4tdt i + dt j + 3t dt k

z {b zd

g

1

F ⋅ dr =

t =0

C

1

=

=

i

i

12t + 8t 2 – 6t 4 + 2t 5 + 3t 6 dt

0

=

d

2 t + 3 4t dt + 2t 5dt + t 4 − 2t 2 3t 2 dt

LM6t N

2

8 6 1 3 + t3 − t5 + t6 + t7 3 5 3 7

288 . 35

OP Q

1 0

}

170


7. Find the work done in moving a particle once around an ellipse C in the xy-plane, if the ellipse has centre at the origin with semi-major axis 4 and semi-minor axis 3 and if the force field is given by

F =

b3x − 4y + 2zg i + d4x + 2y − 3z i j + d2xz − 4y 2

2

i

+ z3 k.

x 2 y2 + = 1 and its 4 2 32 parametric equations are x = 4 cos t, y = 3 sin t. Also t varies from 0 to 2π since C is a curve in the xy-plane, we have z = 0 Solution. Here path of integration C is the ellipse whose equation is

b

g b

g

F = 3x − 4 y i + 4 x + 2 y j

∴

dr = dxi + dyj

and

F ⋅ dr =

b3x − 4 yg i + b4x + 2 yg j ⋅ dxi + dyj

F ⋅ dr = (3x – 4y) dx + (4x + 2y) dy ⇒ dx = – 4 sin t dt ⇒ dy = 3 cos t dt

x = 4 cos t y = 3 sin t ∴ t varies from 0 to 2π

z

z mb zb zb

2π

F ⋅ dr =

C

gb

g b

2π

=

0

IJ K

g

LM N

= 48t +

15 cos 2t 2

= 96 π. 2 8. If F = xyi – z j + x k . Evaluate

OP Q

2π 0

z

F × dr where C is the curve x = t2, y = 2t, z = t3 from

C

t = 0 to 1.

F = xyi – z j + x 2 k and dr = dxi + dy j + dzk i

Hence

sin 2t 2

48 − 15 sin 2 t dt

0

Solution

r

FG3 sin t cos t = H

g

48 − 30 sin t cos t dt

2π

=

g

12 cos t – 12 sin t – 4 sin t dt + 16 cos t + 6 sin t ⋅ 3 cos t dt

0

F × dr =

j

k

xy – z x 2 dx dy

dz

171


F ⋅ dr = x = y = z =

where

– (zdz + t2 ⇒ 2t ⇒ 3 t ⇒

x2dy) dx = dy = dz =

i – (xy dz – x2dx) j + (xydy + zdx) k 2tdt 2dt 3t2dt

t varies from 0 to 1

∴

z

=

{– d3t

i

zd

d

z zd

1

1

i L 3t + 2t OP – 4 j LM t OP –i M N 6 5 Q N6Q

F ⋅ dr = −i

C

1

3t 5 + 2t 4 dt – 4 j t 5dt + k

0

0

6

=

i }

+ 2t 4 i – 4t 5 j + 4t 3 + 2t 4 k dt

5

i

4t 3 + 2t 4 dt

0

5 1

6 1

0

0

L 4t +kM N4

4

2t 5 + 5

OP Q

1

0

–9 2 7 i – j + k. = 10 3 5 9. If φ = 2xyz2, F = xyi − z j + x 2 k and C is the curve: x = t2, y = 2t, z = t3 from t = 0 to t = 1 evaluate the following line integrals. (i)

z

φ ⋅ dr (ii)

C

Solution

dr x y z

∴

dr =

= = = =

z

F × dr .

C

dxi + dy j + dzk t2 ⇒ dx = 2tdt 2t ⇒ dy = 2dt t3 ⇒ dz = 3t2dt

FH 2ti + 2 j + 3t k IK dt 2

φ = 2xyz2 φ = 2.t2.2t.t6 = 4t9 ∴ (i)

z

φ ⋅ dr

= (8t10 i + 8t9 j + 12t11 k) dt =

z

=

φ ⋅ dr

=

C

(ii)

i

8t 10i + 8t 9 j + 12t 11k dt

t=0

C

Thus,

zd 1

φ ⋅ dr

LM 8t OP i + LM8t OP N 11 Q N 10 Q 11 1

10 1

0

0

8 4 i + j+k 11 5

3 3 4 F = 2t i − t j + t k

F × dr =

i 2t 3 2t

j – t3 2

k t4 3t 2

L12t OP k j+M N 12 Q 12 1

0

172


= – (3t5 + 2t4) i – (6t5 –2t5) j + (4t3 + 2t4) k

z

= – (3t5 + 2t4) i – 4t5 j + (4t3 + 2t4) k

z {d 1

F × dr =

C

i

i} L t 2t OP i − 4 LM t OP j + LMt + 2t OP k –M + N2 5 Q N6Q N 5 Q 0

5 1

6

=

z

d

3t 5 + 2t 4 i – 4 t 5 j + 4 t 3 + 2 t 4 k dt

4

0

F × dr = −

C

5 1

6 1

0

0

9 2 7 i – j + k. 10 3 5

EXERCISE 4.1 1. If F = 3xyi – 5z j + 10 xk . Evaluate

z

F ⋅ dr along the curve C given by x = t2 + 1,

C

Ans. 303

y = 2t2, z = t3 from t = 1 to t = 2. 2. Evaluate

z

b

g b

g

F ⋅ dr where F = 2 x + y i + 3 y − x j + yz k and C is the curve x = 2t2,

C

LM Ans. 227 OP N 42 Q

y = t, z = t3 from t = 0 to t = 2. 3. Evaluate

z

F ⋅ dr where F = yzi + zx j + xyk and C is the portion of the curve

C

r = a cos ti + b sin t j + ctk from t = 0 to t = 4. Evaluate

z

π . 2

Ans. 0

F ⋅ dr where F = x 2 i + y 2 j + z 2 k and C is the arc of the curve r = ti + t2

C

Ans. 1

j + t3 k from t = 0 to t = 1.

5. Find the total work done in moving a particle once round a circle C in the xy-plane if the curve has centre at the origin and radius 3 and the force field is given by

F = (2x – y + z) i + (x + y – z2) j + (3x – 2y + 4z) k .

Ans. 8π

6. If F = 2yi − z j + xk and C is the circle x = cos t, y = sin t, z = 2 cos t from t = 0 to t=

π evaluate 2

z C

F × dk .

LMAns. FG 2 – π IJ i + FG π – 1 IJ j OP N H 4 K H 2K Q

173


4.3 INTEGRAL THEOREM 4.3.1 Green’s Theorem in a Plane This theorem gives the relation between the plane, surface and the line integrals. Statement. If R is a closed region in the xy-plane bounded by a simple closed curve C and M (x, y) and N (x, y) are continuous functions having the partial derivatives in R then

z

M dx + N dy =

zz FGH R

C

IJ K

∂N ∂M dx dy. − ∂x ∂y

4.3.2 Surface integral and Volume integral Surface Integral An integral evaluated over a surface is called a surface integral. Consider a surface S and a point P on → it. Let A be a vector function of x, y, z defined and continuous over S. In n^ is the unit outward normal to the surface S → and P then the integral of the normal component of A → at P (i.e., A · n^) over the surface S is called the surface integral written as

zz

Z ^

n s

ds

^

k

O

Y

A ⋅ n ds

R

S

where ds is the small element area. To evaluate integral we have to find the double integral over the orthogonal projection of the surface on one of the coordinate planes.

X

dx dy

Fig. 4.1

Suppose R is the orthogonal projection of S on the XOY plane and n^ is the unit outward normal ^ ^ to S then it should be noted that n^ · k ds (k being the unit vector along z-axis) is the projection of ^ the vectorial area element n ds on the XOY plane and this projection is equal to dx dy which being ^ the area element in the XOY plane. That is to say that n^ · k ds = dx dy. Similarly, we can argue ^ ^ to state that n^ · j ds = dz dx and n^ · i ds = dy dz. All these three results hold good if we write ^ n ds = dy dz i + dz dx j + dx dy k. Sometimes we also write

d s = n ds = ∑ dy dz i Volume Integral

zzz

If V is the volume bounded by a surface and if F (x, y, z) is a single valued function defined over V then the volume integral of F (x, y, z) over V is given by

zzz

F dv . If the volume is divided

V

into sub-elements having sides dx, dy, dz then the volume integral is given by the triple integral

b

g

F x , y , z dx dy dz which can be evaluated by choosing appropriate limits for x, y, z.

174


4.3.3 Stoke’s Theorem Statement. If S is a surface bounded by a simple closed curve C and if F is any continuously differentiable vector function then

z

F ⋅ dr =

C

zz

Curl F ⋅ n ds =

S

z z FH

IK

∇ × F ⋅ n ds

S

4.3.4 Gauss Divergence Theorem Statement. If V is the volume bounded by a surface S and F is a continuously differentiable vector function then

zzz

div F dV =

V

zz

F ⋅ n dS

S

^

where n is the positive unit vector outward drawn normal to S.

WORKED OUT EXAMPLES 1. Verify Green’s theorem in the plane for

zd

i

b

g

3x 2 − 8y 2 dx + 4y − 6xy dy where C is the

C

boundary for the region enclosed by the parabola y2 = x and x2 = y. Solution. We shall find the points of intersection of the parabolas y 2 = x and x2 = y i.e., Equating both, we get

∂M ∂y

Mdx + Ndy

y=x

= x2 ⇒ x = x4

(0,1)

= 0 = 0 = 0, 1 intersection are (0, 0) and (1, 1). = 3x2 – 8y2, N = 4y – 6xy = – 16y

By Green’s theorem,

z

x and y = x2

y =

x or x – x4 x (1 – x3) ∴ x and hence y = 0, 1 the points of Let M

=

zz FGH z zb R

C

L.H.S =

(0,0)

A(1,1)

2

Fig. 4.2

IJ K

∂N ∂M dx dy − ∂x ∂y

OA

z

g b M dx + N dyg = I AO

x

(1,0) y=x

∂N = – 6y ∂x

M dx + N dy +

2

O

Mdx + Ndy

C

=

Y

1

+ I2

175


Along OA:

y = x2 dy = 2xdx, x varies from 0 to 1

zd zd

i

1

I1 =

d

x =0 1

=

i

3x 2 + 8 x 3 − 20 x 4 dx

x=0

= Along AO:

i

3x 2 – 8 x 4 dx + 4 x 2 – 6x 3 2 x dx

x 3 + 2x 4 – 4x5

1 0

= –1

y 2 = x ⇒ dx = 2y dy, y varies from 1 to 0

zd zd 0

I2 =

i

d

y =1 0

=

i

4 y – 22 y 3 + 6 y 5 dy

1

=

LM2 y N

2

–

11 4 y + y6 2

OP Q

0

= 1

Hence,

5 3 L.H.S. = I1 + I2 = – 1 + = 2 2

Also

R.H.S. =

zz FGH IJK z zb z z z LMN OPQ zd i

g

x

1

– 6 y + 16 y dy dx

x =0 y= x

2

x

1

=

5 2


R

=

10 y dy dx

x = 0 y = x2 1

=

10 y 2 2

x =0

x

dx

y = x2

1

x − x 4 dx

= 5

x=0

LM x − x OP N2 5Q L 1 1O 3 5M − P = N 2 5Q 2 2

5 1

= 5

x=0

= ∴

i

3 y 4 – 8 y 2 2 y dy + 4 y − 6 y 3 dy

L.H.S. = R.H.S. =

3 . 2

Hence verified.

176

2. Verify Green’s theorem in the plane for

z {d


i

}

xy + y 2 dx + x 2 dy where C is the closed curve

C

of the region bounded by y = x and y = x2. y

Solution. We shall find the points of intersection of y = x and y = x2. Equating the R.H.S. ∴ x = x2 ⇒ x – x2 = 0 x (1 – x) = 0 x = 0, 1 ∴ y = 0, 1 and hence (0, 0), (1, 1) are the points of intersection. We have Green’s theorem in a plane,

z

M dx + N dy

=

R

C

The line integral,

zd C

zz FGH z {d

i

xy + y 2 dx + x 2 dy

=

Along OA, we have y =

zd zd 1

x

z {d

i

i

1

=

3x 3 + x 4 dx

x =0

=

LM 3x N4

4

+

Along AO, we have y = x ∴ dy = dx x varies from 1 to 0

zd z

x5 5

OP Q

1

= 0

0

I2 =

i

0

=

3 1 19 + = 4 5 20

x ⋅ x + x 2 dx + x 2 dx

1

3x 2 dx = x 3

0 1

= –1

1

L.H.S. = I1 + I2 = R.H.S. =

19 –1 – 1= 20 20

zz FGH R

IJ K


x (1, 0)

i

}

xy + y 2 dx + x 2 dy

AO

x ⋅ x 2 + x 4 dx + x 2 ⋅ 2 x dx

x =0

Also

=

}

xy + y 2 dx + x 2 dy +

2

Fig. 4.3

= I1 + I2 ∴ dy = 2x dx and x varies from 0 to 1.

I1 =

Hence,

y

IJ K

i

(1, 1)

y=x

o (0, 0)


OA

x2,

A

(0, 1)

177


N = x2

where

∂M ∂ y = x + 2y

∂N = 2x ∂x R is the region bounded by y = x2 and y = x

zz R

FG ∂N − ∂M IJ dx dy H ∂x ∂y K

z zb z zb z zd zd

x=0 y=x

2 x – x – 2 y dy dx

2

g

x

1

=

g

x

1

=

x − 2 y dy dx

x = 0 y = x2 1

=

x

xy − y 2

y = x2

x =0

1

i

x 4 − x 3 dx

x=0

LM x − x OP N5 4Q

4 1

5

=

∴

i

x 2 − x 2 – x 3 − x 4 dx

x =0

=

dx

i d

1

=

= xy + y2

M

L.H.S. = R.H.S. =

=

0

–1 . 20

1 1 –1 − = 5 4 20 Hence verified.

3. Apply Green’s theorem in the plane to evaluate

zd

i

d

i

2x 2 − y 2 dx + x 2 + y 2 dy where C is

C

the curve enclosed by the x-axis and the semicircle x2 + y2 = 1. Solution. The region of integration is bounded by AB and the semicircle as shown in the figure. By Green’s theorem,

z

Mdx + Ndy =

C

Given

zd

R

M = 2x2 – y2,

∂M ∂y

i

IJ K


d

...(1)

i

2 x 2 − y 2 dx + x 2 + y 2 dy

C

where

zz FGH

= – 2y

N = x2 + y2

∂N ∂x

Fig. 4.4

= 2x

178


From the equation (1),

zd

i

d

i

2 x 2 – y 2 dx + x 2 + y 2 dy =

C

In the region, x varies from – 1 to 1 and y varies from 0 to = 2

z zb z LMN z LMN 1

1 – x2

x = –1

y=0

1

= 2

1 − x2

g

OP Q

1 − x2

dx y=0

1

= 2

g

2 x + 2 y dx dy

R

x + y dy dx

y2 xy + 2

x = –1

zz b

x 1 − x2 +

x = –1

d

1 1− x 2 2

iOPQ dx

Since, x 1 − x 2 is odd and (1 – x2) is even function

zd 1

= 0+2

i

1 − x 2 dx

0

L xO 2 Mx − P N 3Q

3 1

=

4. Evaluate

zd C

=

i

0

4 . 3

xy − x 2 dx + x 2 y dy where C is the closed curve formed by y = 0, x = 1 and

y = x (i) directly as a line integral (ii) by employing Green’s theorem. Solution (i) Let M = xy – x2, N = x2y

z C

Mdx + Ndy =

zb

OA

Fig. 4.5

g

Mdx + Ndy +

zb

AB

g

Mdx + Ndy +

zb

BO

Mdx + Ndy

g


(a) Along OA:

y = 0

⇒

dy = 0 and x varies from 0 to 1.

(b) Along AB:

x = 1

⇒

dx = 0 and y varies from 0 to 1.

(c) Along BO:

y = x

⇒

dy = dx and x varies from 1 to 0.

z

∴

b M dx + N dyg

z

z

1

=

C

1

– x dx + 2

x =0

=

0

y dy +

y=0 3 1

2 1

0

0

4

0

1

1 1 1 −1 + – = 3 2 4 12

zd i zz FGH zz b g z zb z z

xy − x 2 dx + x 2 ydy =

Thus

C

We have Green’s theorem,

zb

Mdx + Ndy

g

=

R.H.S. =

–1 12

IJ K


R

C

x 3 dx

x =1

Lx O Ly O Lx O –M P +M P +M P N3Q N2Q N4Q

= –

(ii)

z

179

2xy − x dx dy

R

=

g

x

1

2 xy – x dy dx (from the figure)

x=0 y=0 1

=

xy 2 – xy

x =0

x y =0

1

=

x 3 – x 2 dx

x =0

LM x N4

4

=

= ∴

R.H.S. =

−

x3 3

OP Q

1

0

1 1 –1 − = 4 3 12

–1 . 12

5. Verify Stoke’s theorem for the vector F = (x2 + y2) i – 2xyj taken round the rectangle bounded by x = 0, x = a, y = 0, y = b. Solution By Stoke’s theorem :

z C

F ⋅ dr =

zz FH S

IK

∇ × F ⋅ n dS

180


F

= (x2 + y2) i – 2xyj

dr

= dxi + dyj

F ⋅ dr

=

(x2

+

y2)

Y

dx – 2xy dy

(1) Along OP: y = 0, dy = 0, x varies from 0 to a

z z

z z a

F ⋅ dr

=

OP

y=b

a3 x dx = 3

R

2

0

x=a

x=0

(2) Along PQ: x = a, dx = 0 ; y varies from 0 to b b

F ⋅ dr

=

PQ

2ay dy = ab 2

o

y=0

0

Fig. 4.6

(3) Along QR: y = b, dy = 0; x varies from a to 0

z z z

z zb

d x − b i dx = LMN x3 – b xOPQ

0

F ⋅ dr

=

a

QR

3

2

2

0

= ab 2 −

2

a

(4) Along RO: x = 0, dx = 0 ; x varies from b to 0

F ⋅ dr

=

RO

L.H.S. =

F ⋅ dr

C

curl F

Now,

g

0 – 0 dy = 0

a3 a3 2 2 ab ab + + − +0 = 3 3 = 2ab2 i ∂ ∂x x2 − y2

=

j ∂ ∂y 2 xy

For the surface, S ⋅ n = k ∴ R.H.S. =

curl F ⋅ n

zz

Q

= 4y

zz z LMN z a b

curl F ⋅ n dS =

S

4 y dy dx

0 0

a

=

0

y2 4 2

OP Q

a

= 2b

2

0

= 2ab2 L.H.S. = R.H.S. Hence, the Stoke’s theorem is verified.

dx

b

dx 0

k ∂ = 4y k ∂z 0

a3 3

p

X


b

181

g

6. Verify Stoke’s theorem for the vector field F = 2x – y i − yz 2 j − y 2 zk over the upper half surface of

x2

+

y2

+

z2

Solution

= 1 bounded by its projection on the xy-plane.

z

F ⋅ dr =

C

zz

curl F ⋅ n dS

(Stoke’s theorem)

S

x2 + y2 = 1, z = 0 (xy-plane) x = cos t, y = sin t, z = 0

C is the circle: i.e.,

r = xi + y j where 0 ≤ θ ≤ 2π

dr = dxi + dy j F =

where, ∴ L.H.S. =

z

b2 x – yg i − yz

2

j − y 2 zk

F ⋅ dr = (2x – y) dx F ⋅ dr =

C

zb zb zd zd z RST b

( z = 0)

g

2x − y dx

C

2π

=

gb

g

2 cos t – sin t – sin t dt

0

2π

=

i

sin 2 t – 2 cos t sin t dt

0

2π

=

i

sin 2 t – sin 2 t dt

0

2π

=

0

g

UV W

1 1 – cos 2t – sin 2t dt 2

LM t − sin 2t + cos 2t OP N2 4 2 Q FG 1 – 1 IJ + bπ − 0g = π H 2 2K

2π

=

0

= Hence,

Also,

F ⋅ dr = π

...(1)

i ∂ curl F = ∇ × F = ∂x 2x − y

b

j ∂ ∂y – yz 2

k ∂ ∂z – y2z

g bg b g

= i – 2 yz + 2 yz – j 0 + k 0 +1

182


= k

dS = ndS = dydz i + dzdx j + dxdy k

∴ Hence,

R.H.S. =

zz

= curl F ⋅ ndS

S

zz

= π

zz

dx dy ...(2)

dx dy represents the area of the circle x2 + y2 = 1 which is π.

Thus, from (1) and (2) we conclude that the theorem is verified. 7. If F = 3yi − xz j + yz 2 k and S is the surface of the paraboloid 2z = x2 + y2 bounded by z = 2, show by using Stoke’s theorem that

zz

curl F ⋅ n ds = 20 π.

S

Solution. If z = 0 then the given surface becomes x2 + y2 = 4. Hence, C is the circle x2 + y2 = 4 in the plane z = 2 x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ 2π

i.e.,

Hence by Stoke’s theorem, we have

z

F ⋅ dr =

C

L.H.S. put ∴

z z

∴

F = 3 yi − xz j + yz 2 k , dr = dxi + dy j + dzk F ⋅ dr =

z

zd zb

3 y dx – xz dy + yz 2 dz

C

i

z = 2, dz = 0

F ⋅ dr =

C

∴

curl F ⋅ n dS

S

C

where

zz

3 y dx – 2 x dy

g

C

x = 2 cos t

⇒

dx = – 2 sin t dt

y = 2 sin t

⇒

dy = 2 cos t dt

z

b

0

F ⋅ dr =

2π

C

g

b

Since, the surface S lies below the curve C

zd 0

= −

i

12 sin 2 t + 8 cos2 t dt

2π

zd

2π

=

i

12 sin 2 t + 8 cos 2 t dt

0

g

6 sin t –2 sin t dt – 4 cos t 2 cos t dt

183


z

z

π 2

=

48

π 2

sin 2 t dt + 32 cos2 t dt

0

= ∴

=

0

π π 48 ⋅ + 32 ⋅ = 20 π 4 4

zz

curl F ⋅ n dS = 20 π

S

8. Using divergence theorem, evaluate

zz

Hence proved.

where F = 4xzi − y 2 j + yzk and S is the F ⋅ ndS

S

surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. Solution. We have divergence theorem:

zzz

div F dV =

V

zz

F ⋅ n dS

S

div F = ∇ ⋅ F

Now

= =

FGi ∂ + j ∂ + k ∂ IJ d4 xzi – y H ∂x ∂y ∂z K ∂ ∂ ∂ 4 xz g + –y i + b b yzg d ∂x ∂y ∂z 2

= 4z – 2y + y = 4z – y Hence, by divergence theorem, we have

zz

F ⋅ n dS =

S

zzz z z zb zz z zb g z LMN OPQ div F ⋅ dV

V 1

=

1

g

1

4 z – y dz dy dx

x =0 y=0z =0 1

=

1

2 z 2 – yz

x=0 y=0 1

=

1 z=0

1

2 − y dy dx

x=0 y=0 1

=

x =0

y2 2y − 2

1

dx 0

dy dx

2

j + yzk

i

184


z LMN z

OP Q

1

=

1 dx 2

2−

x =0 1

=

x =0

=

3 dx 2

3 x 2

1 0

=


3 · 2

zz

F ⋅ n dS where F = x 2 i + y 2 j + z 2 k and S is the

S

surface of the solid cut off by the plane x + y + z = a from the first octant.

d i

d i

d i

∂ 2 ∂ 2 ∂ 2 x + y + z ∂x ∂y ∂z = 2x + 2y + 2z = 2 (x + y + z) Hence, by divergence theorem, we have

div F = ∇ ⋅ F =

Solution. Now

zz

F ⋅ n dS =

S

zzz zzz b g z z zb g z z LMNb g OPQ zz b g LM b g OP z MN PQ i zd div F ⋅ dV

V

=

2 x + y + z dV

V

a −x a− x− y

a

x + y + z dz dy dx

= 2

x=0 y=0

z=0

a− x

a

1 x + y z + z2 2

= 2

x=0 y =0 a− x

a

= 2

x=0 y =0

1 2 a − x+y 2

a

=

a y− 2

x =0

1 = 3x

zz S

x+ y 3

3

2

dy dx

dx

y=0

2a 3 − 3a 2 x + x 3 dx

=0

1 x2 x4 2a 3 x – 3a 2 + = 3 2 4 1 4 F n ⋅ dS = a . 4

dy dx z=0

a−x

a

LM N

a− x− y

OP Q

a

0

185



zz

F ⋅ n dS where F = x3i + y3j + z3k and S is the

S

surface of the sphere

x2

+

y2

+

z2

=

a2.

d i

d i

d i

∂ ∂ ∂ 3 x3 + y3 + z ∂x ∂y ∂z = 3x2 + 3y2 + 3z2 = 3 (x2 + y2 + z2)

div F =

Solution.

∴by divergence theorem, we get

zz

F ⋅ n d S =

S

zzz zzz

div F dV

V

=

d

i

3 x 2 + y 2 + z 2 dx dy dz

V

...(1)

Since, V is the volume of the sphere we transform the above triple integral into spherical polar coordinates (r, θ, φ). For the spherical polar coordinates (r, θ, φ), we have x2 + y2 + z2 = r2 and dx dy dz = dV ∴ dV = r2 sin θ dr dθ d φ Also, 0 ≤ r ≤ a, 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π Therefore (1) reduces to,

zz

z z zd i z z π

a

F ⋅ n dS = 3

2π

r 2 r 2 sin θ dr dθ dφ

r =0 θ=0 φ=0

S

π

a

= 3

r 4 dr ×

r=0

= 3×

θ=0

LM r OP × – cosθ N5Q × b – cosπ + 1g × 2π 5

11. Evaluate

zz b

3a 5 5

=

12 πa 5 · 5

dφ

φ=0

a

r=0

=

z

2π

sin θ dθ ×

π θ=0

× φ

2π φ=0

g

yzi + zxj + xyk ⋅ n dS where S is the surface of the sphere x2 + y2 + z2 = a2

S

in the first octant. Solution. The given surface is x2 + y2 + z2 = a2, we know that ∇φ is a vector normal to the surface φ (x, y, z) = c. Taking φ (x, y, z) = x2 + y2 + z2

187


=

3 2

z a

d

0

LM a x − x OP N 2 4Q 3 La a O − P M 2 N2 4Q 2 2

3 = 2

4

=

Thus

zz

b yzi + zxj + xyk g ⋅ ndS

S

12. Evaluate

zz b

i

x a 2 − x 2 dx

=

3 a4 8

=

3 a4 . 8

4

a

0

4

g

axi + byj + czk ⋅ n d S where S is the surface of the sphere x2 + y2 + z2 = 1.

S

Solution. Let we have

zz

F = axi + byj + czk F ⋅ n dS =

S

=

zz

div F ⋅ dV

V

div ⋅ F = ∇ ⋅ F =

∴

zzz

FGi ∂ + j ∂ + k ∂ IJ baxi + byj + czk g H ∂x ∂y ∂z K ∂ ∂ ∂ ax g + bby g + bcz g b ∂x ∂y ∂z

= (a + b + c) F ⋅ n dS =

zzz b

g

a + b + c dV

V

S

= (a + b + c) V where V is the volume of the sphere with unit radius and V = Here, since we have r = 1, Thus,

zz S

V =

F ⋅ n dS =

4 π 3

b

g

4π a +b+ c . 3

...(1)

4 3 πr for a sphere of radius r. 3

188


EXERCISE 4.2 1. If F = axi + byj + czk and S is the surface of the sphere x2 + y2 + z2 = 1 by using 4 Ans. π F ⋅ n dS . divergence theorem. Evaluate 3 2. Evaluate

zz

LM N

zz S

OP Q

F ⋅ n dS where F = 4 xyi + yz j − xzk and S is the surface of the cube bounded

S

Ans. 32

by the planes x = 0, x = 2, y = 0, y = 2, z = 0, z = 2. 3. If F = y2z2i + z2x2j + x2y2k, evaluate

zz

F ⋅ n dS where S is the part of the sphere

S

x2 + y2 + z2 = 1 above the xy-plane and bounded by this plane. 4. Use Gauss divergence theorem to evaluate

zz

LM Ans. π OP N 12 Q

F ⋅ n dS where F = (x2 – z2) i + 2xyj

S

+ ( y2 + z2) k where S is the surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, Ans. 3 z = 0, z = 1. 5. Verify Green’s theorem in plane for

zd

b

i

g

3x 2 – 8 y 2 dx + 4 y – 6xy dy , where C is the bound-

C

ary of the triangle formed by the lines x = 0, y = 0 and x + y = 1. 6. Verify Green’s Theorem in the plane for

zd

i

x 2 + y 2 dx − 2 xy dy, where C is the rectangle

C

Ans. – 2ab 2

bounded by the lines x = 0, y = 0, x = a, y = b. 7. Using Green’s theorem, evaluate

zb

LM Ans. 5 OP N 3Q

g

cos x sin y − xy dx + sin x cos y dy , where C is the

C

circle x2 + y2 = 1. 8. Evaluate by Stoke’s theorem

zb

Ans. 0

g

yzdx + xzdy + xydz , where C is the curve x2 + y2 = 1,

C

Ans. 0

z = y2.

9. Verify Stoke’s theorem for the function F = zi + xj + yk, where C is the unit circle in the xy-plane bounding the hemisphere Z =

1 − x2 − y 2 .

Ans. π

189


3 . Evaluate 2

10. If F = yi + z3xj – y3zk and C is the circle x2 + y2 = 4 in the plane z =

z

F ⋅ dr

C

LM Ans. 19π OP N 2Q

by Stoke’s theorem. 11. Evaluate

z

F ⋅ dr by Stoke’s theorem F = y2i + x2j – (x + 2) k and C is the boundary of

C

LM Ans. 1 OP N 3Q

the triangle with vertices (0, 0, 0) (1, 0, 0) and (1, 1, 0).

4.4 ORTHOGONAL CURVILINEAR COORDINATES 4.4.1 Definition Let the coordinates (x, y, z) of any point be expressed as functions of (u1 , u2, u3), so that x = x (u1, u2, u3), y = y (u1, u2, u3), z = z (u1, u2, u3) then u, v, w can be expressed in terms of x, y, z, as u1 = u (x, y, z), u2 = v (x, y, z) and u3 = w (x, y, z). And also if

b

∂ x, y , z

b

g

∂ u1 , u2 , u3

g ≠ 0 then the

system of coordinates (u1, u2, u3) will be an alternative specification of the Cartesian system (x, y, z) and (u1, u2, u3) are called the curvilinear coordinates of the point. If one of the coordinates is kept constant say u1 = c, then and the locus of (x, y, z) is a surface which is called a coordinate surface. Thus, we have three families of coordinate system corresponding to u1 = c, u2 = c, u3 = c. Suppose u1 = c, u2 = c and u3 ≠ c in that case locus obtained is called a coordinate curve and also there are such families. The tangent to the coordinate curves at the point p and the three coordinate axes of the curvilinear systems. The direction of these axes vary from point to point and hence the unit associated with them are not constant. When the coordinate surfaces are mutually perpendicular to each other, the three curves define an orthogonal system and (u, v, w) = (u1, u2, u3) are called orthogonal curvilinear coordinates.

P

Fig. 4.7

4.4.2 Unit Tangent and Unit Normal Vectors →

The position vector of point p (x, y, z) is r = xi1 + yi2 + zi3 where i1, i2, i3 are unit vectors along the tangent to the three coordinate curves. ∴ i1 . i2 = i3 . i2 = i3 . i1 = 0 and ∴

i1 × i2 = i3, i2 × i3 = i1, i3 × i1 = i2

r (u1, u2, u3) = x (u1, u2, u3) i1 + y (u1, u2, u3) i2 + z (u1, u2, u3) i3

190


∴

dr =

∂r ∂r ∂r du1 + du2 + du3 ∂u1 ∂u2 ∂u3

Then r (u1, u2, u3) is a vector point, function of variables u, v, w. The unit tangent vector i1 along the tangent to u-curve at P is

i1 =

If

∂r ∂u1

∂r ∂u1 ∂r ∂u1

∂r = h1 which is called scalar factor, then ∂u = h1i1. 1

Similarly, unit tangent vectors along v-curve and w-curves are

i2 =

∴

∂r ∂ u2

∴

∂r ∂u3

∂r ∂u2

∂r ∂u = 2 h2

= h2 i2

i3 =

∴

∂r ∂u2

∂r ∂u3 ∂r ∂u3

∂r ∂u3 = h3

= h3 i3

∂r ∂r ∂r dr = ∂u du1 + ∂u du2 + ∂u du3 1 2 3 = h1du1i1 + h2du2i2 + h3du3i3

Then length of the arc dS is given by 2 2 2 2 2 2 dS2 = dr ⋅ dr = h1 du1 + h2 du2 + h3 du3

Example 1. Find the square of the element of arc length in cylindrical coordinates and determine the corresponding Lommel constants. Solution. The position vector, r is cylindrical coordinates is

r = r cos θi + r sin θ j + zk


Then

dr = = =

191

∂r ∂r ∂r dr + dθ + dz ∂r ∂θ ∂z

FH cos θ i + sin θ jIK dr + FH – r sin θ i + r cos θ jIK dθ + k dz bcos θ dr − r sin θ dθg i + bsin θ dθ + r cos θ dθg j + k dz

From the relation, dS2 = dr ⋅ dr = h12 du12 + h22 du22 + h32 du32 we have, dS2 = dr ⋅ dr = (cosθ dr – r sinθ dθ)2 + (sinθ dr + r cosθ dθ)2 + dz2 = (dr)2 + r2(dθ)2 + dz2 The Lommel’s constant, also called scale factors are h1 = 1, h2 = r, h3 = 1. Example 2. Find the volume element dv in cylindrical coordinates. Solution. The volume element in orthogonal curvilinear coordinates u1, u2, u3 is given by u1 = r,

u2 = θ,

u3 = z,

In cylindrical coordinates

h1 = 1,

h2 = r,

h3 = 1.

Then

dv = h1h2h3du1du2du3 u1 = r

⇒

du1= dr,

u2 = θ

⇒

du2= dθ,

u3 = z1

⇒

∴

du3 = dz, dv = (1).(r) · (1). dr dθ dz

Then,

dv = r dr dθ dz.

4.4.3 The Differential Operators In this section, we shall express the gradient, divergence and curl in terms of orthogonal curvilinear coordinates u1, u2, u3. Then, the Laplacian can be expressed as the divergence of a gradient by the chain rule, we have

∂f ∂x

=

∂f ∂u1 ∂ f ∂ u2 ∂f ∂u3 + ⋅ + ⋅ ∂u1 ∂x ∂u 2 ∂x ∂u 3 ∂x

∂f ∂f ∂u1 ∂f ∂u2 ∂f ∂u3 ⋅ + ⋅ + ⋅ = ∂u1 ∂y ∂u2 ∂y ∂u3 ∂y ∂y

∂f ∂u1 ∂ f ∂u 2 ∂f ∂u3 ∂f ⋅ + ⋅ + ⋅ = ∂u1 ∂z ∂u 2 ∂z ∂u3 ∂z ∂z In rectangular Cartesian coordinate system ∂f ∂f ∂f H i + j+ k ∇f = ∂x ∂y ∂z where f is a scalar function. In engineering problems this f is usually a potential link

192


Velocity potential or electric potential or gravitational potential. Using chain rule this becomes, ∇f

LM N

∂u ∂u ∂f ∂u1 i+ 1 j+ 1 k ∂u1 ∂x ∂y ∂z

=

LM N

∂f ∂u2 ∂u ∂u i+ 2 j+ 2 k ∂u2 ∂x ∂y ∂z

+

LM N

∂u ∂u ∂f ∂u3 i+ 3 j+ 3 k ∂ u3 ∂ x ∂y ∂z

∇u1

But

OP Q OP Q

OP Q

=

∂f ∂f ∂f ∇ u1 + ∇ u2 + ∇ u3 ∂u1 ∂u 2 ∂u 3

=

1 1 1 e1 , =u = e2 , =u3 = e3 . 2 h h2 h3

Then the gradient of f, in orthogonal curvilinear coordinates, is =f

1 ∂f 1 ∂f 1 ∂f e1 + e2 + e3 . h1 ∂u1 h2 ∂u2 h3 ∂u3

=

WORKED OUT EXAMPLES

LM ∂r ∂r ∂r OP MN ∂u , ∂u , ∂u PQ

1. Show that

1

2

= h1 h 2 h 3 =

3

1 . ∇ u1 , ∇ u2 , ∇ u 3

Solution. By definition of unit vectors, e1

=

1 ∂r 1 ∂r 1 ∂r , e3 = , e2 = h2 ∂u2 h3 ∂u3 h1 ∂u1

...(1)

1 1 1 e1 , =u2 = e2 , =u3 = e3 h1 h2 h3

...(2)

we also know that =u1 = Then using (1), we have

LM ∂r MN ∂u

1

,

∂r ∂r , ∂u2 ∂u3

OP PQ

=

h1 e1 , h2 e2 , h3 e3

=

h1h2 h3 e1e2 e3

= h1 h2 h 3

...(3)

193


Similarly from (2), we have

∇ u1 , ∇ u2 , ∇ u3

=

LM 1 e , 1 e , 1 e OP Nh h h Q 1

1

=

2

2

3

3

1 1 e1 , e2 , e3 = h1h2 h3 h1h2 h3

...(4)

From (3) and (4), we obtain

LM ∂r N ∂u

1

,

∂r ∂r , ∂u2 ∂u3

OP Q

1 = h1h2 h3 = ∇u , ∇ u , ∇ u 1 2 3

as the required solution. 2. Find the =r m.

m , the origin is 2

Solution. The position vector of a point (x, y, z) from

r = xi + yj + zk x2 + y2 + z2

r =

m =rm = mr ⋅

Then

F r I = mr GH r JK

m− 2

2

r = mr m–2 (xi + yj + zk). 3. Find the gradient of f = x2y + zy2 + xz2 in curvilinear coordinates. Solution. In curvilinear coordinates (u1, u2, u3) the given function f takes the form 2 2 2 f = u1 u2 + u3u2 + u1u3

∂f ∂f 2 = 2u1 u2 + u3 , = 2u3u2 + u12 ∂u2 ∂u1 ∂f 2 = u2 + 2u1u3 ∂u 3 The gradient formula is =f = Sub.

1 ∂f 1 ∂f 1 ∂f e1 + e2 + e3 h1 ∂u1 h2 ∂u2 h3 ∂u3

∂f ∂f ∂f , and , we get ∂u 3 ∂u1 ∂u2 =f =

which is the required gradient.

d

i

d

i

d

i

1 1 2 1 2 2u1u2 + u32 e1 + u1 + 2u3u2 e2 + u2 + 2u1u3 e3 h1 h2 h3

194


EXERCISE 4.3 1. Find the gradient of the following functions in cylindrical polar coordinates. (i) xy + yz + zx, (ii) x (y + z) – y (z – x) + z (x + y), (iii) exp (x2 + y2 + z2).

LMAns. (i) br sin 2θ + z sin θ + z cos θg e + d– r sin θ + z cos θ − z sin θi e OP MM PP + r bsin 2θ + cos θg z. MM (ii ) 2 br sin 2θ + z cos θg e + b1 + r g cos θ − z bcos θ + sin θg e PP MN PQ + r bcos θ + sin θg e 2

θ

r

2

θ

r

z

2. Find =f in spherical polar coordinates when (i) f = xy + yz + zx (ii) f = x ( y + z) + y (z – x) + z (x + y)

LM Ans. (i) r sin θ bsin θ sin 2θ + 2 cos θ sin θ + 2r cos θ cos θg e + r bsin θ cos θ sin 2θ + cos 2θ sin θ + cos 2θ cos θg e MM + r bsin θ cos 2θ + r cos θ cos θ − cos θ sin θge . MM MN (ii ) 2r bsin 2θe + cos 2θe gbcos θ + sin θg + cos θ bcos θ − sin θge r

r

y

r

θ

θ

OP PP PP PQ

4.4.4. Divergence of a Vector We now derive the expression for the divergence of a vector. In this orthogonal curvilinear coordinate system (u1, u2, u3) the unit vector F can be expressed as = F1e1 + F2 e2 + F3e3 By the vector equation can be written as

F

F

= h2h3 F1 (=u2 × =u3) + h3h1 F2(=u3 × =u1) + h1h2F3 (=u1 × =u2)

Then the divergence of F is

∇⋅ F

...(1)

= ∇ . [h2h3F1(=u2 × =u3)] + = · [h3h1F2 (=u3 × =u1)] + =u3 [h1h2F3 (=u1 × =u2)] . = =(h2h3F1) (=u2 × =u3) + =(h1h2F2) . (=u3 × =u1) + =(h3h1F2) . (=u1 × =u2) . = (h2h3F1) = (=u2 × =u3) + h3h1F2 = . (=u3 × =u1)

+ h1h2F3 = . (=u1 × =u2) . . . = (=u × =V ) = =V (= × =u) – =u (= × =V ) = 0

...(2)

195


FH

Using (1) and A ⋅ A × B

IK

∇⋅ F

But

b

e1 ⋅ e2 × e3

g

→

∴

=.F

= 0 for any pair of vectors, we have =

b

g

b

∂ h2 h3 F1 ∇u1 ⋅ ∇u2 × ∇u3 ∂u1

g

b

g

b

g

b

g

b

g

+

∂ h3h1 F2 ∇u2 ⋅ ∇u3 × ∇u1 ∂u2

+

∂ h1h2 F2 ∇u3 ⋅ ∇u1 × ∇u2 ⋅ ∂u3

b

g

b

g

= e2 ⋅ e3 × e1 = e3 ⋅ e1 × e2 = h1h2h3 =u1 (=u2 × =u3) = h1h2h3 =u2 (=u3 × =u1) = h1h2h3 =u3 · (=u1 × =u2) = 1 =

1 h1h2 h3

LM ∂ bh h F g + ∂ bh h F g + ∂ bh h F gOP ∂u ∂u N ∂u Q 2 3 1

3 1 2

1

1 2

2

3

3

WORKED OUT EXAMPLES 1. If f and g are continuously differentiable show that ∇f × ∇g is a solenoidal. Solution. A vector F is solenoidal if ∇ ⋅ F = 0. To show that =f × =g is solenoidal. First we show that =f × =g can be expressed as a curl of a vector. We can show this using the identity. = × (f = g) = =f × =g + f = + =g = =f × =g (3 curl glad is zero) Operating divergence on this and using the identity = × =(f = g) = 0 (3 div curl of any vector is zero) This gives =.(=f × =g) = 0 Hence, =f × =g is solenoidal and hence proved. 2. Find =.(= rm)

d

2 2 2 Solution. Let r = xi + yj + zk be the positive vector so that r = x + y + z

m−2 =r m = mr r ∂ ∂ ∂ mr m−2 x + mr m−2 y + mr m−2 z ∴ =.=r m = ∂x ∂y ∂z Differentiating by part, we get

i

1 2

This gives

d

d

∂ mr m− 2 x ∂x

But

i

∂r ∂x

b

i

g

m− 2 ⋅ = m m−2 r

=

d

1 2 x + y2 + z2 2

d

i

∂r x + mr m− 2 ∂x

i

−1 2 2x

d

i

...(1)

196


d

= x x + y +z =

2

2

i

−1 2

x r

d

i

= m(m – 2)r m

d

i

= m(m – 2) r m – 4 y2 + mr m – 2

d

i

= m(m – 2) r m – 4 z2 + mr m

∂ mr m− 2 ⋅ x ∂x

Similarly,

2

∂ mr m − 2 y ∂y

∂ mr m−2 z ∂z

–4

x2 + mr m–2

–2

Equation (1) becomes, then =.(=rm) = m(m – 2) r m – 4 r2 + 3mr m – 2 = m(m + 1) r m – 2.

EXERCISE 4.4 1. Show that (i) ∇ ⋅ r = 3

FH

(iii) ∇ ⋅ a × r

IK

FH IK = 0 ∇ ⋅ FH x × yIK = y ⋅ ∇ x − x ⋅ ∇ × y

(ii) ∇ ⋅ r × a =0

(iv)

2. Prove that (i) = .=f =

∂2 f ∂2 f ∂2 f + 2 + 2 ∂z ∂x 2 ∂y

(ii) ∇ ⋅ ∇ × F = 0

4.4.5 Curl of a Vector Let

F

= F1e1 + F2 e2 + F3e3 = h1F1=u1 + h2F2=u2 + h3F3=u3

Then,

∇× F

= = × (h1F1=u1) + = × (h2F2=u2) + = × (h3F3 =u3) = = (h1F1) × =u1 + h1F1= × =u1 × =(h2F2) × =u2 + h2F2= + =u2 + =(h3F3) × =u3 + h3F3 = × =u3

But curl of a gradient is zero

∇× F But

= =(h1F1) × =u1 + =(h2F2) × =u2 + =(h3F3) × =u3

=(h1F1) × =u1 =

LM ∂ bh F g ⋅ e + ∂ bh F g e h ∂u h N ∂u 1

2

1 1

1

2 2

2

2

+

b g OPQ

e e ∂ h3 F3 3 ⋅ 1 h3 h1 ∂u3

...(1)

197


Using the fact that e1 , e2 and e3 are the orthogonal unit vectors, we have =(h1F1) × =u1 =

b g

b g

b g

b g

b g

b g

1 ∂ 1 ∂ h1 F1 e2 – h1 F1 e2 h1h3 ∂u3 h1h2 ∂u2

Similarly, we can show that

and

=(h2F2) × =u2 =

1 ∂ 1 ∂ h2 F2 ⋅ e3 – h2 F2 ⋅ e1 h1h2 ∂u1 h2 h3 ∂u1

=(h3F3) × =u3 =

1 ∂ 1 ∂ h3 F3 ⋅ e3 – h3 F3 ⋅ e2 h2 h3 ∂u2 h1h3 ∂u2

Then the equation (1) becomes

LM b g N

b gOPQ e + + h 1h LMN ∂∂u bh F g − ∂∂u bh F gOPQ O ∂ 1 L ∂ h F g− h F gP e + b b M h h N ∂u ∂u Q

∂ ∂ 1 ∇ × F = h h ∂u h3 F3 – ∂u h2 F2 1 3 2 3

1

1 1

1 3

1 1

3

1

2 2

1 2

1 = h1h2 h3

h1e1 ∂ ∂u1 h1 F1

h2 e2 ∂ ∂u2 h2 F2

h3 e3 ∂ ∂ u3 h3 F3

WORKED OUT EXAMPLES H H H 1. Show that ∇ × f F = ∇Q × F + f ∇ × f .

d i

Solution. By definition

H ∇× f F

d i

H

=

∑i × ∂x d f F i

=

∑ i × MM ∂x FH f F IK + f

=

∑ i ×

=

∑ i ∂x × F + MM∑ i × ∂x PP f

∂

L∂ N

∂f

∂f F + ∂x

∑ i × f

L N

= ∇f × F + f ∇ × F.

∂F ∂x

OP PQ

∂F ∂x ∂F

O Q

1

1 1

2

3

198


FH

2. Show that ∇ × A × B

IK

= A ∇⋅ B – B ∇⋅ A + B⋅∇ A – A⋅∇ B .

Solution. By definition

H H ∇× A× B

d

i

H H H ∂ B ∂ A i × A × + ×B = ∂x ∂x H H H ∂B H ∂B H i⋅ A– i⋅A ⋅ i ⋅ B + = ∂x ∂x H ∂A H ∂B – i i ⋅ ⋅B– ∂x ∂x H H H ∂A H ∂A H i⋅ B+ i ⋅B – i. A − ∂x ∂x H H H H H H H H = A ∇⋅B – B – B ∇⋅ A + B⋅∇ A – A⋅∇ B

∑

L ∑ MN F GH ∑

OP Q ∑d I JK ∑ F I ∑ GH JK ∑ d

i ∑d i FG IJ H K i

H ∂A ∂x

∑d i

H ∂B ∂x

3. Show that ∇ × ∇ r m = 0. Solution. We know that

e

j

m– 2 x i + y j + z k . Then ∇r m = m r

i ∂ ∂x

j ∂ ∂y

k ∂ ∂z

mr m – 2 x

mr m – 2 y

mr m – 2 z

∇ × ∇r m =

The coefficient of i in this determinant =

d

i

d

∂ ∂ mr m – 2 z – mr m – 2 y ∂y ∂z

b

g

m–3 = m m–2 r

b

i g

yz zy – m m – 2 rm– 3 = 0 r r

∇ × ∇r m = 0.

Hence,

EXERCISE 4.5 H H H 1. If F = x + y + 1 i + j – x + y k then show that F ⋅ ∇ × F = 0. H 2. If F = ∇ (x3 + y3 + z3 – 3xyz), then find ∇ × F. H H H H H H 3. Show that ∇ × r = 0 and ∇ × r × a = – 2a when r is the position vector and a is

b

g

b

g

b

constant vector. H H H H H 4. Show that ∇ × r × a × b = 2b × a .

b

g

g

199


4.4.6. Expression for Laplacian ∇ 2 ψ We now consider the expression for the Laplacian ∇2 ψ. We know that ∇2 ψ = ∇. (∇ ψ) This, using the expression for gradient and divergence, becomes, ∇2 ψ =

LM F MN GH

IJ K

∂ h2 h3 ∂ψ 1 ∂ + ∂u2 h1h2 h3 ∂u1 h1 ∂u1

FG h h Hh

3 1 2

IJ K

FG H

∂ψ ∂ h1h2 ∂ψ + ∂u2 ∂u3 h3 ∂u3

IJ OP K PQ

...(1)

4.4.7. Particular Coordinate System In many practical applications we need differential operator in a particular coordinate system. Some of them will be discussed in this section. (1) Cartesian Coordinates The Cartesian coordinate system form a particular case of the orthogonal curvilinear coordinates (u1 u2 u3) in which u1 = x, u2 = y, u3 = z such that H ∂ ∂r xi + yj + zk = i = ∂x ∂u1 H ∂r ∂r = j and Similarly, = ∂u2 ∂y H ∂r ∂r = k = ∂z ∂ u3

e

j

Then, for Cartesian coordinates system H ∂r = i = 1, h1 = ∂u1 H ∂r = j = 1, ∂u2

h2 = and

H ∂r = 1 ∂ u3

h3 =

So that, the unit tangent (or base) vectors, becomes e^1 = ^i e^2 = ^j

^

e^3 = k

and Note that and

e^1 . e^2 = i^ . j^ = 0 and so on ^

e^1 × e^2 = i^ × j^ = k = e^3 and so on. Hence the Cartesian coordinate system is a right hand orthogonal coordinate system. Then the arc length in this system is dr2 = dx2 + dy2 + dz2

200


and the volume element is dr = dx dy dz The area element in the yz-plane perpendicular to the x-axis is dy dz, in the zx perpendicular to the y-axis is dz dx and in the xy-plane perpendicular to z-axis is dx dy. (2) Cylindrical Polar Coordinates Let (r, θ, z) be the cylindrical coordinates of the point p. The three surfaces through p : r = u1, = constant = c1; θ = u2 = constant = c2 and z = u3 = constant = c3 are respectively, the cylinder through p coaxial with oz, half plane through oz making an angle θ with the coordinate plane xoz; and the planes perpendicular to oz and distance z from it. The coordinates (r, θ, z) are related to the Cartesian coordinates (x, y, z) through the relation x = r cos θ, y = r sin θ, z = z.

e^ r

z O Z

G r X

Fig. 4.8

u1 = r u2 = θ

H position vector r in this is

Hence,

u3 = z and ^ H ^ ^ r = r cos θ i + r sin θ j + z k H H ∂r ∂r ∂ ^ ^ ^ = = (r cos θ i + r sin θ j + z k ) ∂u1 ∂r ∂r ^

H ∂r ∂ u2 L ∂r ∂ u3

^

= cos θ i + sin θ j H ∂r ^ ^ = = – r sin θ i + r cos θ j ∂θ L ∂r = k = ∂z

Then the Lommel constants in this case h1 =

H ∂r ∂r = = cos2 θ + sin 2 θ ∂u1 ∂r

d

e^ G e^ z

The coordinate surface for any set of constants C1, C2 and C3 are orthogonal. Therefore, the cylindrical system is orthogonal. Here,

p(r, G, z)

i

12

= 1

h2 =

H ∂r ∂r 2 2 2 2 1/2 = = (r sin θ + r cos θ) = r ∂u2 ∂θ

h3 =

H H ∂r ∂r = = 1 ∂u3 ∂z

201


In this case the unit vector e^r, e^θ and e^z take the form: H 1 ∂r ^ er = h1 ∂u1 H 1 ∂r e^θ = h2 ∂u2 H 1 ∂r ^ ez = h3 ∂u3

in the increasing direction of (r, θ, z) are respectively

H ∂r = = ∂r H 1 ∂r = 2 ∂θ H ∂r = = ∂z

cos θ i^ + sin θ j^ = – sin θ i^ + cos θ j^

k

From these, it follows that ^

^

^

^

^

^

^

e r · eθ = e θ · e z = e z · e r = 0 ^

^

^

^

e r × eθ = (cos θ i + sin θ j) × (– sin θ i + cos θ j ) 2

2

^

^

^

= (cos θ + sin θ) k = k = e z

^

^

^

^

^

^

^

^

^

^

e r × e z = (sin θ j + cos θ i ) = e r e z × e r = (cos θ j – sin θ i ) = e θ The element of arc dl in cylindrical coordinates will be dl2 = dr2 + r2 dθ2 + dz2. The conditions for e^r, e^θ, and e^z show that the cylindrical polar coordinates system is a right handed orthogonal coordinate system. The differential operators in this coordinate system take the form ∇Q =

∂Q ∂Q 1 ∂Q er + eθ + ez ∂r r ∂θ ∂z

b g

H 1 ∂ 1 ∂Fθ ∂Fz rFr + + ∇⋅ F = r ∂r r ∂θ ∂z H ∇× F =

FG 1 ∂F – ∂F IJ e + FG ∂F H r ∂θ ∂z K H ∂z FG 1 ∂ brF g – 1 ∂F IJ e H r ∂r r ∂θ K θ

z

r

r

θ

r

–

IJ K

∂Fr eθ + ∂θ

θ

Area elements on the coordinate surfaces and the volume elements in this coordinate systems are

e^ r

Z

dSr = rdθ dz, dSθ = dz dr

e^ Q

dSz = rdr dθ, dr = rdr dθ dz.

e^ G

(iii) Spherical Polar Coordinates

G

If (r, θ, Q) are the spherical polar coordinates, then in this system

r

y = r sin θ sin θ

z

Y

B

x = r sin θ cos θ z = r cos θ

r

X

Fig. 4.9

202


with 0 ≤ r, 0 ≤ θ ≤ π and 0 ≤ Q ≤ 2π. Here,

u1 = r u2 = θ

u3 = Q and H ^ ^ r = r sin θ cos φ i + r sin θ cos φ j^ + r cos θ k H ∂r ∂r ^ ^ ^ Then = = sin θ cos φ i + sin θ sin φ j + cos θ k ∂u1 ∂r H ∂r ∂r ^ ^ ^ = = r sin θ cos φ i + r cos θ sin φ j – sin θ k ∂u2 ∂θ H ∂r ∂r ^ ^ = = r sin θ sin φ i + r sin θ cos φ j ∂u3 ∂φ In this case the formed constants are H H ∂r ∂r = = [sin2 θ (cos2 φ + sin2 φ) + cos2 θ]1/2 = 1 h1 = ∂u1 ∂r h2 = h3 =

H H ∂r ∂r = = r [cos2 θ (cos2 φ + sin2 φ) + sin2 φ]1/2 = r ∂ u2 ∂θ

H H ∂r ∂r = = r sin θ (cos2 φ + sin2 φ)1/2 = r sin θ ∂u3 ∂φ

Then the unit vectors are

H H 1 ∂r ∂r ^ ^ ^ = = sin θ cos φ i + sin θ sin φ j + cos k er = h1 ∂u1 ∂r H H 1 ∂r 1 ∂r ^ e^θ = = cos θ cos φ i^ + cos θ sin φ j^ – sin θ k = h2 ∂u2 r ∂θ H 1 ∂r 1 ∂r ^ ^ = e^φ = = – sin φ i + cos φ j h3 ∂u3 r sin θ ∂φ ^

From these, we have e^r . e^θ = sin θ cos θ (cos2 φ + sin2 φ) – cos θ sin θ = 0

e^θ . e^φ = cosθ (– sin φ cos φ + sin φ cos φ) = 0

e^φ . e^r = sinθ (– sin φ cos φ + cos φ sin φ) = 0.

Also e^r × e^θ = cos θ (sin2 θ + cos2 θ) j – sin φ (sin2 θ + cos2 θ) i = e^φ

^

^

^

e^θ × e^φ = cosθ (cos2 φ + sin2 φ) k + sin θ sin φ j + sin θ cos φ i = e^r ^

^

^

e^φ × e^r = – sinθ (sin2 φ + cos2 φ) k + sin φ cosθ j + cos θ cos φ i = e^θ.

203


These conditions show that the spherical polar coordinates system is a right handed orthogonal system. The element of arc length dl is dl 2 = dr 2 + r 2 dθ2 + r 2 sin2 θ dφ2 The differential operators are given by ∇ψ =

∂ψ ∂ψ 1 ∂ψ 1 er + + eθ + eφ ∂r r ∂θ r sin θ ∂φ

d

i

d

i

∇.F =

∂Fφ 1 ∂ 2 1 1 ∂ sin θ Fφ + r Fr + 2 r sin θ ∂θ r sin θ ∂φ r ∂r

∇2 ψ =

1 1 ∂ 2 ψ 2 ∂ψ ∂ψ ∂2θ . . . sin θ + + ∂θ r 2 sin 2 θ ∂φ2 ∂r 2 r ∂r r 2 sin θ

→

∇× F =

LM d N

i

OP Q

∂F 1 ∂ sin θ Fφ – θ er + r sin θ ∂θ ∂φ

LM d N ∂F O 1L∂ rF g – e b M ∂θ PQ r N ∂r

1 1 ∂Fr ∂ rFφ – r sin θ ∂φ ∂r θ

θ

iOPQ e

θ

+

z

WORKED OUT EXAMPLES 1. Verify the equations (i) ∇θ + ∇ × (R log r) = 0 (ii) ∇ log r + ∇ × Rθ = 0 for cylindrical coordinates. Solution. (i) In cylindrical coordinates ∇r = e^1 ∇θ = Then

1 er r

∇ × (R log r) = (∇ log r) × R =

1 ∇r × R r

=

1 ^ e × k r 1

204


= ∴

–1 ^ e 2 = – ∇θ r

∇ × R log r + ∇θ = 0 Hence the result.

(ii) By definition, ∇ log r = Similarly,

∴

1 ∇r = r

FG 1IJ e H rK

1

∇ × (R θ) = ∇ θ × R =

1 ^ e ×R r 2

=

–1 e1 = –∇ log r r

∇ log r + ∇ × kθ = 0 Hence proved.

Example 2. Determine the scale factors for spherical coordinates. Also find the arc and volume elements. Solution. For spherical coordinates x = r sin θ cos φ

Hence, and

= = = =

r sin θ sin φ r cos θ dr sin θ cos φ + d θ r cos θ cos φ – d φ r sin θ sin φ dr sin θ sin φ + d θ r cos θ sin φ + d φ r sin θ cos φ

dz = dr cos θ – d θ r sin θ Squaring and adding (1), (2) and (3), we have dx2 + dy2 + dz2 = ds2 = dr2 + r2 d θ2 + r2 sin2 θ d φ2 Hence,

or

y z dx dy

h12 = 1, ⇒ h1 = 1, h2 = r, h3 = r sin θ

hr = 1, hθ = r, hϕ = r sin θ The arc element ds =

dr 2 + r 2 d θ 2 + r 2 sin 2 θ d φ 2

Similarly the volume element dV = h1 h2 h3 du1 du2 du3 = r2 sin θ dr d θ dz. Example 3. Prove that the spherical coordinate system is orthogonal. Solution. If R be the position vector of a point P (x, y, z), then

R =

xi+ y j+z k

...(1) ...(2) ...(3)

205


Substituting the values of x, y, z we have

R = r sin θ cos φ i + r sin θ sin φ j + r cos θ k ∂R = sin θ cos φ i + sin θ sin φ j + cos θ k ∂r

∴ ∴

e1 = er

=

Similarly,

∂R ∂r ∂R ∂r

= sin θ cos φ i + cos θ sin φ j + cos θ k

...(1)

= cos θ cos φ i + cos θ sin φ j – sin θ k

...(2)

= − sin φ i + cos φ j

...(3)

e2 = eθ

=

e3 = eϕ =

∂R ∂φ ∂R ∂θ ∂R ∂θ ∂R ∂φ

From (1), (2) and (3) e1 ⋅ e2 = e2 ⋅ e3 = e3 ⋅ e1 = 0 Hence the spherical system is orthogonal in which case

er = Er , eθ = Eθ and eϕ = E ϕ . Example 4. Obtain expression for grad α, Div A and curl A in spherical coordinates. Solution. For spherical coordinates u2 = θ,

u1 = r, Let

u3 = φ,

A = A1 e1 + A2 e 2 + A3 e 3

(i) Expression for grad α, grad α =

e1 ∂ α e 2 ∂ α e 3 ∂ α + + h1 ∂ u1 h2 ∂ u2 h3 ∂ u3

Substituting for h1 h2 h3 and u1, u2, u3 we have grad α = e r

eφ ∂ α ∂ α eθ ∂ α + + r ∂ θ r sin θ ∂ φ ∂r

h1 = 1,

h2 = r,

h3 = r sin θ

206

ENGINEERING MATHEMATICS—II —

(ii) Expression for Div A —

—

Div A = ∆ · A =

=

LM ∂ b A h h g + ∂ b A h h g + ∂ b A h h gOP ∂u ∂u N∂ u Q LM ∂ d A r sin θi + ∂ b A r sin θg + ∂ b A rgOP 1 r sin θ N ∂ r ∂θ ∂φ Q LMsin θ ∂ d A r i + r ∂ b A sin θg + r ∂ A OP 1 ∂r ∂θ ∂φ Q r sin θ N 1 h1 h2 h3

1

2

3

2

1

3

1

3

2

1

2

3

2

1

2

3

2

=

2

3

1

2

2

—

(iii) Expression for curl A

1 — curl A = h1 h2 h3

h1 e1 ∂ ∂ u1 A1 h1

h2 e 2 ∂ ∂ u2 A2 h2

h3 e 3 1 ∂ = 2 ∂ u3 r sin θ A3 h3

er ∂ ∂r A1

r eθ ∂ ∂θ A2 r

r sin θ e φ ∂ ∂φ A3 r sin θ

Example 5. Show that for spherical polar coordinates (r, θ, φ) curl (cos θ grad φ) = grad —

—

—

FG 1IJ ⋅ H rK

Solution. We know curl (φ A ) = φ curl A + grad φ × A . Hence curl {(cos θ) (grad φ)} = cos θ curl grad φ + grad (cos θ) × grad φ. But ∴

curl grad φ = 0 L.H.S. = grad (cos θ) × grad φ

We know grad α =

e1 ∂ α e 2 ∂ α e 3 ∂ α + + h1 ∂ u1 h2 ∂ u2 h3 ∂ u3

For spherical coordinates u1 = r, u2 = θ u3 = φ, h1 = 1, h2 = r, h3 = sin θ ∴

grad (cos θ) = e1

= –

Since

b

∂ cos θ ∂r

g

=

b

g

b g

b

e2 ∂ e3 ∂ ∂ cos θ + cos θ + cos θ r ∂θ r sin θ ∂ φ ∂r e2 sin θ r

b

g ...(1)

g

∂ cos θ = 0 ∂φ

207


Similarly,

grad φ = e1

=

bg

FG H

bg

IJ K

∂ ∂ e2 ∂ e3 φ + φ + φ ∂r r ∂θ r sin θ ∂ φ

e3 r sin θ

...(2)

Hence L.H.S. on using (1) and (2) gives

–

e2 e3 e1 sin θ × = – 2 r r sin θ r

Since

e 2 × e 3 = e1

Similarly

R.H.S. = grad

e1 ∴

FG IJ HK

FG IJ HK

– e1 ∂ 1 e3 ∂ 1 ∂ 1 + e2 + = 2 ∂r r ∂ θ r r sin θ ∂ φ r r

L.H.S. = R.H.S.

Example 6. Find J

FG x, y, z IJ H u ,u ,u K 1

sin φ,

FG 1IJ H rK

2

in spherical coordinates.

3

Solution. For spherical coordinates u1 = r, u2 = θ, u3 = φ and x = r sin θ cos φ, y = r sin φ z = r cos θ Hence,

J

FG x, y, z IJ Hu ,u ,u K 1

2

= J

3

=

FG x, y, z IJ H r , θ, φ K

∂x ∂r ∂x ∂θ ∂x ∂φ

∂y ∂r ∂y ∂θ ∂y ∂φ

∂z ∂r sin θ cos φ ∂z = r cos θ cos φ ∂θ – r sin θ sin φ ∂z ∂φ

sin θ cos φ r cos θ sin φ r sin θ cos φ

cos φ – r sin θ 0

= sin θ cos φ (r2 sin2 θ cos φ) – sin θ sin φ (– r2 sin2 θ sin φ) + cos θ (r2 sin θ cos θ cos2 φ + r2 cos θ sin θ sin2 φ) = r2 sin3 θ cos2 φ + r2 sin3 θ sin2 φ + cos2 θ sin θ r2 = r2 sin3 θ + r2 sin θ cos2 θ = r2 sin θ.

208


EXERCISE 4.6 —

1. Represent A = z i − 2 x j + y k in cylindrical coordinates. Hence obtain its components in

Ans. Ar = z cos θ – r sin 2 θ, Aθ = – z sin θ – 2 r cos2 θ, Az = r sin θ

that system.

2. Prove that for cylindrical system

d eθ dt

=

dθ er . dt

3. Obtain expression for velocity v and acceleration a in cylindrical coordinates. [Hint : r = x i + y j + z k . Substitute for x, y, z and i , j , k ⋅ dr dν and a = dt dt 2 a = r − r θ e r + 2r θ + r θ ′′ e θ + z ′′ e z

V =

d

i

d

i

Ans. V = r e r + r θ e θ + z e z

Where dots denote differentiation with respect to time t. 4. Obtain an expression for ∆2 χ in (i) cylindrical (ii) spherical systems.

LMAns. (i ) 1 ∂ F r ∂ x I + 1 ∂ χ + ∂ χ G J r ∂ r H ∂ rK r ∂ θ ∂z MN ∂ χI ∂ χO ∂ F ∂ χI 1 ∂ F 1 + + r sin θ sin θ P G J G J ∂r H ∂rK r ∂θ H ∂ θ K r sin θ ∂ φ PQ 2

2

(ii )

1 r2

2

2

2

2

2

2

2

2

2

5. For spherical coordinates prove that (i) er′ = θ e θ + sin θ φ e φ (ii) eθ′ = − θ′ e r + cos θ φ e φ (iii) e φ′ = – sin θ φ e r − cos θ φ e θ

ADDITIONAL PROBLEMS 1. Verify Green’s theorem for

zd

i

xy + y 2 dx + x 2 dy where C is the closed curve of the region

c

bounded by the line y = x and the parabola y = x2. Solution. Refer page no. 176, Example 2. 2. Using the divergence theorem evaluate

zz

F ⋅ n ds where F = x3 i + y3 j + z3 k and S is

S

the surface of the sphere

x2

+

y2

+

z2

= a2.

Solution. Refer page no. 185, Example 10. 3. Prove that Cylindrical coordinate system is orthogonal. Solution. Refer page no. 200.

209


4. Evaluate

z

xy dx + xy 2 dy by Stoke’s theorem where C is the square in the x – y plane with

C

vertices (1, 0), (– 1, 0), (0, 1), (0, – 1). Solution. We have Stoke’s theorem

z

F ⋅ dr =

C

zz

B (0, 1)

curl F ⋅ ds

S

2 F = xy i + xy j

since, Hence,

z

A

C (–1, 0)

From the given integral it is evident that

(1, 0)

dr = dx i + dy j + dz k xy dx + xy dy = 2

z

D (0, –1)

F ⋅ dr

C

C

Fig. 4.10

Which is to be evaluated by applying Stoke’s theorem.

Now,

curl F = ∆ × F =

i.e.,

Curl F =

Further

zz

∴

i ∂ ∂x xy

dy

2

k ∂ ∂z 0

j ∂ ∂y xy 2

i

− x k , on expanding the determinant

ds = dy dz i + dz dx j + dx dy k curl F ⋅ ds =

S

zz d

i

y 2 − x dx dy

It can be clearly seen from the figure that – 1 ≤ x ≤ 1 and – 1 ≤ y ≤ 1

zz

Now,

z zd z LMN z LMNFGH 1

curl F ⋅ ds =

1

i

y 2 − x dy dx

x =–1 y=–1

S

1

=

x =–1 1

=

x =–1

y3 − xy 3

OP Q

1

dx y = –1

IJ b gOP K Q

1 1 + − x 1 + 1 dx = 3 3

LM 2 x − x OP N3 Q 2 b1 + 1g − b1 − 1g = 3 1

=

2

x = −1

Thus,

z C

=

xy dx + x 2 dy =

4 3

4 3

z 1

x =–1

FG 2 − 2 xIJ dx H3 K

210


OBJECTIVE QUESTIONS

bg

1. If F t has a constant magnitude then (a)

bg

d F t =0 dt

bg

(c) F t ×

b g=0

dF t

bg

F t ⋅

(d )

F t –

bg

bg

dF t =0 dt

(b)

b g =0

dF t

dt dt 2. A unit vector normal to the surface xy3z2 = 4 at the point (– 1, – 1, 2) is

FH IK FH i + 3 j + k IK

(a) –

1 i + 3j − k 11

(b)

(c) –

1 11

(d )

1 11 1 11

FH i + 3 j − k IK FH i − 3 j + k IK

Ans. b

Ans. a

3. The greatest rate of increase of u = x2 + yz2 at the point (1, – 1, 3) is (a)

79

(b) 2 79

(d ) 4 7 Ans. c 89 4. The vector grad φ at the point (1, 1, 2) where φ is the level surface xy2 z2 = 4 is along (c)

(a) normal to the surface at (1, 1, 2)

(b) tangent to the surface at (1, 1, 2)

(c) Z-axis

(d ) i + j + 2 k

5. Directional derivative is maximum along (a) tangent to the surface (c) any unit vector

Ans. a

(b) normal to the surface

Ans. b

(d ) coordinate ones

6. If for a vector function F , div F = 0 then F is called (a) irrotational

(b) conservative

(c) solenoidal

(d ) rotational

Ans. c

7. For a vector function F , there exists a scalar potential only when

FH

IK

(a) div F = 0

(b) grad div F = 0

(c) curl F = 0

(d ) F curl F = 0

Ans. c

FH

8. If a is a constant vector and r = x i + y j + z k , then ∆ × a × r (a) 0

(b) a

(c) 2 a

(d ) – 2 a

IK

is equal to

Ans. c


211

9. Which of the following is true

FH IK (c) div FH A ⋅ BIK = div A . div B

(a) curl A ⋅ B = curl A + curl B

(b) div curl A = ∇ A (d ) div curl A = 0

Ans. d

10. Using the following integral, work done by a force F can be calculated: (a) Line integral (b) Surface integral (c) Volume integral

(d ) None of these

11. If F is the velocity of a fluid particle then

z

Ans. a

F ⋅ dr represents

C

(a) work done

(b) circulation

(c) flux

(d ) conservative field

Ans. b

12. The well-known equations of poisson and Laplace hold good for every (a) rotational field (b) solenoidal field (c) irrotational field

(d ) compressible field

Ans. c

13. If the vector functions F and G are irrotational, then F × G is (a) irrotational

(b) solenoidal

(c) both irrotational and solenoidal

(d ) none of these

Ans. b

14. The gradient of a differentiable scalas field is (a) irrotational (b) solenoidal (c) both irrotational and solenoidal

(d ) none of these

Ans. a

15. Gauss Divergence theorem is a relation between (a) a line integral and a surface integral (b) a surface integral and a volume integral (c) a line integral and a volume integral

Ans. b

(d ) two volume integrals 16. Green’s theorem in the plane is applicable to (a) xy-plane (b) yz-plane (c) zx-plane

(d ) all of these

Ans. d

17. If all the surfaces are closed in a region containing volume V then the following theorem is applicable (a) Stoke’s theorem (b) Green’s theorem (c) Gauss divergence theorem

(d ) only (a) and (b)

[Ans. c]

212

18. The value of

zb


g

b

g

2 x − y dx + x + 3 y dy where C : x2 + 4y2 = 4 is

C

(a) 0

(b) 4

(c) 2π

(d ) 4π

Ans. d

19. The magnitude of the vector drawn in a direction perpendicular to the surface x2 + 2y2 + z2 = 7 at the point (1, – 1, 2) is (a)

2 3

(b)

(c) 3

3 2

Ans. d

(d ) 6

20. A unit normal to the surface z = 2xy at the point (2, 1, 4) is (a) 2i + 4j – k (b) 2j + 4j + k

b

g

b

g

1 1 2i + 4 j − k 4i + 2 j − k (d ) Ans. c 21 21 21. The maximum value of the directional derivative φ = x2 – 2y2 + 4z2 at the point (1, 1, – 1) is (c)

7 3

(a)

(b) 84

7 3

(c) 6

(d ) 3

3 7

Ans. c

FG 1 , 1 , 0IJ H 2 2 K 1 F – i + j IK 2H

22. The unit normal to the surface x2 + y2 + z2 = 1 at the point (a)

FH FHi

IK + j IK

1 i −j 2

(b)

is

1 (d ) None of these Ans. c 2 23. The unit vector tangent to the curve x = t, y = t2, z = t3 at the point (– 1, 1, – 1) is (c)

b

g

1 i + 2 j + 3k 14 1 i+ j+k (c) 3 24. The value of curl (grad f ), where f = 2x2 (a) 4x – 6y + 8z (a)

b

(c) 0

g

(b)

b

1 i – 2 j + 3k 14

b

g

1 i– j+k 3 – 3y2 + 4z2 is (b) 4xi – 6yj + 8zk (d )

(d ) 3

g Ans. a

Ans. c

25. The curl of the gradient of a scalar function 4 is (a) 1 (b) ∆2 4 (c) ∇ 4

(d ) 0

Ans. d

213


26. A force field F is said to be conservative if (a) curl F = 0

(b) grad F = 0

(c) div F = 0

(d ) curl grad F

27. The value of the line integral (a) – 1

z

b

FH

g

IK

=0

Ans. d

grad x + y − z dr from (0, 1, – 1) to (1, 2, 0) is (b) 3

(c) 0

Ans. b

(d ) No obtainable

28. A necessary and sufficient condition that line integral

z

A ⋅ dr = 0 for every closed curve C

C

is that (a) div A = 0

(b) curl A = 0

(c) div A ≠ 0

(d ) curl A ≠ 0

29. The value of the line integral – 1 ≤ x ≤ 1, – 1 ≤ y ≤ 1 is (a) 0

zd

i

y 2 dx + x 2 dy where C is the boundary of the square

C

(c) 4 30. The value of the surface integral

Ans. b

(b) 2 (x + y)

zz b

(d )

4 3

Ans. a

g

yz dy dz + zx dzdx + xy dx dy where S is the surface of

S

the sphere x2 + y2 + z2 = 1 is (a)

4π 3

(b) 0

(c) 4π (d ) 12π Ans. b 31. Let S be a closed orientable surface enclosing a unit volume. Then the magnitude of the surface integral

z

r ⋅ n ds, where r = x i + y j + z k and n is the unit normal to the surface

S

S, equals. (a) 1

(b) 2

(c) 3

(d ) 4

32. If f = ax i + by j + cz k , a, b, c constants then sphere, is (a) 0 (c)

(b)

b

4 π a +b +c 3

g

2

zz

Ans. c f ⋅ dS where S is the surface of a unit

b

4 π a+b+c 3

g

(d ) None of these

Ans. b GGG

UNIT

8

Differential Equations-I 5.1 INTRODUCTION We have studied methods of solving ordinary differential equations of first order and first degree, in chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders. Differential equations of second order arise very often in physical problems, especially in connection with mechanical vibrations and electric circuits.

5.2

LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS

A differential equation of the form

dny d n –1y dn–2y a a + + + ... + an y = X ...(1) 2 1 dx n dx n – 1 dx n – 2 where X is a function of x and a1, a2 ..., an are constants is called a linear differential equation of nth order with constant coefficients. Since the highest order of the derivative appearing in (1) is n, it is called a differential equation of nth order and it is called linear. Using the familiar notation of differential operators:

d d2 d3 dn 3 n , D2 = , ..., D D = = dx dx 2 dx 3 dx n Then (1) can be written in the form {Dn + a1 Dn – 1 + a2 Dn – 2 + ... an} y = X i.e., f (D) y = X where f (D) = Dn + a1 Dn – 1 + a2 Dn – 2 + ... an. Here f (D) is a polynomial of degree n in D If x = 0, the equation f (D) y = 0 is called a homogeneous equation. If x ≠ 0 then the Eqn. (2) is called a non-homogeneous equation. D =

214

...(2)

215

DIFFERENTIAL EQUATIONS–I

5.3

SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR DIFFERENTIAL EQUATION

We consider the homogeneous equation

d2y dy +p + qy = 0 dx dx 2 where p and q are constants (D2 + pD + q) y = 0 The Auxiliary equations (A.E. ) put D = m m2 + pm + q = 0 Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) quadratic in m, will have two roots in general. There are three cases. Case (i): Roots are real and distinct The roots are real and distinct, say m1 and m2 i.e., m1 ≠ m2 Hence, the general solution of eqn. (1) is y = C1 em1x + C2 em2x where C1 and C2 are arbitrary constant. Case (ii): Roots are equal The roots are equal i.e., m1 = m2 = m. Hence, the general solution of eqn. (1) is y = (C1 + C2 x) emx where C1 and C2 are arbitrary constant. Case (iii): Roots are complex The Roots are complex, say α ± iβ Hence, the general solution is y = eαx (C1 cos β x + C2 sin β x) where C1 and C2 are arbitrary constants. Note. Complementary Function (C.F.) which itself is the general solution of the D.E.

WORKED OUT EXAMPLES d2y dy –5 + 6y = 0. 2 dx dx Solution. Given equation is (D2 – 5D + 6) y = 0 A.E. is m2 – 5m + 6 = 0 i.e., (m – 2) (m – 3) = 0 i.e., m = 2, 3 ∴ m 1 = 2, m2 = 3 ∴ The roots are real and distinct. 1. Solve

...(1)

...(2) ...(3) being

216


∴ The general solution of the equation is y = C1 e2x + C2 e3x.

d3y d2y dy – 2 –4 + 4y = 0. 3 dx dx dx Solution. Given equation is (D3 – D2 – 4D + 4) y = 0

2. Solve

A.E. is m3 – m2 – 4m + 4 = 0 m2 (m – 1) – 4 (m – 1) = 0 (m – 1) (m2 – 4) = 0 m = 1, m = ± 2 m 1 = 1, m2 = 2, m3 = – 2 ∴ The general solution of the given equation is y = C1 ex + C2 e2x + C3 e–2x.

d 2 y dy – – 6y = 0. dx 2 dx Solution. The D.E. can be written as (D2 – D – 6) y = 0 3. Solve

A.E. is ∴ ∴

m2 – m – 6 = 0 (m – 3) (m + 2) = 0 m = 3, – 2

∴ The general solution is y = C1 e3x + C2 e–2x.

d2y dy +8 + 16y = 0. 2 dx dx Solution. The D.E. can be written as 4. Solve

(D2 + 8D + 16) y = 0 A.E. is ∴

m2 + 8m + 16 = 0 (m + 4)2 = 0 (m + 4) (m + 4) = 0 m = – 4, – 4

∴ The general solution is y = (C1 + C2 x) e– 4x.

d2y + w 2 y = 0. 2 dx Solution. Equation can be written as 5. Solve

(D2 + w2) y = 0 A.E. is

m2 + w2 = 0

217


m 2 = – w2 = w2i2 (i2 = – 1) m = ± wi This is the form α ± iβ where α = 0, β = w. ∴ The general solution is y = e0t (C1 cos wt + C2 sin wt) ∴ y = C1 cos wt + C2 sin wt.

d2y dy +4 + 13y = 0. 2 dx dx Solution. The equation can be written as

6. Solve

(D2 + 4D + 13) y = 0 A.E. is

m2 + 4m + 13 = 0 m =

– 4 ± 16 – 52

2 = – 2 ± 3i (of the form α ± iβ)

∴ The general solution is y = e–2x (C1 cos 3x + C2 sin 3x).

d 2 y dy – = 0 given y′(0) = 0, y(0) = 1. dx 2 dx Solution. Equation is (D2 – D) y = 0

7. Solve

A.E. is m2 – m = 0 m (m – 1) = 0 ⇒

m = 0, 1

∴ The general solution is y = C1 e0x + C2 ex i.e., when

y = C1 + C2 ex x = 0, y = 0 (Given) y (0) = 0

⇒

0 = C1 + C2 y′(x) = C2e

...(1)

x

Given, when x = 0, y′ = 1 y (0) = 1 ⇒

1 = C2 e0

⇒

C2 = 1

From (1) and (2) ⇒

C 1 = –1.

∴ The general solution is y = ex – 1.

...(2)

218


8. Solve (4D4 – 4D3 – 23D2 + 12D + 36) y = 0. Solution. A.E. is 4m4 – 4m3 – 23m2 + 12m + 36 = 0 If

m = 2, 64 – 32 – 92 + 24 + 36 = 0

⇒ m = 2 is a root of inspection. By synthetic division,

4

–4

– 23

12

36

8

8

– 30

– 36

4

– 15

– 18

0

2 4 i.e., 4m3 + 4m2 – 15m – 18 = 0 If

m = 2 32 + 16 – 30 – 18 = 0

Again m = 2 is a root. By synthetic division 4

4

–15

–18

8

24

18

12

9

0

2 4

4m2 + 12m + 9 = 0 (2m + 3)2 = 0 m =

–3 –3 , 2 2

–3 –3 , . 2 2 + (C3 + C4 x) e–3x/2.

∴ The roots of the A.E. are 2, 2, Thus, y = (C1 + C2 x) e2x

9. Solve (D5 – D4 – D + 1) y = 0. Solution. A.E. is m5 – m4 – m + 1 = 0 m4(m – 1) – 1 (m – 1) = 0

i.e.,

(m – 1) (m4 – 1) = 0 (m – 1) (m2 – 1) (m2 + 1) = 0 (m – 1) (m – 1) (m + 1) (m2 + 1) = 0 ∴ The roots of the A.E. are 1, 1, –1, ± i. Thus y = (C1 + C2 x) ex + C3 e–x + (C4 cos x + C5 sin x). 10. Solve y″ + 4y′ + 4y = 0 given that y = 0, y′ = –1 at x = 1. Solution. We have (D2 + 4D + 4) y = 0 A.E. is

m2 + 4m + 4 = 0

219


∴

(m + 2)2 = 0

⇒

m = – 2, – 2

Hence,

y = (C1 + C2 x) e– 2x y′ = (C1 + C2 x) (– 2 e

Consider

...(1) – 2x

– 2x

) + C2 e

...(2)

y = 0 at x = 1

Hence, Eqn. (1) becomes 0 = (C1 + C2) e– 2 i.e.,

0 = (C1 + C2)

⇒

FG 1 IJ He K 2

C1 + C2 = 0

Also

y′ = 1 at x = 1

Hence, Eqn. (2) becomes – 1 = (C1 + C2) (– 2 e– 2) + C2 e– 2 C1 + C2 = 0

But

– 1 = C2 e–2

i.e.,

C2 = – e+ 2

or ∴

C 1 = – C2 = e2

Substituting these values in Eqn. (1), we get y = e2 (1 – x) e– 2x = (1 – x) e2 (1 – x).

EXERCISE 5.1 Solve the following differential equations: 1.

d2y dy –2 + 3 y = 0. 2 dx dx

2. 6y″ – y′ – y = 0.

2

3. (2D – D – 6) y = 0. 4. (D2 + 4D + 4) y = 0. 5. 9y″ – 6y′ + y = 0. 6. y″ + 9y = 0. 7. (D2 – 2D + 2) y = 0.

Ans. y = C1 e 3 x + C2 e – x

LMAns. y = C e + C e MN LMAns. y = C e + C e MN Ans. y = bC + C x g e LMAns. y = bC + C xg e MN 1

1 x 2

2

1 − x 3

−

2x

1

2

3 x 2

–2 x

1

1

2

2

1 x 3

OP PQ OP PQ OP PQ

Ans. y = C1 cos 3x + C2 sin 3x

b

Ans. y = e x C1 cos x + C2 sin x

g

220


LMAns. y = e F C cos GH MN

I OP JK PQ Ans. y = e bC cos x + C sin x g Ans. y = C e + e eC cos 3 x + C sin 3 x j −

2

8. (D + D + 1) y = 0.

1 x 2

1

3 3 x + C2 sin x 2 2

−2 x

9. y″ + 4y′ + 5y = 0.

1

2x

10. (D3 – 8) y = 0.

2

–x

1

2

2

Ans. y = C1 e – x + C2 e –2 x + C3 e –3 x

11. (D3 + 6D2 + 11D + 6) y = 0.

b

g

Ans. y = C1 + C2 x e x + C3 e 2 x

12. (D3 – 4D2 + 5D – 2) y = 0.

d

i

Ans. y = C1 + C2 x + C3 x 2 e –2 x

13. (D3 + 6D2 + 12D + 8) y = 0. 14. (D4 – 2D3 + 5D2 – 8D + 4) y = 0.

b

g

Ans. y = C1 + C2 x e x + C3 cos 2 x + C4 sin 2 x

b

g

b

g

x 15. (D4 – 4D3 + 8D2 – 8D + 4) y = 0. Ans. y = e C1 + C2 x cos x + C3 + C4 x sin x

16. (D4 – D3 – 9D2 – 11D – 4) y = 0.

d

i

Ans. y = C1 + C2 x + C3 x 2 e – x + C4 e 4 x

5.4 INVERSE DIFFERENTIAL OPERATOR AND PARTICULAR INTEGRAL Consider a differential equation f (D) y = x Define

1 f D

b g

...(1)

such that

b gRS| f b1Dg UV| x T W

f D

= x

...(2)

Here f (D) is called the inverse differential operator. Hence from Eqn. (1), we obtain y =

1 x f D

b g

...(3)

Since this satisfies the Eqn. (1) hence the particular integral of Eqn. (1) is given by Eqn. (3) Thus, particular Integral (P.I.) =

1 x f D

The inverse differential operator

1 f D

i.e.,

b g lax

1 f D

1

+ bx 2

q

= a

b g

b g

is linear.

1 1 x1 + b x2 f D f D

b g

b g

where a, b are constants and x1 and x2 are some functions of x.

221


5.5 SPECIAL FORMS OF X e ax f D

b g

Type 1: P.I. of the form

We have the equation f (D) y = eax Let We have ∴

f (D) = D2 + a1 D + a2 D (eax) = a eax, D2 (eax) = a2 eax and so on. f (D) eax = (D2 + a1 D + a2) eax = a2 eax + a1 . aeax + a2 eax = (a2 + a1 . a + a2) eax = f (a) eax

Thus f (b) eax = f (a) eax Operating with

1 on both sides f ( D)

b g f b1D g . e

eax = f a .

We get, or

P.I. =

ax

1 e ax e ax = f D f D

b g

b g

In particular if f (D) = D – a, then using the general formula. We get,

i.e., ∴

1 e ax = D–a e ax f D

b g

=

b

e ax e ax 1 . = D–a φ D D–a φ a

g b g

bg

z

1 1 e ax 1 . d x = . x e ax φ a φ a

bg

f′(a) = 0 + φ (a)

bg

f′(a) = φ (a)

or

Thus, Eqn. (1) becomes

e ax f D

b g

= x.

where

f (a) = 0

and

f′ (a) ≠ 0

e ax f′ D

b g

This result can be extended further also if f′(a) = 0,

Type 2: P.I. of the form We have

e ax e ax 2 = x . f D f ′′ a

b g

sin ax cos ax , f D f D

b g

b g

D (sin ax) = a cos ax

bg

and so on.

…(1)

222


D2 (sin ax) = – a2 sin ax D3 (sin ax) = – a3 cos ax D4 (sin ax) = a4 sin ax = (– a2)2 sin ax and so on. Therefore, if f (D2) is a rational integral function of D2 then f (D2) sin ax = f (– a2) sin ax. Hence

{ f d D i sin ax} = f d D i f d – a i sin ax f dD i 1

1

2

1

d i

sin ax = f (– a2)

i.e.,

1

d i

i.e.,

f D2

sin ax =

f D2

sin ax

sin ax

d i

f – a2

f (– a2) ≠ 0

Provided

Similarly, we can prove that 1 cos ax = f D2

d i

if

1

d i

In general,

f D

if

1

2

b

cos ax =

g

=

g

=

cos ax + b

d i

f D

d

f – a2

b

i

cos ax

d i

f – a2

f (– a2) ≠ 0

1

2

cos ax

f (– a ) ≠ 0

d i 2

...(1)

2

sin ax + b

f D and

2

2

2

1

d i

f – a2 1

d i

f – a2

b

g

b

g

sin ax + b

cos ax + b

...(2)

These formula can be easily remembered as follows.

x 1 sin ax = 2 2 D +a 2

1 x cos ax = 2 2 D +a 2

bg b g

z z

sin ax dx =

–x cos ax 2a

cos ax ax =

x sin ax. 2a

φ x where φ (x) is a polynomial in x, we seeking the polynomial f D Eqn. as the particular solution of f (D)y = φ (x) Type 3: P.I. of the form

where

φ (x) = a0 xn + a1 xn – 1 + ... an – 1 x + an

Hence P.I. is found by divisor. By writing φ (x) in descending powers of x and f (D) in ascending powers of D. The division get completed without any remainder. The quotient so obtained in the process of division will be particular integral.

223


Type 4: P.I. of the form

e ax V where V is a function of x. f D

b g

We shall prove that

1 1 e ax V = e ax V. f D f D+a

Consider

ax

b g

D (e

b

DV + Va e

ax

(D + a) V

ax

D2 V + a eax DV + a2 eax V + a eax DV

V) = e

= e 2

and

ax

D (e

g

ax

V) = e

ax

= eax (D2 V + 2a DV + a2 V ) = eax (D + a)2 V Similarly,

D3 (eax V ) = eax (D + a)3 V and so on.

∴

f (D) eax V = eax f (D + a) V

Let

f (D + a) V = U, so that V =

…(1)

1 U f D+a

b

g

Hence (1) reduces to 1 U = eax U f (D) eax f D+a

b

g

Operating both sides by e ax

i.e.,

1 f D

b g

1 U = f D+a

b g 1 e U f b Dg ax

we get, 1 e ax U f D

b g

ax = e

1 U f D+a

b

g

Replacing U by V, we get the required result. Type 5: P.I. of the form

xV xn V , where V is a function of x. f D f D

b g b g

By Leibniz’s theorem, we have Dn (x V ) = x Dn V + n·1 Dn – 1·V = x Dn V + ∴

RS d D UV V T dD W n

f (D) x V = x f (D) V + f ′ (D) V

Eqn. (1) reduces to xV f D

b g

=

LM x – f ′b Dg OP V MN f bDg PQ f b Dg

...(1)

...(2)

This is formula for finding the particular integral of the functions of the xV. By repeated application of this formula, we can find P.I. as x2 V, x3 V ...... .

224


WORKED OUT EXAMPLES Type 1 d2y dy –5 + 6y = e 5x. 1. Solve dx dx 2 Solution. We have (D2 – 5D + 6) y = e5x A.E. is m2 – 5m + 6 = 0 i.e., (m – 2) (m – 3) = 0 ⇒ m = 2, 3 Hence the complementary function is ∴ C.F. = C1 e2x + C2 e3x Particular Integral (P.I.) is P.I. = =

1 e5x D – 5D + 6

(D → 5)

2

1 e5 x 5x e = · 6 52 – 5 × 5 + 6

∴ The general solution is given by y = C.F. + P.I.

e5x · 6

= C1 e2x + C2 e3x +

d2y dy − 3 + 2y = 10e 3x. 2. Solve 2 dx dx Solution. We have (D2 – 3D + 2) y = 10 e3x 2 A.E. is m – 3m + 2 = 0 i.e., (m – 2) (m – 1) = 0 m = 2, 1 C.F. = C1 e2x + C2 ex P.I. = =

P.I. =

1 10e 3 x D – 3D + 2 2

1 10e 3 x 3 – 3×3+ 2 2

10 e 3 x 2

∴ The general solution is y = C.F. + P.I. = C1 e2x + C2 ex +

10 e 3 x . 2

(D → 3)

225


d2y dy –4 + 4y = e 2x. 2 dx dx Solution. Given equation is 3. Solve

(D2 – 4D + 4)y = e2x A.E. is i.e.,

m2 – 4m + 4 = 0 (m – 2) (m – 2) = 0 m = 2, 2 C.F. = (C1 + C2) e2x P.I. =

1 e2x D2 – 4 D + 4

(D = 2)

=

1 e2 x 2 –4 2 +4

(Dr = 0)

bg

2

Differentiate the denominator and multiply ‘x’ = x.

1 e2 x 2D – 4

(D → 2)

= x.

1 e2x 2 2 –4

(Dr = 0)

bg

Again differentiate denominator and multiply ‘x’ 2 = x

P.I. =

1 2x e 2

x 2 e2 x 2

y = C.F. + P.I. = (C1 + C2 x) e2x +

x 2 e2 x · 2

d4x + 4x = cosh t . dt 4 Solution. We have 4. Solve

(D4 + 4) x = cosh t A.E. is i.e.,

m4 + 4 = 0 (m2 + 2)2 – 4m2 = 0

or [(m2 + 2) – 2m] [m2 + 2 + 2m] = 0 m2 – 2m + 2 = 0; m2 + 2m + 2 = 0 ∴

m = m =

2± 4–8 2

; m=

–2± 4–8

2 ± 2i – 2 ± 2i ;m= 2 2

2

226


m = 1 ± i, m = – 1 ± i t

C.F. = e (C1 cos t + C2 sin t) + e–t (C3 cos t + C4 sin t) P.I. =

where cos h t =

et + e–t 2 =

cosh t D4 + 4

LM OP N Q OP 1L e O 1L e + M M P 2 N D + 4Q 2 N D + 4Q 1 et + e –t 2 D4 + 4 t

=

–t

4

LM N

4

D→1

D → –1

1 et e–t + = 2 5 5

OP = 1 . 1 de Q 52

t

+ e–t

i = 15 cos h t

y = C.F. + P.I. = et (C1 cos t + C2 sin t) + e–t (C3 cos t + C4 sin t) +

d

i

2 d3y – y = ex + 1 . 3 dx Solution. Given equation can be written as

5. Solve

(D3 – 1) y = e2x + 2ex + 1 A.E. is

m3 – 1 = 0

i.e., (m – 1) (m2 + m + 1) = 0 Hence,

m = 1 D =

–1± 3 i 2

x C.F. = C1 e + e

P.I. = =

−

1 x 2

FC GH

2

cos

3 3 x + C3 sin x 2 2

1 (e2x + 2ex + 1) D –1 3

1 1 1 2 ex + 3 e2x + 3 e0 D –1 D –1 D –1 3

= P.I.1 + P.I.2 + P.I.3 P.I.1 =

e2 x D3 – 1

D→ 2

I JK

1 cos ht. 5

227


e2 x e2 x = = 3 7 2 –1 P.I.2 = =

1 2 ex D –1

(D → 1)

1 2 ex 1 –1

(Dr = 0)

3

3

Differentiate the Dr and multiply x =

2 xe x 3 D2

(D → 1)

2 ex 3 1 e0 P.I.3 = D3 – 1 P.I.2 = x ⋅

(D → 1)

= –1 P.I. =

e2 x 2x e x –1 + 7 3

∴ The general solution is y = C.F. + P.I. x = C1 e + e

−

1 x 2

LMC MN

2

cos

OP PQ

3 3 e2 x 2 + x ex – 1 x + C3 sin x + 2 2 7 3

6. Solve (D3 + 2D2 – D – 2) y = 2 cos h x. Solution. Given equation is (D3 + 2D2 – D – 2) y = 2 cosh x A.E. is m3 + 2m2 – m – 2 = 0 m2 (m + 2) – 1 (m + 2) = 0 (m + 2) (m2 – 1) = 0 (m + 2) (m + 1) (m – 1) = 0 m = – 2, – 1, 1 C.F. = C1 e–2x + C2 e–x + C3 ex P.I. =

1 2 cos h x 3 D + 2 D2 – D – 2

F GH

I JK

F where cos h x = e GH

=

e x + e– x 2 2 D3 + 2 D2 – D – 2

=

ex e– x + D3 + 2 D2 – D – 2 D3 + 2 D2 – D – 2

= P.I.1 + P.I.2

x

+ e− x 2

I JK

228


ex P.I.1 = D 3+ 2 D 2 – D – 2 =

=

1

bg

2

1 + 2 1 –1– 2 3

(D → 1)

ex

(Dr = 0)

x ex 3 D2 + 4 D – 1

(D → 1)

x ex = 6 P.I.2 = =

1 e– x D + 2D2 – D – 2

(D → – 1)

3

1

e– x

b g b g b g 3

2

–1 + 2 –1 – –1 – 2

=

x e– x 3D + 4 D – 1

=

– x e– x 2

(Dr = 0) (D → – 1)

2

x e x x e− x – 6 2 y = C.F. + P.I.

P.I. = ∴

= C1 e–2x + C2 e–x + C3 ex +

x ex x e−x – ⋅ 6 2

Type 2 1. Solve (D2 + 9) y = cos 4x. Solution. Given equation is (D2 + 9) y = cos 4x A.E. is i.e.,

m2 + 9 = 0 m = ± 3i C.F. = C1 cos 3x + C2 sin 3x P.I. = =

1 cos 4 x D +9 2

(D2 → – 42 = – 16)

1 1 cos 4 x = – cos 4 x – 16 + 9 7

∴ The general solution is y = C.F. + P.I. = C1 cos 3x + C2 sin 3x –

1 cos 4x. 7

229


2. Solve (D2 + D + 1) y = sin 2x. Solution. The A.E. is m2 + m + 1 = 0 i.e.,

m =

– 1± 1– 4 2

Hence the C.F. is C.F. = e P.I. =

−

LMC cos MN

x 2

1

=

– 1± 3 i 2

3 3 x + C2 sin x 2 2

1 sin 2 x D2 + D + 1

OP PQ (D2 → – 22)

=

1 sin 2 x – 2 + D+1

=

1 sin 2 x D–3

2

Multiplying and dividing by (D + 3)

∴ y = C.F. + P.I. = e

−x 2

=

b D + 3g sin 2 x

=

b D + 3g sin 2 x

D2 – 9 2

LMC cos MN 1

–2 –9 3 x + C2 sin 2

3. Solve (D2 + 5D + 6) y = cos x + e–2x.

=

b g 3 O 1 x P – (2 cos 2x + 3 sin 2x). 2 PQ 3 –1 2 cos 2 x + 3 sin 2 x 13

Solution. The A.E. is m2 + 5m + 6 = 0 i.e.,

(m + 2) (m + 3) = 0 m = – 2, – 3 C.F. = C1 e–2x + C2 e–3x 1 ⋅ [cos x + e–2x] P.I. = 2 D + 5D + 6 =

cos x e –2 x + D 2 + 5D + 6 D 2 + 5D + 6

= P.I.1 + P.I.2 cos x P.I.1 = 2 D + 5D + 6 =

cos x cos x = 5 D+5 – 1 + 5D + 6 2

(D2 = – 12)

230


=

=

b g b gb g 1 b D – 1g cos x 1 5

cos x D – 1

5

D2 – 1

D +1 D –1

=

1 – sin x – cos x 5 – 12 – 1

=

– 1 sin x + cos x 5 –2

=

1 sin x + cos x 10

P.I.2 = =

b

g

e– 2x D 2 + 5D + 6

b – 2g

(D → – 2)

e –2 x 2

(Dr = 0)

+ 5× – 2 + 6

Differential and multiply ‘x’ =

x e –2 x 2D + 5

=

x e –2 x x e –2 x = = x e –2 x 2 –2 +5 1

P.I. =

(D → – 2)

b g

1 (sin x + cos x) + x e– 2x 10

∴ The general solution is y = C.F. + P.I. y = C1 e–2x + C2 e–3x + 4. Solve (D2 + 3D + 2) y = cos2 x. Solution. The A.E. is m2 + 3m + 2 = 0 i.e., ∴

(m + 1) (m + 2) = 0 m = – 1, – 2 C.F. = C1 e–x + C2 e–2x P.I. =

where

cos2 x =

1 . cos 2 x D + 3D + 2 2

1 + cos 2 x 2

1 (sin x + cos x) + x e–2x. 10

231


LM N

OP Q

=

1 + cos 2 x 1 2 D + 3D + 2

=

1 1 1 e0x + 2 cos 2 x 2 D 2 + 3D + 2 D + 3D + 2

=

1 P.I.1 + P.I.2 2

P.I.1 =

=

P.I.2 =

2

LM N

e0x D 2 + 3D + 2

(D → 0)

1 e0x = 2 2 cos 2 x D + 3D + 2

(D2 → – 22)

2

=

cos 2 x – 2 + 3D + 2

=

cos 2 x 3D + 2 × 3 D – 2 3D + 2

=

=

2

b3D + 2g cos 2 x

(D2 → – 22)

9 D2 – 4

bg 9 b – 2g – 4

– 3 sin 2 x ⋅ 2 + 2 cos 2 x 2

=

– 6 sin 2 x + 2 cos 2 x – 40

=

6 sin 2 x – 2 cos 2 x 3 sin 2 x – cos 2 x = 40 20

P.I. =

=

OP Q

LM N

1 1 3 sin 2 x – cos 2 x + 2 2 20

OP Q

1 3 sin 2 x – cos 2 x + 4 40

y = C.F. + P.I. = C1 e–x + C2 e–2x +

1 1 + (3 sin 2x – cos 2x). 4 40

232


5. Solve (D3 + D2 – D – 1) y = cos 2x. Solution. The A.E. is m3 + m2 – m – 1 = 0 i.e., m2 (m + 1) – 1 (m + 1) = 0 (m + 1) (m2 – 1) = 0 m = – 1, m2 = 1 m = – 1, m = ± 1 ∴

m = – 1, – 1, 1 C.F. = C1 ex + (C2 + C3 x) e–x P.I. = =

=

1 3

D + D2 – D – 1 1

b D + 1g d D – 1i 2

(D2 → – 22)

cos 2 x

cos 2 x

1

b D + 1g d – 2 – 1i cos 2 x 2

=

–1 1 cos 2 x 5 D +1

=

– 1 cos 2 x D – 1 × 5 D +1 D –1

=

– 1 D – 1 cos 2 x 5 D2 – 1

=

– 1 – 2 sin 2 x – cos 2 x 5 – 22 – 1

=

–1 (2 sin 2x + cos 2x) 25

b

g

LM N

(D2 → – 22)

OP Q

∴ The general solution is y = C.F. + P.I. = C1 ex + (C2 + C3 x) e–x – 6. Solve (D3 + 1) y = sin 3x – cos2 (1/2) x. Solution. The A.E. is m3 + 1 = 0 i.e., (m + 1) (m2 – m + 1) = 0 m + 1 = 0, m2 – m + 1 = 0 m = –1

1 (2 sin 2x + cos 2x). 25

233


m = m =

1± 1– 4 2 1± 3 i 2

LM MN

C.F. = C1 e–x + ex/2 C2 cos P.I. = = =

3 3 x + C3 sin x 2 2

FG IJ H K 1 F 1 + cos x I sin 3x – G JK H 2 D +1

OP PQ

1 1 sin 3x – cos2 x 2 D +1 3

3

1 1 1 1 1 sin 3x – . e0x – cos x 3 3 2 D +1 2 D +1 D +1 3

= P.I.1 – P.I.2 – P.I.3 P.I.1 =

1 sin 3x D3 + 1

=

1 sin 3x D ⋅D +1

=

1 sin 3x – 9D + 1

=

1 + 9D 1 sin 3x × 1 – 9D 1 + 9D

= =

2

b1 + 9 Dg sin 3x 1 – 81 D 2

(D2 → – 32)

(D2 → – 32)

sin 3x + 27 cos 3x

d i

1 – 81 – 32

P.I.1 =

1 (sin 3x + 27 cos 3x) 730

P.I.2 =

1 e0x 2 D3 + 1

(D → 0)

1 2 1 1 . cos x P.I.3 = 3 2 D +1 =

=

1 1 cos x 2 2 D ⋅D +1

(D2 → – 12)

234


=

1 1 cos x 2 – D +1

=

1+ D 1 1 cos x × 2 1– D 1+ D

=

1 cos x – sin x 2 1 – D2

LM N

OP Q

(D2 → – 12)

1 cos x – sin x 2 2 1 P.I.3 = (cos x – sin x) 4 1 1 1 P.I. = (sin 3x + 27 cos 3x) − – (cos x – sin x). 730 2 4 7. Solve: (D2 + 4) y = sin2 x. Solution. The A.E. is m2 + 4 = 0 i.e., m2 = – 4 m = ± 2i C.F. = C1 cos 2x + C2 sin 2x

=

P.I. = where sin2 x =

1 sin 2 x D +4 2

1 – cos 2 x 2

LM N

OP Q

=

1 1 – cos 2 x 2 D +4

=

1 1 (1 – cos 2x) 2 2 D +4

2

LM N

cos 2 x 1 e0x – 2 = 2 2 D +4 D +4 = P.I.1 = = P.I.2 = =

OP Q

1 P.I.1 – P.I.2 2

e0 x D2 + 4

(D → 0)

1 4 1 cos 2 x D +4

(D2 → – 22)

1 cos 2 x – 22 + 4

(Dr = 0)

2

235


Differentiate the Dr and multiplying by ‘x’ =

1 D x cos 2 x × 2D D

=

– x sin 2 x ⋅ 2 2 D2

=

= P.I. = =

(D2 → – 22)

– 2 x sin 2 x

d i

2 – 22

2 x sin 2 x x sin 2 x = 8 4

LM N

1 1 x sin 2 x – 2 4 4

OP Q

1 x sin 2 x – 8 8

∴ The general solution is y = C.F. + P.I. y = C1 cos 2x + C2 sin 2x +

1 x – sin 2x. 8 8

8. Solve y″ + 9y = cos 2x · cos x. Solution. We have (D2 + 9) y = cos 2x cos x A.E. is

m2 + 9 = 0 m2 = – 32 m = ± 3i C.F. = C1 cos 3x + C2 sin 3x P.I. =

cos 2 x ⋅ cos x D2 + 9

where cos 2x · cos x = 1/2 (cos x + cos 3x)

LM N

OP Q

=

1 cos x + cos 3x 2 D2 + 9

=

1 cos x 1 cos 3x + 2 D2 + 9 2 D2 + 9

= P.I.1 + P.I.2 P.I.1 =

1 cos x 2 D2 + 9

(D2 → –12)

236


1 cos x 1 = cos x 2 8 16

=

1 cos 3x 2 D2 + 9 1 cos 3x = 2 – 32 + 9

(D2 → – 32)

P.I.2 =

(Dr = 0)

Differentiate the Dr and multiplying by ‘x’ =

1 x cos 3 x D × 2 2D D

=

1 x – sin 3x ⋅ 3 2 2 D2

b

d i

4 –32

– 3x sin 3x – 36

=

x sin 3x 12

P.I.2 = P.I. =

(D2 → – 32)

– 3x sin 3x

=

∴

g

1 1 cos x + x sin 3x 16 12

∴ The general solution is y = C.F. + P.I. = C1 cos 3x + C2 sin 3x +

1 1 cos x + x sin 3x. 16 12

EXERCISE 5.2 Solve the following equations: 1. (D2 + 1) y = sin 2x.

LM Ans. y = C cos x + C sin x – 1 sin 2 x OP 3 N Q LM Ans. y = C e + C e – 1 sin 2 x − 1 cos 3x OP 8 13 N Q LM Ans. y = C cos 3x + C sin 3x – x cos 3x OP 6 N Q LM Ans. y = C cos 4 x + C sin 4 x + x sin 4 x OP 8 N Q LM Ans. y = C cos x + C sin x + 1 b4 x sin x + cos 3xgOP 16 N Q 1

2x

2. (D2 – 4) y = sin 2x + cos 3x. 3. (D2 + 9) y = sin 3x. 4. (D2 + 16) y = cos 4x. 5. (D2 + 1) y = sin x sin 2x.

–2 x

1

1

2

2

1

2

1

2

2

237


LM Ans. y = C e N

gOPQ LM Ans. y = C e + C e + xe – 1 bcos x − sin xgOP + 5D + 6) y = sin x + e . 10 N Q LM Ans. y = C e + C e + 1 + 1 b3 sin 2 x − cos 2 x gOP + 3D + 2) y = 4 cos x. 10 N Q LMAns. y = C e + C e – 1 FG 1 cos 3x + 1 cos xIJ OP – 4) y = cos x cos 2x. KQ 2 H 13 5 N LM Ans. y = bC + C xge + 1 bcos 2 x − sin 2 x g OP – 4D + 4) y = sin 2x + cos 2x. 8 N Q

6. (D2 – 2D – 8) y = 4 cos 2x. 7. (D2

8. (D2 9. (D2 10. (D2

+ C2 e –2 x –

4x

1

–3x

–2x

–2 x

1

–2 x

2

–x

2

b

1 3 cos 2 x + sin 2 x 10

–2 x

1

2

2x

–2 x

1

2

2x

1

2

b

g

Ans. y = e –4 x C1 cos 3x + C2 sin 3x + 2 cos x

11. (D2 + 8D + 25) y = 48 cos x – 16 sin x. 12. (D3 + D2 + D + 1) y = sin 3x.

LM Ans. y = C e + C cos x + C sin x + 1 b3cos 3x − sin 3xgOP 80 N Q LM Ans. y = bC + C xg e + bC + C xg e + 1 cos xOP 13. (D – 2D + 1) y = cos x. 4 N Q 1 L O cos 2 x + sin 2 x gP + e e C cos 3 x + C sin 3x j + b 14. (D + 8) y = sin 2x. M Ans. y = C e 16 N Q 1 L O x cos 2 x P 15. (D – 16) y = sin x cos x. M Ans. y = C e + C e + C cos 2 x + C sin 2 x + 64 N Q LM Ans. y = dC + C x + C x ie – 1 – 1 b2 sin 2 x − 11cos 2 x g OP 16. (D – 1) y = cos x. 2 250 N Q LM Ans. y = bC + C x ge + C e + 1 b7 cos 2 x + sin 2 xgOP 17. (D – 3D + 2) y = sin 2x. 100 N Q –x

1

4

2

3

x

2

1

–2 x

3

–x

3

4

x

1

2

3

2x

4

–2 x

1

3

2

2

3

x

2

2

1

2

4

3

x

3

1

–2 x

2

3

18. (D2 + 1)(D2 + 4) y = cos 2x + sin x.

LM Ans. y = C cos x + C sin x + C cos 2 x + C sin 2 x – 1 x sin 2 x – 1 x cos x OP 12 6 N Q 1

19.

2

d 3y + y = 65 cos (2x + 1) + e–x. dx 3

LMAns. y = C e MN 1

x –x

R|S |T

+ e 2 C2 cos

3

U|V |W

4

b

g

b

g

3 3 1 x + C3 sin x + cos 2 x + 1 – 8 sin 2 x + 1 + xe – x 2 2 3

OP PQ

238


Type 3 1. Solve y″ + 3y′ + 2y = 12x2. Solution. We have (D2 + 3D + 2) y = 12x2 A.E. is i.e., ⇒

m2 + 3m + 2 = 0 (m + 1) (m + 2) = 0 m = –1, – 2 C.F. = C1e–x + C2e–2x P.I. =

12 x 2 D 2 + 3D + 2

We need to divide for obtaining the P.I. 6x2 – 18x + 21 2 + 3D + D2 12x2 12x2 + 36x + 12 – 36x – 12 – 36x – 54 42 42 0 Hence, P.I. = 6x2 – 18x + 21 ∴ The general solution is y = C.F. + P.I. y = C1e–x + C2e–2x + 6x2 – 18x + 21. 2. Solve

d2y dy +2 + y = 2x + x 2 . 2 dx dx

Solution. We have (D2 + 2D + 1) y = 2x + x2 A.E. is

m2 + 2m + 1 = 0

i.e.,

(m + 1)2 = 0

i.e.,

(m + 1) (m + 1) = 0

⇒

m = – 1, – 1 C.F. = (C1 + C2 x) e– x P.I. =

2x + x2 x2 + 2x = D2 + 2 D + 1 1 + 2 D + D2

Note: 3D(6x2) = 36x D2(6x2) = 12

239


x2 – 2x + 2 1 + 2D + D2

x2 + 2x x2 + 4x + 2 – 2x – 2 – 2x – 4 2 2 0

∴

P.I. = x2 – 2x + 2

∴

y = C.F. + P.I. = (C1 + C2 x) e–x + (x2 – 2x + 2).

d2y dy +5 + 6y = x 2 . 2 dx dx Solution. We have 3. Solve

(D2 + 5D + 6) y = x2 A.E. is

m2 + 5m + 6 = 0 (m + 2) (m + 3) = 0

i.e.,

m = – 2, – 3 C.F. = C1e–2x + C2e–3x P.I. =

x2 x2 = 6 + 5D + D 2 D 2 + 5D + 6

P.I. is found by division method

x 2 5x 19 – + 6 18 108 6 + 5D + D2

F x I = 5x GH 6 JK 3 Fx I 1 D G J = H 6K 3 2

x2

x2 +

5x 1 + 3 3

5x 1 – – 3 3 −5x 25 – 3 18

19 18 19 18

0

5D

2

2

5D

FG – 5x IJ H 18 K

=

– 25 18

240


∴

P.I. = =

∴

x 2 5x 19 – + 6 18 108 1 (18x2 – 30x + 19) 108

y = C.F. + P.I. = C1 e– 2x + C2 e– 3x +

1 (18x2 – 30x + 19). 108

4. Solve (D3 + 2D2 + D) y = x3. Solution. A.E. is m3 + 2m2 + m = 0 i.e.,

m (m2 + 2m + 1) = 0

i.e.,

m (m + 1)2 = 0

⇒

m = 0, –1, –1 C.F. = C1 + (C2 + C3 x) e– x

x3 x3 = P.I. = D3 + 2 D2 + D D + 2 D2 + D3 x4/4 – 2x3 + 9x2 – 24x D + 2D2 + D3

x3 x3 + 6x2 + 6x

Note:

2 D2

2

– 6x – 6x – 6x2 – 24x – 12

D3

18x + 12 – 24 0 4

∴

x 4

4

z

x 3 dx =

4

= 6x

z z zb

– 6x 2 dx = – 2x3

18x dx = 9 x 2

y = C.F. + P.I. = C1 + (C2 + C3 x) e

g

– 24 dx = – 24 x

– 2x3 + 9x2 – 24x

–x

x4 4

= 6x2

18x = D – 24 = D

– 24

P.I. =

Fx I GH 4 JK Fx I GH 4 JK

– 6x2 = D

18x + 36

∴

x3 = D

x4 + – 2x3 + 9x2 – 24x. 4

241


EXERCISE 5.3 Solve the following equations:

LM Ans. y = C e N

gOPQ LM Ans. y = C cos 2 x + C sin 2 x + 1 d2 x – 2 x – 1iOP + 4) y = x – x. 8 N Q LM Ans. y = bC + C xg e + 1 d2 x + 4 x + 3iOP – 4D + 4) y = x . 8 N Q LM Ans. y = C e + C e – 1 b6x + 1gOP + D – 6) y = x. 36 N Q Ans. y = bC + C x g e + x – 2 x + 2 + 2D + 1) y = x + 2x. LM Ans. y = e eC cos 2 x + C sin 2 x j + 1 d9 x + 12 x + 2iOP – 2D + 3) y = x . 27 N Q LM Ans. y = C e + C e + 1 d9 x + 24 x – 1iOP – 5D + 6) y = x + x – 2. 54 N Q LMAns. y = C e + e F C cos 3 x + C sin 3 xI + x – 6OP + 1) y = x . GH 2 2 JK MN PQ LMAns. y = C + C x + C e – x b x – 2gOP – D ) y = x – 3x + 1. 12 N Q + 8) y = x + 2x + 1. LM Ans. y = C e + e eC cos 3x + C sin 3x j + 1 d x – x + 1iOP 8 N Q

1. (D2 – 9) y = 2x – 1. 2. (D2 3. (D2 4. (D2 5. (D2 6. (D2 7. (D2

8. (D3

3x

1

b

1 2x – 1 9

+ C2 e – 3 x –

2

2

1

2

2x

2

1

2

2

2x

– 3x

1

2

–x

2

1

2

2

x

2

2

1

2

2x

2

3x

1

x

–x

3

1

2

2

2

3

2

3

3

3

9. (D

10. (D3

2

x

2

1

– 2x

x

4

2

Type 4 d2y dy – 3y = e x cos x. +2 2 dx dx Solution. We have (D2 + 2D – 3) y = ex cos x 1. Solve

i.e., i.e.,

3

4

1

A.E. is

2

m2 + 2m – 3 = 0 (m + 3) (m – 1) = 0 m = – 3, 1 C.F. = C1 e– 3x + C2 ex P.I. =

1 e x cos x D + 2D – 3 2

Taking ex outside the operator and changing D to D + 1

3

242

ENGINEERING MATHEMATICS—II x = e

b D + 1g

1 2

b

g

+ 2 D +1 – 3

x = e

1 cos x D + 4D

x = e

1 cos x – 1 + 4D

(D2 → – 12)

2

LM cos x × 4 D + 1OP N 4 D – 1 4 D + 1Q LM – 4 sin x + cos x OP N 16 D – 1 Q LM – 4 sin x + cos x OP N – 17 Q

x = e

x = e

(D2 → – 12)

2

x = e

cos x

ex (4 sin x – cos x) 17 y = C.F. + P.I. =

∴

y = C1 e

–3x

ex + C2 e + (4 sin x – cos x). 17 x

2. Solve (D3 + 1) y = 5ex x2. Solution. A.E. is m3 + 1 = 0 i.e., (m + 1) (m2 – m + 1) = 0 (m + 1) = 0, m2 – m + 1 = 0 m = –1 m =

1 ± 3i 2 x

F GH

–x C.F. = C1 e + e 2 C2 cos

P.I. =

3 3 x + C3 sin x 2 2

1 5e x x 2 D +1 3

Taking ex outside the operator and changing D to D + 1 x = e

x = e

1

b D + 1g

3

+1

⋅ 5x 2

5x 2 D 3 + 3D 2 + 3D + 2

I JK

243


=

LM N

5e x 2x2 2 2 + 3D + 3 D 2 + D 3

OP Q

(For a convenient division we have multiplied and divided by 2) x2 – 3x + 2 + 3D + 3D2 + D3

3 2

2x2 2x2 + 6x + 6 – 6x – 6 – 6x – 9 3 3

∴

P.I. =

Fx H

0

− 3x +

2

I K

3 5e x ⋅ 2 2

5e x (2x2 – 6x + 3) 4 y = C.F. + P.I. =

= C1 e 3. Solve (D2 – 4D + 3) y = Solution. A.E. is m2 – 4m + 3 i.e., (m – 1) (m – 3) m C.F.

–x

+

x 2 e

e2x sin 3x. = = = =

P.I. =

|RSC |T

2

cos

|UV |W

3 3 5e x x + C3 sin x + (2x2 – 6x + 3). 2 2 4

0 0 1, 3 C1 ex + C2 e3x 1 e 2 x sin 3x D – 4D + 3 2

Taking e2x outside the operator and changing D to D + 2 2x = e

2x = e

1

b D + 2 g – 4 b D + 2g + 3 LM 1 OP sin 3x N D – 1Q 2

⋅ sin 3x

2

1 2x e sin 3x 10 y = C.F. + P.I. = –

= C1 ex + C2 e3x –

1 2x e sin 3x. 10

(D2 → – 32)

244


d2y + 4y = 2e x sin 2 4. Solve 2 dx Solution. We have (D2 + 4) y = A.E. is m2 + 4 = i.e., m2 = m = C.F. = P.I. = where sin2 x =

x. 2ex sin2 x 0 –4 ± 2i C1 cos 2x + C2 sin 2x 1 2 e x sin 2 x D +4 2

1 – cos 2 x 2

=

1 ex (1 – cos 2x) D2 + 4

=

1 (ex – ex cos 2x) D +4 2

ex e x cos 2 x – D2 + 4 D2 + 4 = P.I.1 – P.I.2

=

P.I.1 = = P.I.2 =

1 ex D +4

(D → 1)

2

ex 5 e x cos 2 x D2 + 4

Taking ex outside the operator and changing D to D + 1 x = e

x = e

x = e

x = e

1

b D + 1g + 4 LM OP cos 2 x 1 N D + 2D + 1 + 4 Q LM 1 OP cos 2 x N D + 2 D + 5Q 2

cos 2 x

2

2

1 cos 2 x – 22 + 2D + 5

x = e .

2D – 1 1 cos 2 x × 2D + 1 2D – 1

(D2 → – 22)

245


LM – 4 sin 2 x – cos 2 x OP N 4D – 1 Q LM – 4 sin 2 x – cos 2 x OP – 17 N Q

x = e

(D2 → – 22)

2

x = e

ex (4 sin 2x + cos 2x) 17 y = C.F. + P.I. =

∴

y = C1 cos 2x + C2 sin 2x +

ex (4 sin 2x + cos 2x). 17

5. Solve (D2 – 4D + 3) y = 2x e3x. Solution. A.E. is m2 – 4m + 3 = 0 i.e.,

(m – 1) (m – 3) = 0

i.e.,

m = 1, 3 C.F. = C1 ex + C2 e3x P.I. =

1 2 x e3x D – 4D + 3 2

Taking e3x outside the operator and changing D → D + 3 3x = e .

3x = e

2x

b D + 3g – 4 b D + 3g + 3 LM 2 x OP N D + 2D Q 2

2

By division method x2/2 – x/2 2D + D2

2x 2x + 1 –1 –1

2x = 2D

Note:

0

Fx GH 2

2

3x P.I. = e

–

x 2

–1 = 2D

I JK

y = C.F. + P.I.

Fx GH 2

2

x

3x

= C1 e + C2 e

3x + e

–

I JK

x · 2

z

z

x dx =

x2 2

–1 x dx = − 2 2

246


Type 5 d2y + 4y = x sin x. dx 2 Solution. We have 1. Solve

(D2 + 4) y = x sin x m2 + 4 = 0

A.E. is

m2 = – 4 m = ± 2i C.F. = C1 cos 2x + C2 sin 2x P.I. = xV f D

b g

Let us use

=

x sin x = D2 + 4

=

1 x sin x D +4 2

LM x – f ′b Dg OP V MN f b Dg PQ f b Dg LM x – 2 D OP sin x N D + 4Q D + 4 x sin x 2 D b sin x g – D + 4 d D + 4i 2

2

2

2

=

x sin x 2 cos x – 3 32

=

x sin x 2 cos x – 3 9

P.I. =

2

(D2 → – 12)

(D2 → – 12)

1 (3x sin x – 2 cos x) 9

y = C.F. + P.I. = C1 cos 2x + C2 sin 2x + 2. Solve (D2 + 2D + 1) y = x cos x. Solution. A.E. is m2 + 2m + 1 = 0 i.e.,

(m + 1)2 = 0 m = – 1, – 1 C.F. = (C1 + C2 x) e–x P.I. =

x cos x . D + 2D + 1 2

1 (3x sin x – 2 cos x). 9

247


Let us we have

xV f D

b g

=

=

=

LM x – f ′b Dg OP . V MN f b Dg PQ f b Dg LM x – 2 D + 2 OP . cos x N D + 2 D + 1Q D + 2 D + 1 b2 D + 2g cos x x cos x – D + 2 D + 1 d D + 2 D + 1i 2

2

2

2

2

= P.I.1 – P.I.2 P.I.1 =

x cos x D + 2D + 1

=

x cos x × D 2D × D

=

– x sin x 2 D2

P.I.1 = P.I.2 =

=

(D2 → – 12)

2

(D2 → – 12)

x sin x 2

b2 D + 2g cos x d D + 2 D + 1i

– 2 sin x + 2 cos x

b 2 Dg

2

=

– 2 sin x + 2 cos x 4 D2

=

2 sin x − 2 cos x 4

=

1 (sin x – cos x) 2

P.I. =

(D2 → – 12)

2

2

(D2 = – 12)

1 1 x sin x – (sin x – cos x) 2 2

1 (x sin x – sin x + cos x) 2 y = C.F. + P.I. =

y = (C1 + C2 x) e– x +

1 (x sin x – sin x + cos x). 2

248


3. Solve

d2y – y = x ex sin x. dx 2

Solution. We have the equation (D2 – 1) y = x ex sin x A.E. is m2 – 1 = 0 m2 = 1 m = + 1, – 1 C.F. = C1ex + C2e–x P.I. =

1 x e x sin x D2 – 1

Taking ex outside the operator and changing D → D + 1

1

x = e

=

Let us use

xV f D

b g

=

=

=

=

= = = P.I. =

b D + 1g – 1 L x sin x OP e M N D + 2D Q LM x – f ′b Dg OP V MN f bDg PQ f b Dg L 2 D + 2 OP . sin x e Mx – (D → – 1 ) N D + 2D Q D + 2D R| x sin x 2 b D + 1g U| – sin x e S |T 2 D – 1 D b D + 2g V|W R| x b2 D + 1g sin x U| 2 b D + 1g e S – sin x V |T 4 D – 1 – 1 dD + 4 D + 4i |W R 2 x cos x + x sin x + 2 b D + 1gb4 D – 3g sin xUV e S –5 16 D – 9 T W R – 1 x b2 cos x + sin xg – 2 d4 D + D – 3i sin x UV e S 25 T5 W R – 1 x FG 2 cos x + sin x – 2 b– 4 sin x + cos x – 3 sin xgUV e S 25 T5 H W 2

x sin x

x

2

x

2

2

2

2

x

2

2

x

2

2

x

2

x

2

x

–1 x e {(5x – 14) sin x + 2 (5x + 1) cos x} 25

∴ The general solution is y = C.F. + P.I. = C1ex + C2 e–x –

1 x e {(5x – 14) sin x + 2 (5x + 1) cos x}. 25

249


d2y dy –4 + 4y = 3x2 e2x sin 2x. 2 dx dx Solution. We have 4. Solve

(D2 – 4D + 4) y = 3x2 e2x sin 2x m2 – 4m + 4 = 0

A.E. is

(m – 2)2 = 0

i.e.,

m = 2, 2 C.F. = (C1 + C2 x) e2x P.I. = 2x

Taking e

1

b D – 2g

We shall find

1 2 x sin 2 x D2

1 2 x sin 2x Integrating twice D2

1 (x2 sin 2x) = D Applying Integration by parts

z

x 2 sin 2 x dx

2 = x

d

1 x 2 sin 2 x D2

i

FG – cos 2 x IJ – 2 x FG – sin 2 x IJ + 2 FG cos 2 x IJ H 2 K H 4 K H 8 K

=

– x 2 cos 2 x x sin 2 x cos 2 x + + 2 2 4

=

–1 2 1 x cos 2 x dx + 2 2

=

= ∴

. 3x2 e2x sin 2x

outside the operator and changing D to D + 2 2x = 3e

Hence,

2

P.I. =

z

z

x sin 2 x dx +

1 4

z

cos 2 x dx

LM FG IJ FG IJ OP H K H KQ N 1 L F – cos 2 x I F – sin 2 x IJ OP + 1 . sin 2 x + Mx G – 1G J H K H 4 KQ 4 2 2N 2 – 1 2 sin 2 x – cos 2 x – sin 2 x – 2x x +2 2 2 4 8

–1 (2x2 sin 2x + 4x cos 2x – 3 sin 2x) 8 – 3 2x e (2x2 sin 2x + 4x cos 2x – 3 sin 2x) 8

∴ The general solution is y = C.F. + P.I. = (C1 + C2 x) e 2 x –

3 2x e (2x2 sin 2x + 4x cos 2x – 3 sin 2x). 8

250


5. Solve

d2y dy –2 + y = x ex sin x. 2 dx dx

Solution. We have (D2 – 2D + 1) y = xex sin x A.E. is i.e.,

m2 – 2m + 1 = 0 (m – 1)2 = 0 m = 1, 1 C.F. = (C1 + C2 x) ex P.I. =

=

1 x e x sin x D – 2D + 1 2

1

b D – 1g

Since

b

1 x sin x D2

g

x e x sin x

(D → D + 1)

1 x sin x D2

x = e

Now

2

z

=

1 ⋅ x sin x dx D

=

1 x – cos x – 1 – sin x D

=

1 (– x cos x + sin x) D

1 is the Integration D

b

g b

g

On integration by parts

z

z

= – x cos x dx + sin x dx

m

b

= – x sin x – 1 – cos x

gr – cos x

= – x sin x – 2 cos x = – (x sin x + 2 cos x) Hence

P.I. = – ex (x sin x + 2 cos x).

∴ The complete solution is y = C.F. + P.I. = (C1 + C2x) ex – ex (x sin x + 2 cos x).

251


EXERCISE 5.4 Solve the following equations:

LM Ans. y = C cos 3x + C sin 3x + 1 b4 x cos x + sin xgOP 32 N Q LM Ans. y = bC + C xg e + 1 bsin x + cos x – 1gOP (D – 2D + 1) y = x sin x. 2 N Q LM Ans. y = C e + C e – 1 b5x sin 3x + 3 cos 3x gOP (D – 1) y = x sin 3x. 50 N Q (D – 3D + 2) y = x cos 2x. LM Ans. y = C e + C e – 1 x (3 sin 2 x + cos 2 x) – 1 b7 sin 2 x + 24 cos 2 xgOP 20 200 N Q 1 L O d y ax sin ax + x cos ax iP d + a y = x cos ax. M Ans. y = C cos ax + C sin ax + 4a N Q dx LM Ans. y = C e + C e – 1 d x – 1i cos x – 2 x sin x OP d y – y = x cos x. 2 N Q dx

1. (D2 + 9) y = x cos x. 2. 3. 4.

1

2

–x

2

1

2

x

2

–x

1

2

2

x

1

2x

2

2

5.

2

2

1

2

2

2

6.

2

x

2

–x

1

2

2

2

7. (D2 – 4D + 4) y = 4x2 e2x cos 2x.

LM Ans. y = bC + C xg e N 1

8.

2

2x

–

1 2x e 2

d2 x

2

i

– 3 cos 2 x – 4 x sin 2 x

d4y – y = x sin x. dx 4

LM Ans. y = C cos x + C N 1

2

sin x + C3 e x + C4 e – x +

d

i

1 2 x + 2 cos 5x – 3x sin x 8

OP Q OP Q

5.6 METHOD OF UNDETERMINED COEFFICIENTS Here we consider a method of finding the particular integral of the equation f (D) y = φ(x). When φ(x) is of some special forms. In this method first we assume that the particular integral is of certain form with some coefficients. Then substituting the value of this particular integral in the given equation and comparing the coefficients, we get the value of these “undetermined” coefficients. Therefore, the particular integral can be obtained. This method is applicable only when the equation is with constant coefficients. In the following cases we give the forms of the particular integral corresponding to a special form of φ(x). Case (i): If φ(x) is polynomial of degree n i.e.,

φ(x) = a0 + a1 x + ....... an xn

Then particular integral is of the form y p = a0 + a1 x + a2 x2 + ........ + an xn

252


φ (x) = 3x2

Example:

y p = a + bx + cx2 Case (ii): If φ (x) = emx then the particular y p = a emx

integral is

Example: 5ex is the φ (x) ∴

φ (x) = 5ex

∴ Particular Integral

y p = a e x.

Case (iii): If φ (x) = sin ax or cos ax or p sin ax + q cos ax then P.I. is of the form y p = A cos ax + B sin ax Example 1: ∴

φ (x) = 2 sin 5x y p = a cos 5x + b sin 5x

Example 2: ∴

φ (x) = 3 cos 2x y p = a cos 2x + b sin 2x

Case (iv): If

φ (x) = emx sin bx

or

φ (x) = emx cos bx

or

φ (x) = emx (a sin bx + b sin bx) then Particular Integral of the form y p = emx (a sin bx + b cos bx) Example:

φ (x) = 2e2x cos 3x y p = e2x (a cos 3x + b sin 3x).

WORKED OUT EXAMPLES 1. Solve by the method of undetermined coefficients, y″ – 3y′ + 2y = x2 + x + 1. Solution. For the given equation is (D2 – 3D + 2) y = x2 + x + 1 A.E. is i.e.,

m2 – 3m + 2 = 0 (m – 1) (m – 2) = 0 m = 1, 2 C.F. = C1 ex + C2 e2x

Here

φ (x) = x2 + x + 1 and 0 is not a root of the A.E.

We assume for P.I. in the form y p = a + bx + cx2

...(1)

We have to find a, b and c such that y″p – 3y′p + 2yp = x2 + x + 1

...(2)

253


From Eqn. (1)

y′p = b + 2cx y″p = 2c

Now Eqn. (2) becomes 2c – 3 (b + 2cx) + 2 (a + bx + cx2) = x2 + x + 1 (2c – 3b + 2a) + (– 6c + 2b) x + (2c) x2 = x2 + x + 1 Equating the coefficients 2c – 3b + 2a = 1

...(3)

– 6c + 2b = 1

...(4)

2c = 1

...(5)

1 2

Eqn. (5) ⇒

c =

Eqn. (4) ⇒

–6 + 2b = 1 2 2b = 1 + 3

b = 2 Eqn. (3) ⇒

2 – 3 × 2 + 2a = 1 2 2a = 1 – 1 + 6 a = 3

Eqn. (1) becomes P.I. = yp = 3 + 2x +

1 2 x 2

∴ The general solution is y = C.F. + P.I. y = C1 ex + C2 e2x +

FG 3 + 2 x + 1 x IJ · H 2 K 2

2. Solve by the method of undetermined coefficients y″ – 2y′ + 5y = e2x. Solution. We have (D2 – 2D + 5) y = e2x A.E. is

m2 – 2m + 5 = 0 m = 1 ± 2i C.F. = ex (C1 cos 2x + C2 sin 2x)

Here φ (x) = e2x and 2 is not a root of the A.E. We assume for P.I. in the form yp = ae2x

...(1)

We have to find ‘a’ such that y″p – 2y′p + 5yp = e2x

...(2)

254


y′p = 2a e2x

From Eqn. (1)

y″p = 4a e2x Eqn. (2) ⇒ 4a e2x – 4a e2x + 5a e2x = e2x 5a e2x = e2x Equating the coefficients 5a = 1 a =

1 5

1 2x e 5 y = C.F. + P.I.

Eqn. (1),

yp =

∴

y = ex (C1 cos 2x + C2 sin 2x) +

1 2x e · 5

3. Solve using the method of undetermined coefficients y″ – 5y′ + 6y = sin 2x. Solution. We have (D2 – 5D + 6) y = sin 2x A.E. is i.e.,

m2 – 5m + 6 = 0 (m – 2) (m – 3) = 0 m = 2, 3

∴

C.F. = C1 e2x + C2 e3x φ (x) = sin 2x and 0 is not a root of the A.E.

We assume for P.I. in the form y p = a cos 2x + b sin 2x

...(1)

We have to find ‘a’ and ‘b’ such that y″p – 5y′p + 6yp = sin 2x From Eqn. (1),

...(2)

y′p = – 2a sin 2x + 2b cos 2x y″p = – 4a cos 2x – 4b sin 2x

Eqn. (2) becomes – 4a cos 2x – 4b sin 2x – 5 (– 2a sin 2x + 2b cos 2x) + 6 (a cos 2x + b sin 2x) = sin 2x (10a + 2b) sin 2x + (2a – 10b) cos 2x = sin 2x Comparing the coefficients, we get 10a + 2b = 1 and

2a – 10b = 0 Solving we get,

a =

5 52

b =

1 52

255


Eqn. (1) becomes 5 1 cos 2x + sin 2x 52 52

P.I. = yp =

y = C.F. + P.I. = C1 e2x + C2 e3x +

1 (cos 2x + sin 2x). 52

4. Solve by the method of undetermined coefficients y″ + y′ – 2y = x + sin x. Solution. We have (D2 + D – 2) y = x + sin x A.E. is m2 + m – 2 = 0 i.e., (m – 1) (m + 2) = 0, m = 1, – 2 C.F. = C1ex + C2e– 2x φ (x) = x + sin x and 0 is not root of the A.E. We assume for P.I. in the form ...(1) y p = a + bx + c cos x + d sin x. We have to find a, b, c, and d such that ...(2) y″p + y′p – 2yp = x + sin x From (1), y′p = b – c sin x + d cos x y″p = – c cos x – d sin x Eqn. (2), becomes – cos x – d sin x + b – c sin x + d cos x – 2 (a + bx + c cos x + d sin x) = x + sin x (– 2a + b) – 2bx + (– 3c – d ) sin x + (c – 3d ) cos x = x + sin x Comparing the coefficients, we get – 2a + b = 0, – 2b = 1, – 3c – d = 1, c – 3d = 0 Solving, we get a =

–1 –3 –1 –1 ,b= ,c= ,d= 2 10 10 4

Eqn. (1), becomes yp = − = −

1 1 3 1 − x − sin x − cos x 4 2 10 10

b

g

b

1 1 2x + 1 − 3 sin x + cos x 4 10

g

∴ The general solution is y = C.F. + P.I. = C1e x + C2 e –2 x –

b

g

b

g

1 1 2x + 1 – 3 sin x + cos x . 4 10

256


5. Solve by the method of undetermined coefficients y″ + 4y = x2 + e–x. Solution. We have

(D2 + 4) y = x2 + e–x m2 + 4 = 0

A.E. is

m2 = – 4

i.e., i.e.,

m = ± 2i C.F. = C1 cos 2x + C2 sin 2x φ (x) = x2 + e–x

where 0 and – 1 is not a root of the A.E. We assume for P.I. in the form yp = a + bx + cx2 + de–x

...(1)

We have to find a, b, c, and d such that y″p + 4yp = x2 + e–x From (1),

...(2)

y′p = b + 2cx –

de–x

y″p = 2c + de–x Eqn. (2) ⇒ 2c + de–x + 4(a + bx + cx2 + de–x) = x2 + e–x (2c + 4a) + 4bx + 4cx2 + 5de–x = x2 + e–x Equating the coefficients 2c + 4a = 0, 4b = 0, 4c = 1, 5d = 1 Solving we get a =

–1 1 1 , b = 0, c = , d = 8 4 5

Equation (1) becomes P.I. = yp =

d

i

–1 1 2 1 − x 1 1 + x + e = 2 x 2 − 1 + e− x 8 4 5 8 5

∴ The general solution is y = C.F. + P.I. = c1 cos 2 x + c2 sin 2 x + 6. Solve by the method of undetermined coefficients y″ + y′ – 4y = x + cos 2x. Solution. We have (D2 – D – 4) y = x + cos 2x A.E. is m2 – m – 4 = 0 ∴

m =

∴

C.F. =

+1± 1+ 4 2

=

1+ 5 1– 5 , 2 2

1± 5 2

d

i

1 1 2x2 − 1 + e−x . 8 5

257


C.F. =

F1+ 5I x F 1– 5I x GH 2 JK G 2 JK Ce + C eH 1

2

We assume for P.I. in the form y p = a + bx + c cos 2x + d sin 2x

...(1)

Since 0, ± i are not roots of the A.E. We have to find a, b, c and d such that y″p – y′p – 4yp = x + cos 2x

...(2)

From Eqn. (1) y′p = b – 2c sin 2x + 2d cos 2x y″p = – 4c cos 2x – 4d sin 2x Now Eqn. (2) becomes, – 4c cos 2x – 4d sin 2x – (b – 2c sin 2x + 2d cos 2x) – [4 (a + bx + c cos 2x + d sin 2x)] = x + cos 2x Comparing the coefficients, we have – 4a – b = 0, – 4b = 1 and – 8c – 2d = 1, 2c – 8d = 0 Solving these we get, a =

1 −1 −2 −1 , b = , c = , d = 16 4 17 34

Therefore P.I. = =

1 1 2 1 − x− cos 2 x − sin 2 x 16 4 17 34

b

g

b

1 1 1 − 4x − 4 cos 2 x + sin 2 x 16 34

g

Hence the complete solution is y = C.F. + P.I. = C1

F1+ 5I x F1− 5I x GH 2 JK G 2 JK e + C eH + 2

b

g

b

g

1 1 1 − 4x − 4 cos x + sin 2 x . 16 34

7. Solve by the method of undetermined coefficients (D2 + 1) y = sin x. Solution. Here A.E. is m2 + 1 = 0 and its roots are m = ± i Hence

C.F. = C1cos x + C2 sin x

Note that sin x is common in the C.F. and the R.H.S. of the given equation. (± i is the root of the A.E.) Therefore P.I. is y the form y p = x (a cos x + b sin x) Since ± i is root of the A.E.

...(1)

258


We have to find a and b such that y″p + yp = sin x

...(2)

y′p = x (– a sin x + b cos x) + (a cos x + b sin x)

From Eqn. (1)

y″p = x (– a cos x – b sin x) + (– a sin x + b cos x) – a sin x + b cos x = x (– a cos x – b sin x) – 2a sin x + 2b cos x Then the given equation reduces to using the Eqn. (1) x (– a cos x – b sin x) – 2a sin x + 2b cos x + x (a cos x + b sin x) = sin x Equating the coefficients, we get i.e., – 2a sin x + 2b cos x = sin x – 2a = 1, a =

2b = 0

–1 , 2

b=0

–1 x cos x 2 y = C.F. + P.I.

Thus P.I. is

yp =

∴

1 x sin x . 2 8. Solve by the method of undetermined coefficients

= C1 cos x + C2 sin x –

(D2 + 1) y = 4x – 2 sin x. Solution. A.E. is m2 + 1 = 0 ⇒ m = ± i ∴

C.F. = C1cos x + C2sin x φ (x) = 4x – 2 sin x. We assume for P.I in the form y p = a + bx + x (c cos x + d sin x)

...(1)

Since 0 is not the root of the A.E. and ± i is the root of A.E. We have to find a and b such that y″p + yp = 4x – 2sin x From Eqn. (1)

...(2)

y′p = b + x (– c sin x + d cos x) + (c cos x + d sin x) y″p = x (– c cos x – d sin x) + (– c sin x + d cos x) + (– c sin x + d cos x)

Eqn. (2), becomes x (– c cos x – d sin x) + (– c sin x + d cos x) + (– c sin x + d cos x) a + bx + x (cos x + d sin x) = 4x – 2 sin x i.e.,a + bx + (– 2c sin x + 2d cos x) = 4x – 2 sin x Comparing the coefficients, we get, i.e.,

a = 0,

b = 4,

– 2c = – 2,

2d = 0

a = 0,

b = 4,

c = 1,

d=0

259


Therefore the required P.I. (using Eqn. (1) From Eqn. (1) ⇒

P.I. = 4x + x cos x

∴

y = C.F. + P.I. = C1 cos x + C2 sin x + 4x + x cos x.

9. Solve by the method of undetermined coefficients y″ + 3y′ + 2y = x2 + ex. Solution. A.E. is m2 – 3m + 2 = 0 i.e.,

(m – 1) (m – 2) = 0 ⇒ m = 1, 2

∴

C.F. = C1ex + C2e2x φ(x) = x2 + ex

with reference to the form x2 we assume the P.I. in the form a + bx + cx2. Since 0 is not a root of the A.E. and with reference to ex. We assume the P.I. to be d xex, since Eqn. (1) is a root of the A.E. We assume for P.I. in the form y p = a + bx + cx2 + d xex

...(1)

We have to find a, b, c and d such that y″p – 3y′p+ 2yp = x2 + ex

...(2)

y′p = b + 2cx + d (xex + ex)

From Eqn. (1)

y″p = 2c + d (xex + 2ex) Now Eqn. (2) becomes, (2c + dxex + 2dex) – 3 (b + 2cx + dxex + dex) + 2 (a + bx + cx2 + dxex) = x2 + ex (2a – 3b + 2c) + (2b – 6c) x + 2cx2 – dex = x2 + ex

i.e.,

Comparing the coefficients, we get, 2a – 3b + 2c = 0, ∴

2b – 6c = 0,

d = – 1,

c=

1 , 2

7 3 1 + x + x 2 – xe x 4 2 2 1 7 + 6x + 2 x 2 – 4 xe x P.I. = yp = 4 y = C.F. + P.I.

Hence from Eqn. (1)

∴

b=

3 7 ,a= 2 4

yp =

d

∴

2c = 1, – d = 1

i

1 (7 + 6x + 2x2 – 4xex). 4 10. Solve by the method of undetermined coefficients

y = C1ex + C2e2x +

(D2 + 1) y = 4x cos x – 2 sin x. Solution. A.E. is m2 + 1 = 0 m = ±i

260


C.F. = C1cos x + C2 sin x Note that both cos x and sin x are common in the C.F. and the R.H.S. of the given equation. Now P.I. corresponding to 4x cos x is of the form y 1 = x {(a + bx) cos x + (c + dx) sin x} and P.I. corresponding to 2 sin x is y 2 = x {e cos x + f sin x} Hence, we take P.I. in the form y p = y1 + y2 = (ax + xe + bx2) cos x + (cx + fx + dx2) sin x = {(a + e) x + bx2} cos x + {(c + f ) x + dx2} sin x P.I. must be of the form a + e = c1, b = c2 c + f = c3, d = c4 y p = (c1x + c2x2) cos x + (c3x + c4x2) sin x

...(1)

Differentiating, we get y′p = (c1x + c2x2) (– sin x) + cos x (c1 + 2xc2) + (c3x + c4x2) cos x + sin x (c3 + 2xc4) y″p = (c1x +

c2x2)(–

cos x) + (– sin x) (c1 + 2xc2) + cos x (2c2)

+ (c1 + 2xc2) (– sin x) + (c3x + c4x2) (– sin x) + (cos x) (c3 + 2xc4) + sin x (2c4) + (c3 + 2xc4) cos x y″p = [(– 2c1 + 2c4) + (– 4c2 – c3) x – c4x2] sin x + [(2c2 + 2c3) + (4c4 – c1) x – c2x2] cos x Substituting these values in the given equation and simplifying, we get {(2c2 + 2c3) + 4c4x}cos x + {(– 2c1 + 2c4) – 4c2x} sin x = 4x cos x – 2 sin x Comparing the coefficients, we obtain, 2c2 + 2c3 = 0, 4c4 = 4, 2c1 + 2c4 = – 2, 4c2 = 0 Solving we get c 2 = 0, c4 = 1 and hence c 3 = 0, c1 = 2 From Eqn. (1), the required P.I. is y p = (2x + 0) cos x + (0x + x2) sin x = 2x cos x + x2 sin x = 2x cos x + x2 sin x ∴ The general solution is y = C.F. + P.I. = C1 cos x + C2 sin x + 2x cos x + x2 sin x. 11. Solve by the method of undetermined coefficients (D3 + 3D2 + 2D)y = x2 + 4x + 8. Solution. A.E. is m3 + 3m2 + 2m = 0 i.e., m(m2 + 3m + 2) = 0

261


m(m + 1)(m + 2) = 0 Hence the roots are m = 0, – 1, – 2 ∴ C.F. = C1 + C2e–x + C3e–2x Since the constant term appears in both C.F. and the R.H.S. of the given equation we take P.I. in the form, y p = x (a0 + a1x + a2x2) ...(1) = a0x + a1x2 + a2x3 We have to find a0, a1, and a2 such that 2 ...(2) y′″ p + 3y″p + 2y′p = x + 4x + 8 2 From Eqn. (1) ⇒ y′p = a0 + a1(2x) + 3x a2 y″p = 2a1 + 6a2x

y ′′′ = 6a2 p Eqn. (2) becomes, 6a2 + 3 (2a1 + 6a2x) + 2 (a0 + 2a1x + 3a2x2) = x2 + 4x + 8 i.e., 6a2x2 + (4a1 + 18a2) x + 2a0 + 6a1 + 6a2 = x2 + 4x + 8 Comparing the coefficients, we get 6a2 = 1, 4a1 + 18a2 = 4, 2a0 + 6a1 + 6a2 = 8 Solving these equations, we get 11 1 1 , a1 = , a2 = a0 = 4 4 6 Eqn. (1) becomes yp = x ∴ The general solution is

FG 11 + 1 x + 1 x IJ H4 4 6 K 2

y = C.F. + P.I.

11 x 2 x3 + . x+ 4 4 6 12. Solve by the method of undetermined coefficients; (D3 + 2D2 – D –2) y = x2 + ex. Solution. A.E. is m3 + 2m2 – m – 2 = 0 i.e., (m + 2) (m – 1)(m + 1) = 0 So that m = ± 1, – 2 Hence C.F. is C.F. = C1ex + C2e–x + C3e–2x Note that ex is common in C.F. and the R.H.S. of the given equation. Therefore P.I. is of the form y p = a + bx + cx2 + dxex We have to find a, b, c and d such that y ′′′+ 2 y ′′p – y ′p − 2 y p = x2 + ex p −x –2 x = C1 + C2 e + C3e +

...(1) ...(2)

262


Eqn. (1), we get y′p = b + 2cx + dxex + dex y″p = 2c + dxex + dex + dex = 2c + dxex + 2dex

y ′′′ = dxex + dex + 2dex p = dxex + 3dex Substituting these values is in the eqn. (2) we get (– 2a – b + 4c) – 2 (b + c) x – 2cx2 + 6dex = x2 + ex Comparing the coefficients, we get – 2a – b + 4c = 0, – 2 (b + c) = 0, – 2c = 1, 6d = 1 Solving these equations, we get,

1 −1 1 −5 ,b= ,c= ,d= 6 2 2 4 Using these values in Eqn. (1), we get required P.I.. is a =

5 1 1 1 + x − x2 + x ex 4 2 2 6 y = C.F. + P.I.

yp = − Therefore,

= C1 x + C2 e − x + C3 e –2 x −

d

i

1 1 2 x 2 – 2 x + 5 + xe x. 4 6

EXERCISE 5.5 Solve the following equations by the method of undetermined coefficients: 1. y″ + 9y = x2.

2. y″ + 4y′ = x2 + x. 3. y″ – 6y′+ 10y = x2. 4. y″ – 2y′ + y = x2 – 1. 5. y″ + 4y = e–2x.

6. y″ + y′ – 12y = e2x.

LM Ans. y = C cos 3x + C sin 3x + 1 d9 x − 2iOP 81 N Q LM Ans. y = C e + C e – 1 d2 x + 2 x + 1iOP 8 N Q LM Ans. y = e bC cos x + C sin x g + 1 d5x + 6x + 35iOP 50 N Q Ans. y = bC + C x g e + x + 4 x + 5 LM Ans. y = C cos 2 x + C sin 2 x + 1 e OP 8 N Q LM Ans. y = C e + C e – 1 e OP 6 N Q 2

1

2

2x

–2 x

1

2

2

3x

2

1

2

x

1

2

2

–2 x

1

2

3x

1

–4x

2

2x

263


LM Ans. y = C e N

7. y″ + y′ – 6y = e–3x.

–3 x

1

+ C2 e –2 x –

1 –3 x xe 5

OP Q

Ans. y = C1e – x + C2 e –2 x – xe –2 x

8. y″ + 3y′ + 2y = e–2x.

LM Ans. y = bC + C xge + 1 d2 x − 4 x + 3i + 1 e OP 8 9 N Q LM Ans. y = C e + C e + 1 b6x + 5g – xe OP 36 N Q LM Ans. y = C cos x + C sin x + dx − 2 xi + 1 e OP 5 N Q LM Ans. y = C cos5x + C sin 5x + 1 cos 2 x OP 21 N Q LM Ans. y = bC + C xge – 1 b4 cos x – 3sin xgOP 25 N Q LM Ans. y = C e + C e – 1 bcos 3x + 2 sin 3xgOP 30 N Q LM Ans. y = C cos 3x + C sin 3x – 1 x cos 3x OP 6 N Q –2 x

9. y″ + 4y′ + 4y = x2 + ex.

1

2x

10. y″ – 5y′ + 6y = x + e2x.

x

2

2

3x

1

2x

2

2x

2

11. y″ + y = x2 + e2x.

1

12. y″ + 25y = cos 2x.

2

1

2

–2 x

13. y″ + 4y′ + 4y = sin x.

1

2

x

14. y″ + 4y′ – 3y = sin 3x.

–3 x

1

15. y″ + 9y = sin 3x.

2

1

2

Ans. y = C1e x + C2 e – x – 5 + cos 2 x

16. y″ – y = 10 sin2 x.

LM Ans. y = C e + C e – 1 b2 x + 1g – 1 bsin x − 3cos xgOP 4 10 N Q 1 1 L O 18. y″ – 2y′ – 3y = e + 5 cos 2x. M Ans. y = C e + C e – xe – b4 sin 2 x + 7 cos 2 x gP 4 5 N Q LM Ans. y = C cos 2 x + C sin 2 x + 1 d2 x – 1i – 1 x cos 2 x OP 19. y″ + 4y = x + sin 2x. 8 4 N Q x

17. y″ + y′ – 2y = x + cos x.

–2 x

1

2

–x

–x

3x

1

–x

2

2

2

1

2

20. y″ + 3y′ + 2y = x2 + cos x.

LM Ans. y = C e + C e N LM Ans. y = C e 21. y″ + y′ – 6y = cos 2x + e . N –x

1

–2 x

2

2x

2x

1

+

d

i

b

1 1 2 x 2 – 6x + 7 + 3 sin x + cos x 4 10

+ C2 e –3 x +

b

g

gOPQ

1 1 sin 2 x – 5 cos 2 x + xe 2 x 52 5

OP Q

264


LM Ans. y = bC + C x ge N

OP Q LM Ans. y = C cos x + C sin x + 1 bx − 1g e OP 2 N Q LM Ans. y = C + C e – 1 e sin xOP 2 N Q

22. y″ + 2y′ + y = ex cos 2x.

1

2

−x

–

1 x e cos 2 x 8 2x

23. y″ + y = xe2x.

1

2

2x

24. y″ – 2y′ = ex sin x.

1

x

2

25. y″ – 2y′ + 3y = x3 + sin x.

LM Ans. y = e eC cos N x

1

j

3x + C2 sin 3 x +

26. (D3 – 3D + 2) y = x2 + e–2x.

LM Ans. y = C e N

i b

d

1 1 9 x 3 + 18 x 2 + 6 x – 8 + sin x + cos x 27 4

b

g

d

i OPQ

d

i OPQ

+ C2 + C3 x e x –

1 −x 1 e + 2 x 2 + 6x + 9 2 4

+ C2 e − x + C3 e –2 x +

1 −x 1 e − 2x2 − 2x + 9 6 4

−2 x

1

27. (D3 + 2D2 – D – 2) y = ex + x2.

LM Ans. y = C e N 1

x

28. (D3 + 2D2 + 1) y = e2x + sin 2x.

LM Ans. = + MN y C bC 1

2

g

b

2x

+ C3 x e − x +

O g PP Q + d x + 2 x + 5i

1 ex + 3 cos 2 x − 4 sin 2 x 18 50

b

g

Ans. y = C1 + C2 x e x + C3e − x

29. (D3 – D2 – D + 1) y = x2 + 1.

2

30. (D3 – D2 + 3D + 5) y = ex cos 2x.

LM Ans. y = C e N

–x

1

31. (D3 – 6D2 + 11D – 6) y = 2xe–x.

32. (D3 – D2 – 2D) y = x2 + x + 1.

gOPQ

b

g

+ e x C2 cos 2 x + C3 sin 2 x +

b

1 x xe sin 2 x − cos 2 x 16

g OPQ

LM Ans. y = C e + C e + C e – 1 b12 x + 13ge OP 144 N Q LM Ans. y = C + C e + C e – 1 x d x + 6iOP 6 N Q x

1

2x

3x

2

–x

3

–x

1

2

2x

2

3

5.7 SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS Let us suppose that x and y are functions of an independent variable ‘t’ connected by a system of d first order equation with D = dt

265


f1(D) x + f2(D) y = φ1(t)

...(1)

g1(D) x + g2(D) y = φ2(t)

...(2)

By solving a system of linear algebraic equations in cancelling either of the dependent variables (x or y) operating (1) with g1 (D) and (2) with f1 (D), x cancels out by subtraction. We obtain a second order differential equation in y. Which can be solved x can be obtained independently by cancelling y or by substituting the obtained y (t) in a suitable equation.

WORKED OUT EXAMPLES dx dx − 7x + y = 0, − 2x − 5y = 0. dt dt d , we have the system of equations Solution. Taking D = dt (D – 7) x + y = 0 – 2x + (D – 5) y = 0 Multiply (1) by 2 and operate (2) by (D – 7) i.e., 2 (D – 7) x + 2y = 0 – 2 (D – 7) x + (D – 5) (D – 7) y = 0 1. Solve

Adding A.E. is or ⇒ Thus

[(D – 5)(D – 7) + 2] y (D2 – 12D + 37) y m2 – 12m + 37 (m – 6)2 + 1 m– 6 m y

By considering

∴

0 or 0 0 0 ±i 6±i e6t (C1 cos t + C2 sin t)

dy − 2 x − 5 y = 0, we get dt 1 dy − 5y x = 2 dt x = =

x = Thus

= = = = = = =

x =

...(1) ...(2)

FG H 1Rd S e 2 T dt o b

6t

...(3)

IJ K

bC cos t + C sin t g − 5e bC cos t + C sin t gUVW 6t

1

2

1

g

2

b

g

1 6t e – C1 sin t + C2 cos t + 6e 6t C1 cos t + C2 sin t 2 – 5e 6 t C1 cos t + C2 sin t

o b 1 obC + C g e 2

g

b

b

1 6t e – C1 sin t + C2 cos t + e 6t C1 cos t + C2 sin t 2 1

2

6t

b

g

cos t + C2 − C1 e 6t sin t

t

(3) and (4) represents the complete solution of the given system of equations.

gt

gt

...(4)

266


2. Solve:

dx dy = 2x – 3y, = y – 2x given x (0) = 8 and y (0) = 3. dt dt

Solution. Taking D =

i.e.,

d we have the system of equations. dt Dx = 2x – 3y ; Dy = y – 2x

(D – 2) x + 3y = 0

...(1)

2x + (D – 1) y = 0

...(2)

Multiplying (1) by 2 and (2) by (D – 2), we get 2 (D – 2) x + 6y = 0 2 (D – 2) x + (D – 1) (D – 2) y = 0 Subtracting, we get (D2 – 3D – 4) y = 0 m2 – 3m – 4 = 0

A.E. is or

⇒

(m – 4) (m + 1) = 0 ∴

m = 4, – 1

y = C1 e4t + C2 e–t

By considering

dy = y – 2x, we get dt x =

i.e.,

x = =

...(3)

RS UV T W 1 {C e + C e − d4C e 2 1 d– 3C e + 2C e i 2 1 dy y− 2 dt 4t

–t

1

2

1

4t

1

4t

− C2 e – t

i}

–t

2

...(4)

We have conditions x = 8, y = 3 at t = 0

3C1 + C2 = 8. 2 Solving these equations, we get C 2 = 5, C1 = –2

Hence (3) and (4) become C1 + C2 = 3 and −

x = 3e4t + 5e–t

Thus

y = – 2e4t + 5e–t is the required solution. 3. Solve:

dx dy − 2y = cos 2t, + 2x = sin 2t given that x = 1, y = 0 at t = 0. dt dt

Solution. Taking D =

d we have the system of equations dt Dx – 2y = cos 2t

2x + Dy = sin 2t Multiplying (1) by D and (2) by 2, we have

...(1) ...(2)

267


D2x – 2Dy = D(cos 2t) = – 2 sin 2t 4x + 2Dy = 2 sin 2t Adding, we get (D2 + 4) x = 0 ⇒

m2 + 4 = 0

A.E. is ∴

m = ± 2i

x = C1 cos 2t + C2 sin 2t

By considering

...(3)

dx − 2 y = cos 2t, we get dt y =

i.e.,

y = =

LM OP N Q 1 Ld O C cos 2t + C sin 2t g − cos 2t P b M 2 N dt Q 1 dx − cos 2t 2 dt 1

2

1 −2C1 sin 2t + 2C2 cos 2t − cos 2t 2

FG H

y = − C1 sin 2t + C2 −

IJ K

1 cos 2 t 2

...(4)

Equation (3) and (4) represents the general solution Applying the given conditions x = 1 at t = 0 Hence (3) becomes,

1 = C1 + 0

⇒ C1 = 1

y = 0 at t = 0 Hence (4) becomes,

FG H

0 = 0 + C2 −

1 2

IJ K

⇒ C2 =

1 2

Substituting these values in (3) and (4), we get x = cos 2 t +

1 sin 2t 2

y = – sin 2t Which is the required solution.

EXERCISE 5.6 1. Solve

2. Solve

dx dy + 2 y = − sin t , − 2 x = cos t . dt dt dx dy + y = et , − x = e−t. dt dt

dx dy 3. Solve + x − y = et , + y − x = 0. dt dt

LM Ans. x = − C sin 2t + C cos 2t − cos t OP N y = C cos 2t + C sin 2t − sin t Q LMAns. x = − C sin t + C cos t + sin h t OP N y = C cos t + C sin t + sin h t Q LM Ans. x = C + C e + 2e OP MN y = C − C e + e PQ 1

2

1

2

1

1

2

2

−2 t

1

2

1

2

−2 t

t 3

t 3

268


ADDITIONAL PROBLEMS (From Previous Years VTU Exams.) 1. Solve (D3 –1) y = (ex + 1)2. Solution. Refer page no. 226. Example 5.

d2y dy +4 − 12y = e2x – 3 sin 2x. 2. 2 dx dx Solution. We have (D2 + 4D – 12) y = e2x – 3 sin 2x A.E. is

m2 + 4m – 12 = 0 (m + 6) (m – 2) = 0 ⇒ m = – 6 and 2

∴

C.F. = C1 e–6x + C2 e2x

3 sin 2 x e2 x − 2 2 D + 4 D − 12 D + 4 D − 12 = P.I.1 – P.I.2

P.I. =

Now,

e2 x e2 x = P.I.1 = D 2 + 4 D − 12 2 2 + 4 ⋅ 2 − 12

(D → 2) Dr = 0

Differential denominator and multiply ‘x’

x e2 x x e2 x x e2 x = = 2D + 4 2 ⋅ 2 + 4 8

(D → 2)

P.I.2 =

3 sin 2 x D + 4 D − 12

(D2 → – 22 = – 4)

P.I.2 =

3 sin 2 x 3 sin 2 x D+4 = × – 4 + 4 D − 12 4 D − 4 D+4

=

=

= ∴

P.I.2 =

2

b

g

b

g

3 D + 4 sin 2 x

d

i

4 D − 16

b

2

3 2 cos 2 x + 4 sin 2 x

b

b g

4 −20

3 cos 2 x + 2 sin 2 x

g

(D2 → – 4)

g

– 40

∴ The general solution: y = C.F. + P.I. y = C1 e –6 x + C2 e 2 x + 3. Solve (D2 + 5D + 6) y = cos x + e–2x. Solution. Refer page no. 229. Example 3.

b

x e 2 x 3 cos 2 x + 2 sin 2 x + 8 40

g

269


d3y dy −7 + 6y = 1 – x + x2. 4. Solve 3 dx dx Solution. We have (D3 – 7D + 6) y = 1 – x + x2 A.E. is

m3 – 7m + 6 = 0

m = 1 is a root by inspection. We now apply the method of synthetic division.

1

1

0

−7

6

0

1

1

−6

1

1

−6

0

m2 + m – 6 = 0 or (m + 3) (m – 2) = 0 ⇒

m = – 3, 2

m = 1, 2, – 3 are the roots of A.E. ∴

C.F. = C1 ex + C2 e2x + C3 e–3x P.I. =

1 − x + x2 x2 − x + 1 = D3 − 7 D + 6 6 − 7 D + D3

P.I. is found by division

x 2 2 x 23 + + 6 9 54 6 – 7D + D3

x2 – x + 1 7x x2 – + 0 3

4x + 1 3 4x 14 – 3 9 23 9 23 9 0 ∴

x 2 2 x 23 1 + + = P.I. = (9x2 + 12x + 23) 6 9 54 54

∴The general solution is y = C.F. + P.I. y = C1 e x + C2 e 2 x + C3 e –3 x +

d

i

1 9 x 2 + 12 x + 23 54

270


d2y + 4y = x2 + cos 2x + 2–x. 2 dx Solution. We have

5. Solve

(D2 + 4) y = x2 + cos 2x + 2–x A.E. is ∴

m2 + 4 = 0 ⇒ m = ± 2i C.F. = C1 cos 2x + C2 sin 2x

cos 2 x 2– x x2 + + D2 + 4 D2 + 4 D2 + 4 P.I. = P.I.1 + P.I.2 + P.I.3

P.I. =

∴

P.I.1 =

x2 D2 + 4

P.I.1 is found by division

x2 1 − 4 8 4 + D2

x2 x2 +

1 2

x4 1 − 4 8

P.I.1 =

1 2 1 – 2 –

=

d

i

1 2x2 − 1 8

0 ∴

P.I.1 =

d

i

1 2x2 − 1 8

cos2 x D2 + 4 Differentiate the denominator and multiply ‘x’

(D2 → – 22 = – 4) (Dr = 0)

P.I.2 =

= x = P.I.2 =

cos2 x D × 2D D

b

g

x – sin 2 x ⋅ 2 2D

2

(D2 → – 4)

x sin 2 x 4

d i d i −x

e log 2 e – log 2⋅ x 2−x e – log 2⋅ x = = P.I.3 = = 2 D2 + 4 D2 + 4 D2 + 4 – log 2 + 4

b

g

271


P.I.3 =

2– x

b– log 2g

2

+4

∴The general solution is y = C.F. + P.I. y = C1 cos 2x + C2 sin 2 x +

d

i

1 2– x x sin x 2x2 − 1 + + ⋅ 3 8 4 log 2 − 4

b g

d2y dy −4 + 4y = e2x + cos 2x + 4. 2 dx dx Solution. We have (D2 + 4D + 4) y = e2x + cos 2x + 4

6. Solve

A.E. is i.e., ∴

m2 – 4m + 4 = 0 (m – 2)2 = 0 ⇒ m = 2, 2 C.F. = (C1 + C2x) e2x

cos 2 x 4 e2 x + 2 + 2 P.I. = 2 D − 4D + 4 D − 4D + 4 D − 4D + 4 ∴ Now,

= P.I.1 + P.I.2 + P.I.3

e2 x D2 − 4 D + 4

(D → 2)

=

e2 x 4−8+4

(Dr = 0)

=

x e2 x 2D − 4

(Dr = 0 as D → 0)

P.I.1 =

x 2 e2 x P.I.1 = 2 P.I.2 =

cos2 x D − 4D + 4 2

(D2 → – 22 = – 4)

cos 2 x D = – 4D × D = ∴

– sin 2 x ⋅ 2 4 D2

P.I.2 =

– sin 2 x 8

P.I.3 =

4 ⋅ e0 x D2 − 4 D + 4

(D2 → – 4)

(D → 0)

272


=

4 e0 x 4 = =1 0–0+4 4

∴The general solution is y = C.F. + P.I.

b

∴

g

y = C1 + C2 x e 2 x +

x 2 e 2 x sin 2 x − + 1. 2 8

d3y d2y dy + + 4 + 4y = x2 – 4x – 6. dx dx 3 dx 2 Solution. We have

7. Solve

(D3 + D2 + 4D + 4) y = x2 – 4x – 6 A.E. is m3 + m2 + 4m + 4 = 0 m2 (m + 1) + 4 (m + 1) = 0 (m + 1) (m2 + 4) = 0 m = – 1, m = ± 2i ∴

C.F. = C1 e–x + C2 cos 2x + C3 sin 2x P.I. =

x2 − 4x − 6 D3 + D2 + 4 D + 4

P.I. is found by division

x 2 3x 1 − − 4 2 8 4 + 4D + D2 + D3

x2 – 4x – 6 1 x2 + 2x + 2

13 2 – 6x – 6 – 6x –

–

1 2

–

1 2 0

Hence,

P.I. =

x 2 3x 1 − − 4 2 8

d

i

1 2 x 2 − 12 x − 1 8 ∴The general solution is y = C.F. + P.I.

=

y = C1 e – x + C2 cos 2 x + C3 sin 2 x +

d

i

1 2 x 2 − 12 x – 1 . 8

273


d2y dy –6 + 25y = e2x + sin x + x. 8. Solve 2 dx dx Solution. We have (D2 – 6D + 25) y = e2x + sin x + x A.E. is

m2 – 6m + 25 = 0 6 ± 36 − 100 6 ± 8i = = 3 ± 4i 2 2 C.F. = e3x (C1 cos 4x + C2 sin 4x)

m =

sin x e2 x x + 2 + 2 2 D − 6D + 25 D − 6 D + 25 D − 6 D + 25 P.I. = P.I.1 + P.I.2 + P.I.3

P.I. =

P.I.1 = =

e2 x D 2 − 6 D + 25

(D → 2)

e2x e2 x = 2 2 − 6 2 + 25 17

bg

sin x (D → – 12 = – 1) D − 6 D + 25 sin x sin x sin x 4+ D = = × = – 1 − 6 D + 25 24 − 6 D 6 4 − D 4 + D

P.I.2 =

2

=

b4 + Dg sin x 6 d4 − D i

=

4 sin x + cos x 6 17

∴

P.I.2 =

∴

P.I.3 =

4 sin x + cos x 102 x

2

2

b g

25 − 6 D + D 2

P.I.3 is found by division.

25 – 6D + D2

x 6 + 25 625 x 6 x– 25 6 25 6 25 0

b

g

(D2 = – 1)

274


x 6 + 25 625 = 625 (25x + 6)

P.I.3 =

∴ The general solution is y = C.F. + P.I.

b

g

y = e3x C1 cos 4 x + C2 sin 4 x +

b

g

1 e 2 x 4 sin x + cos x 25x + 6 . + + 17 102 625

d4x + 4x = cos ht. dt 4 Solution. Refer page no. 225. Example 4.

9. Solve

10. Solve the differential equation

d2y dy −5 + 6y = e2x + sin x by the method of undetermined 2 dx dx

coefficients. Solution. We have (D2 – 5D + 6) y = e2x + sin x A.E. is

m2 – 5m + 6 = 0 (m – 2) (m – 3) = 0

⇒

m = 2 and 3

∴

C.F. = C1 e2x + C2 e3x

Let

φ (x) = e2x + sin x and we assume the P.I. in the form yp = ax e2x + b cos x + c sin x

...(1)

Since 2 is root of the A.E. We have to find a, b, c such that 2x + sin x y ′′′ p − 5 y ′p + 6 y p = e

...(2)

From (1), we obtain yp′ = a (2xe2x + e2x) – b sin x + c cos x

y ′′p = a (4xe2x + 4e2x) – b cos x – c sin x Now (2) becomes (4ax e2x + 4ae2x – b cos x – c sin x) – (10ax e2x + 5a e2x – 5b sin x + 5c cos x) + (6axe2x + 6b cos x + 6c sin x) = e2x + sin x i.e., – ae2x + (5b – 5c) cos x + (5c + 5b) sin x = e2x + sin x ⇒

– a = 1, 5b – 5c = 0, 5c + 5b = 1

∴

a = – 1, b = c, 5c + 5b = 1

∴

a = – 1, b =

we have

yp = – xe 2 x +

1 1 ,c= by solving using these values in (1) 10 10

b

1 cos x + sin x 10

g

275


Complete solution is

y = C.F. + yp

b

11. Using the method of undetermined coefficients solve

g

1 cos x + sin x . 10

y = C1 e 2 x + C2 e 3 x − xe 2 x +

d2y dy +2 + 4y = 2x2 + 3e–x. 2 dx dx

Solution. We have (D2 + 2D + 4) y = 2x2 + 3e–x A.E. is

m2 + 2m + 4 = 0 – 2 ± 4 − 16 – 2 ± –12 – 2 ± 2i 3 = = 2 2 2

∴

m =

∴

m = –1± i 3 –x C.F. = e

∴

{eC cose 3xj + sine 3xjj} 1

φ (x) = 2x2 + 3e–x and we assume for P.I. of the form

Let

yp = a + bx + cx2 + de–x

y ′p = b + 2cx – de–x

Now,

y ′′p = 2c + de–2x We find a, b, c, d such that

y ′′p + 2 y ′p + 4 y p = 2x2 + 3e–x i.e., (2c + de–x) + (2b + 4cx – 2de–x) + (4a + 4bx + 4cx2 + 4de–x) = 2x2 + 3e–x

z

(2c + 2b + 4a) + (4c + 4b) x + 4cx2 + 3de–x = 2x2 + 3e–x

Comparing both sides, we have 2c + 2b + 4a = 0

or

2a + b + c = 0

4c + 4b = 0

or

b + c= 0

4c = 2

∴

c=

3d = 3

∴

d= 1

Hence, we also obtain a = 0, b =

yp =

1 2

–1 using the values of a, b, c, d in (1), we have 2

–1 1 x + x2 + e–x 2 2

∴ The complete solution is y = C.F. + yp

{

e j

e 3xj} + 2x b x − 1g + e

−x = e C1 cos 3x + C2 sin

–x

.

276


d2y dy –2 + 3y = x2 + cos x. 12. Using the method of undetermined coefficients solve 2 dx dx Solution. We have (D2 – 2D + 3) y = x2 + cos x A.E. is

m2 – 2m + 3 = 0

∴

2±

m =

4 − 12 2 ± –8 = = 1±i 2 2

e

–x C.F. = e C1 cos 2 x + C2 sin 2 x

∴

φ (x) =

x2

2

j

+ cos x and we assume for P.I. in the form

yp = a + bx + cx2 + d cos x + e sin x

...(1)

We have to find a, b, c, d, e such that

y ′′p − 2 y ′p + 3 y p = x2 cos x

...(2)

From (1), we obtain

y ′p = b + 2cx – d sin x + e cos x y ′′p = 2c – d cos x – e sin x Hence (2) becomes, (2c – d cos x – e sin x) – (2b + 4cx – 2d sin x + 2e cos x) + (3a + 3bx + 3cx2 + 3d cos x + 3e sin x) = x2 + cos x Comparing both sides, we obtain 2c – 2b + 3a = 0

...(3)

– 4c + 3b = 0

...(4)

3c = 1

...(5)

2d – 2e = 1

...(6)

2d + 2e = 0

...(7)

Solving these equations, we get

1, 1 –1, 2 , 4 d = , e= a = b= · 3 4 4 27 9 Substituting these values in (1), we get c =

2 4 1 1 1 + x + x 2 + cos x − sin x 27 9 3 4 4 ∴The complete solution is yp =

y = C.F. + P.I.

e

j

−x C1 cos 2 x + C2 sin 2 x + y = e

d

i b

g

1 1 2 + 12 x + 9 x 2 + cos x – sin x . 27 4

277


OBJECTIVE QUESTIONS 1. The solution of a differential equation which is not obtained from the general solution is known as (a) Particular solution

(b) Singular solution

(c) Complete solution

(d ) Auxiliary solution

Ans. b

2. The differential equation formed by y = a cos x + b sin x + 4 where a and b are arbitrary constants is (a)

d2y +y=0 dx 2

(b)

d2y – y=0 dx 2

d2y d2y 4 y + = = y+4 (d ) dx 2 dx 2 3. The differential equation representing the family given by r = a + b cos θ is (b) r1 = r2 cot θ (a) r3 + r1 = 0 (c)

(c) r2 = r1 cot θ 4. The differential equation

(d ) r22 = r12 tan 2 θ

Ans. c

Ans. c

dy = y 2 is dx

(a) Linear

(b) Non-linear

(c) Quasilinear

(d ) None of these

Ans. b

5. xdx + ydy + zdz = 0 is the first order differential equation of a (a) Sphere

(b) Right circular cone

(c) Cylinder

(d ) Ellipsoid

−

1 x

(a) c = 1, d = 0

dy y = 2 , when dx x (b) c = 2, d = 0

(c) c = 1, d = 1

(d ) c = 2, d = 2

6. y = ce

+ d is a solution of

Ans. a

Ans. b

7. For the differential equation (y + 3x) dx + xdy = 0, the particular solution when x = 1, y = 3 is (a) 3y2 + 2xy = 9

(b) 3x2 + 2y2 = 21

(c) 3x2 + 2y = 9

(d ) 3x2 + 2xy = 9

8. The expression (xdy – ydx)/(x2 + y2) is equal to (a) d tan –1

(c) d

LM MN d x

FG x IJ H yK

OP + y iP Q 1

2

2

(b) d tan –1

(d ) d cot –1

FG y IJ H xK FG y IJ H xK

Ans. d

Ans. b

278


9. The general solution of an nth order differential equation contains (a) At least n independent constants

(b) At most n independent constants

(c) Exactly n independent constants

(d ) Exactly n dependent constants

10. The particular integral of (a)

Ans. c

d2y + y = cos x is dx 2

1 sin x 2

(b)

1 cos x 2

1 1 x cos x x sin x (d ) 2 2 11. The general differential equation (D2 + 1)2 y = 0 is (c)

Ans. d

(a) C1 cos x + C2 sin x (b) (C1 + C2 x) cos x + (C3 + C4 x) sin x (c) C1 cos x + C2 sin x + C3 cos x + C4 sin x

Ans. b

(d ) (C1 cos x + C2 sin x) (C3 cos x + C4 sin x) 12. The particular integral of the differential equation

(D2

+ D + 1) y = sin 2x is

(a)

2 sin 2 x + 3 cos 2 x 13

(b)

2 cos 2 x + 3 sin 2 x – 13

(c)

2 cos 2 x + 3 sin 2 x – 13

(d )

2 cos 2 x – 3 sin 2 x 13

13. The roots of

d4y d2y 10 + + 25 y = 0 is dx 4 dx 2

(a) ± i 5

(b)

(c) ± 10 14. The particular integral of (a)

x sin 2 x 2

5

(d ) + 5 , − 5 (D2

(b)

–x cos 2 x 4

x cos 2 x (d ) None of these 2 15. The particular integral of (D3 + 2D2 + D) y = e2x is

1 x e 18

(b)

Ans. d

1 2x e 9

1 2x e2 x e (d ) 18 3 2 2 16. The particular integral of (D – 2D + 1) y = x is (b) x3 + 4x2 + 6 (a) x2 + 4x + 6 (c)

(c) x2 – 4x – 6

Ans. d

– 4) y = sin 2x is

(c)

(a)

Ans. b

(d ) None of these

Ans. c

Ans. a

279


17. The complementary function of (a) C1 sin 2x + C2 sin 3x

d2y + 4 y = 5 is dx 2 (b) C1 cos 2x + C2 sin 2x

(c) C1 cos 2x – C2 sin 2x

(d ) None of these

Ans. b

18. By the method of undetermined coefficients y ′′ + 3 y ′ + 2 y = 12 x 2, the yp is (a) a + bx + cx2

(b) a + bx

(c) ax + bx2 + cx3

(d ) None of these

bg

Ans. a

bg

19. The particular integral of x ′′′ t – 8x t = 1 is (a)

1 8

(b)

1 4

(c)

–1 8

(d )

1 10

20. The particular integral of

Ans. c

d2y + 9 y = cos 3x is dx 2

(a)

cos x 8

(b)

(c)

cos3x 9

(d ) None of these

cos x 16

Ans. b GGG

UNIT

81

Differential Equations-II 6.1 METHOD OF VARIATION OF PARAMETERS Consider a linear differential equation of second order

d2y dy + a1 + a2 y = φ(x) ...(1) dx dx 2 where a1, a2 are functions of ‘x’. If the complimentary function of this equation is known then we can find the particular integral by using the method known as the method of variation of parameters. Suppose the complimentary function of the Eqn. (1) is C.F. = C1 y1 + C2 y2 where C1 and C2 are constants and y1 and y2 are the complementary solutions of Eqn. (1) The Eqn. (1) implies that y1′′+ a1 y1′ + a 2 y1 = 0

...(2)

y2′′ + a1 y2′ + a 2 y2 = 0

...(3)

We replace the arbitrary constants C1, C2 present in C.F. by functions of x, say A, B respectively, ∴

y = Ay1 + By2

...(4)

is the complete solution of the given equation. The procedure to determine A and B is as follows. From Eqn. (4)

y′ =

b Ay′ + By ′ g + b A′y + B′y g 1

2

1

2

...(5)

We shall choose A and B such that A′y1 + B′y2 = 0

...(6)

Thus Eqn. (5) becomes y′1 = Ay1′ + By 2′

...(7)

Differentiating Eqn. (7) w.r.t. ‘x’ again, we have y″ =

b Ay ′′+ Ay′′ g + b A′y ′ + B′y′ g 1

2

280

1

2

...(8)

281

DIFFERENTIAL EQUATIONS–II

Thus, Eqn. (1) as a consequence of (4), (7) and (8) becomes A ′ y1′ + B ′ y 2′ = φ(x)

...(9)

Let us consider equations (6) and (9) for solving

A′y1 + B′ y 2 = 0

...(6)

A′y1′ + B′ y 2′ = φ(x)

...(9)

Solving A′ and B′ by cross multiplication, we get A′ =

bg

Find A and B A = –

Integrating,

B =

where W =

y1 y1′

bg

– y2 φ x y φx , B′ = 1 W W

z

z

...(10)

y2 φ ( x) dx + k1 W

y1 φ( x ) dx + k 2 W

y2 = y1 y 2′ – y 2 y1′ y 2′

Substituting the expressions of A and B y = Ay1 + By2 is the complete solution.

WORKED OUT EXAMPLES 1. Solve by the method of variation of parameters

d2y + y = cosec x. dx 2 Solution. We have (D2 + 1) y = cosec x A.E. is

m2 + 1 = 0

⇒

m2 = – 1

⇒

m = ± i

Hence the C.F. is given by ∴

yc = C1 cos x + C2 sin x

...(1)

y = A cos x + B sin x

...(2)

be the complete solution of the given equation where A and B are to be found. The general solution is We have

y = Ay1 + By2 y 1 = cos x and y2 = sin x y1′ = – sin x and y2′ = cos x W = y1 y2′ − y2 y1′ = cos x . cos x + sin x . sin x = cos2x + sin2x = 1

282


A′ =

bg

– y2 φ x , W

B′ =

– sin x ⋅ cosec x , 1 1 , = – sin x ⋅ sin x

=

A′ = – 1, On integrating, we get A = B =

bg

y1 φ x W

cos x ⋅ cosec x 1 1 B′ = cos x ⋅ sin x B′ =

B′ = cot x

zb g z

–1 dx + C1 , i.e., A = – x + C1

cot x dx + C2 , i.e., B = log sin x + C2

Hence the general solution of the given Eqn. (2) is y = (– x + C1) cos x + (log sin x + C2) sin x i.e.,

y = C1 cos x + C2 sin x – x cos x + sin x log sin x.

2. Solve by the method of variation of parameters

d2 y + 4y = 4 tan 2x. dx 2 Solution. We have (D2 + 4) y = 4 tan 2x A.E. is

m2 + 4 = 0

where φ(x) = 4 tan 2x. i.e., i.e.,

m2 = – 4 m = ± 2i

Hence the complementary function is given by yc = C1 cos 2x + C2 sin 2x y = A cos 2x + B sin 2x

...(1)

be the complete solution of the given equation where A and B are to be found We have

Then

y 1 = cos 2x

and y2 = sin 2x

y1′ = – 2 sin 2x

and y2′ = 2 cos 2x

W = y1 y2′ − y2 y1′ = cos 2x . 2 cos 2x + 2 sin 2x . sin 2x = 2 (cos22x + sin22x)

Also,

= 2 φ(x) = 4 tan 2x A′ =

bg

– y2 φ x W

and B′ =

bg

y1φ x W

283


A′ =

– sin 2 x ⋅ 4 tan 2 x – cos 2 x ⋅ 4 tan 2 x , B′ = 2 2

A′ =

– 2 sin 2 2 x , B′ = 2 sin 2x cos 2 x

On integrating, we get

z

z z zl

sin 2 2 x dx , B = 2 sin 2 x dx cos 2 x

A = –2

1 − cos2 2 x dx cos 2 x

= –2 = −2

= –2

z

q

sec 2 x − cos 2 x dx

RS 1 log bsec 2 x + tan 2 xg – 1 sin 2 xUV 2 T2 W

A = – log (sec 2x + tan 2x) + sin 2x + C1 B = 2

sin 2 x dx

b

g

2 – cos 2 x + C2 2 B = – cos 2x + C2 Substituting these values of A and B in Eqn. (1), we get y = C1 cos 2x + C2 sin 2x – cos 2x log (sec 2x + tan 2x) which is the required general solution. =


d2y + a 2 y = sec ax. dx 2 Solution. We have (D2 + a2) y = sec ax A.E. is

⇒ m = ± ai

m2 + a2 = 0

C.F. = yc = C1 cos ax + C2 sin ax y = A cos ax + B sin ax

...(1)

be the complete solution of the given equation where A and B are to be found. We have,

y 1 = cos ax,

y2 = sin ax

y1′ = – a sin ax, y2′ = a cos ax W = y1 y2′ − y2 y1′ = a. Also, φ (x) = sec ax

bg

– y2 φ x , and W – sin ax ⋅ sec ax , A′ = a

A′ =

bg

y1 φ x W cos ax ⋅ sec ax B′ = a

B′ =

284


A′ =

– tan ax , a

A = – A =

1 a

z

1 a 1 dx + c2 B= a x + C2 B= a

B′ =

tan ax dx + C1 ,

b

– log sec ax 2

g+C

z

1

a Substituting these values of A and B in Eqn. (1), we get Thus,

y = C1 cos ax + C2 sin ax –

cos ax log (sec ax ) x sin ax + ⋅ a a2


d2y dy +2 + 2y = e–x sec3x. 2 dx dx Solution. We have (D2 + 2D + 2) y = e–x sec3x A.E. is i.e.,

m2 + 2m + 2 = 0 m =

–2± 4−8

2 m = –1±i ∴The complementary function (C.F.) is C.F. = yc = e–x(C1cos x + C2 sin x)

∴

y = C1 e–x cos x + C2 e–x sin x

⇒

y = A e–x cos x + B e–x sin x

Thus

–x

y1 = e

–x

cos x and y2 = e

...(1)

sin x

y1′ = – e– x(sin x + cos x)

y2′ = e–x (cos x – sin x)

W = y1 y2′ – y2 y1′ = e–2x Also,

φ(x) = e–x sec3x A′ = =

bg

– e – x sin x ⋅ e – x sec 3 x = – tan x sec 2 x e −2 x

A = – A = – and

B′ =

bg

– y2 φ x – y1 φ x , B′ = W W

z

tan x sec 2 x dx

1 tan 2 x + C1 2

bg

y1 φ x W

285


e – x cos x ⋅ e − x sec 3 x e −2 x 2 B′ = sec x =

B =

z

sec 2 x dx

B = tan x + C2 Substituting these values of A and B in Eqn. (1), we get y = =

RS – 1 tan x + C UV e cos x + ltan x + C q e T2 W 1 e bC cos x + C sin x g + e sin x sec x 2 −x

2

1

−x

–x

1

–x

2

sin x

2

2

This is general solution of the given solution. 5. Solve by the method of variation of parameters 2 d2y . –y = 2 1 + ex dx

Solution. We have (D2 – 1) y = A.E. is i.e., Hence C.F. is

2 1 + ex

m2 – 1 = 0 m2 = 1

⇒m =±1 x

yc = C1 e + C2 e–x y = Aex + Be–x

...(1)


y 1 = ex

y2 = e–x

and

y1′ = ex

y2′ = – e–x

W = y1 y2′ − y2 y1′

also

LM N

φ(x) =

2 1 + ex

= ex (– e–x) – e–x – e–x. ex – x = e0 = 1 = –1–1=–2 e–x + x = e0 = 1 – y2 φ x A′ = W 2 – e– x ⋅ 1 + ex e− x 1 = + = = x –2 1 + ex e 1+ ex

bg

d

A′ =

A =

1

e

z

x

d1 + e i x

d

1

e 1 + ex x

i

dx + C1

i

OP Q

286


Substitute

Hence,

ex = t, then ex dx = dt

A = =

z b z LMN t

dx =

dt t

1

g dt + C 1 1 1 O – + P dt + C , t 1+ tQ t

2

1

1+ t

1

2

using partial fractions.

1 – log t + log (1 + t ) + C1 t A = – e–x – x + log (1 + ex) + C1 = –

B′ =

bg

y1 φ x W ex ⋅

= B = –

z

2 1 + ex ex =– –2 1 + ex

ex dx + C2 1 + ex

B = – log (1 + ex) + C2 Substituting these values of A and B in eqn. (1), we get y =

d

i

d

i

– e − x − x + log 1 + e x + C1 e x + – log 1 + e x + C2 e – x

= C1 ex + C2 e–x – 1 – xex + ex log (1 + ex) – e–x log (1 + ex) = C1 ex + C2 e–x – 1 – xex + (ex – e–x) log (1 + ex) This is the complete solution of the given equation. 6. Solve by the method of variation of parameters y″ + y = tan x. Solution. We have (D2 + 1) y = tan x A.E. is m2 + 1 = 0 i.e., m2 = – 1 i.e., m = ±i C.F. is C.F. = yc = C1cos x + C2 sin x ∴ y = A cos x + B sin x be the complete solution of the given equation where A and B are to be found We have y 1 = cos x and y2 = sin x

y1′ = – sin x

y2′ = cos x

W = y1 y2′ − y2 y1′ = cos x . cos x + sin x . sin x = cos2 x + sin2 x = 1

...(1)

287


φ (x) = tan x

Also,

bg

– y2 φ x W – sin x ⋅ tan x = 1

A′ =

A′ =

– sin 2 x cos x

A = – = – = –

z z zb

sin 2 x dx + C1 cos x

1 − cos2 x dx + C1 cos x

g

sec x − cos x dx + C1

A = – [log (sec x + tan x) – sin x] + C1 B′ =

bg

y1 φ x W

cos x ⋅ tan x 1 B′ = sin x

=

B =

z

sin x dx + C2

B = – cos x + C2 Substitute these values of A and B in Eqn. (1), we get y = {– log (sec x + tan x) + sin x + C1} cos x + {– cos x + C2} sin x y = C1 cos x + C2 sin x – cos x log (sec x + tan x) This is the complete solution. 7. Solve by the method of variation of parameters

d2y dy –2 + 2y = ex tan x. 2 dx dx Solution. We have (D2 – 2D + 2) y = ex tan x A.E. is

m2 – 2m + 2 = 0

i.e.,

m =

2± 4−8 2

m = 1±i Therefore C.F. is yc = ex (C1 cos x + C2 sin x) ∴

y = ex (A cos x + B sin x)

be the complete solution of the given equation

...(1)

288


where A and B are to be found y 1 = ex cos x

We have

and

y2 = ex sin x

. y1′ = ex (cos x – sin x), y2′ = ex (sin x + cos x) W = y1 y2′ − y2 y1′ = e2x Also, φ(x)= ex tan x A′ = = A′ =

bg

– y2 φ x W

– e x sin x ⋅ e x tan x e2 x – sin 2 x cos x

A = – = –

z zb

sin 2 x dx = – cos x

z

g

1 − cos2 x dx cos x

sec x − cos x dx

A = – log (sec x + tan x) + sin x + C1 B′ =

bg

y1 ⋅ φ x W

e x cos x ⋅ e x tan x e2 x B′ = sin x =

B =

z

sin x dx + C2

B = – cos x + C2 Substituting these values of A and B in Eqn. (1), we get y = {– log (sec x + tan x) + sin x + C1}ex cos x + {– cos x + C2} ex sin x y = ex (C1 cos x + C2 sin x) – ex cos x log (sec x + tan x) This is the complete solution of the given equation. 8. Solve by the method of variation of parameters

d2y dy –2 + y = ex log x. 2 dx dx Solution. We have (D2 – 2D + 1) y = ex log x A.E. is i.e., i.e.,

m2 – 2m + 1 = 0 (m – 1)2 = 0 m = 1, 1

Hence C.F. is yc = (C1 + C2 x) ex = C1 ex + C2 x ex ∴

y = Aex + Bxex

...(1)

289


y 1 = ex and y2 = xex

We have

y1′ = ex,

y2′ = xex + ex

W = y1 y2′ − y2 y1′ = xe2x + e2x – xe2x = e2x Also

φ(x) = ex log x A′ =

bg

– y2 φ x , W

B′ =

– xe x ⋅ e x log x e x ⋅ e x log x = e2 x e2 x A′ = – x . log x,

bg

– y1 φ x W

=

A= –

z

log x ⋅ x dx

LM N

x2 – 2

Integrating both these terms by parts, we get A = – log x ⋅

z

B′ = log x

OP Q

x2 1 ⋅ dx + C1 2 x

– x 2 log x x 2 + + C1 , A= 2 4 Substituting these values of A and B in Eqn. (1), we get

Thus,

2

x

1

y=

bC + C x g e

x

–

x 2 e x log x x 2 e x + + x 2 log x e x – x 2 e x 2 4

=

bC + C x g e

x

+

x 2 log x ⋅ e x 3 2 x – x e 2 4

y=

2

1

1

b

2

g

C1 + C2 x e x +

y″ + 4y′ + 4y = 4 +

b

m2 + 4m + 4 (m + 2)2 m yc

= = = =

g

x 2e x 2 log x – 3 . 4

e –2x ⋅ x

Solution. We have (D2 + 4D + 4) y = 4 +

C.F. is

x

2


A.E. is i.e.,

z

e –2 x x

0 0 – 2, – 2 e–2x (C1 + C2 x)

x⋅

1 dx + C2 x

B = x log x – x + C2

RS – x log x + x + C UV e + b x log x − x + C g xe 4 T 2 W 2

y=

B = log x ⋅ x –

290


y = C1e–2x + C2x e–2x ∴

y = Ae–2x + Bxe–2x

...(1)


y 1 = e–2x

y 2 = xe–2x

and

y1′ = – 2e–2x

y2′ = e–2x (1 – 2x)

W = y1 y2′ − y2 y1′ = e–2x · e–2x (1 – 2x) + xe–2x · 2 · e–2x W = e–4x A′ =

Also,

bg

– y2 φ x W

FG H

– xe −2 x ⋅ 4 + =

e −2 x x

e −4 x

IJ K

A = –

z

LM N

d4 xe

A = − 4x ⋅

2x

i

bg

F GH

e –2 x ⋅ 4 + =

F GH

+ 1 dx + C1 ,

OP Q

e2 x e2x – x + C1 , − 4⋅ 2 4

A = – 2xe2x + e2x – x + C1

B =

=

z z FGH

e –2 x x

e –4 x

2x 4+ B′ = e

A′ = – (4xe2x + 1) , On integrating, we get

e –2 x x – y1 φ x B′ = W

φ (x) = 4 +

e –2 x x

F GH

e2 x 4 + 4e2 x +

I JK

I JK I JK

e –2 x dx + C2 x

IJ K

1 dx + C2 x

B = 2e2x + log x + C2

Substituting these values of A and B in Eqn. (1), we get y = (– 2x e2x + e2x – x + C1) e–2x + (2e2x + log x + C2) xe–2x = (C1 + C2x) e–2x – 2x + 1 – xe–2x + 2x + xe–2x log x y = (C1 + C2x) e–2x + 1 + xe–2x(log x – 1). 10. Solve by the method of variation of parameters y″ + 2y′ + 2y = e–x sec3x. Solution. We have (D2 + 2D + 2) y = e–x sec3x A.E.. is

m2 + 2m + 2 = 0

–2± 4−8

i.e.,

m =

∴ C.F. is

yc = e–x

∴

= –1± i 2 (C1cos x + C2 sin x)

y = Ae–x cos x + Be–x sin x

...(1)

be the complete solution of the given equation where A and B are functions of x to be found

291


We have

y 1 = e–x cos x

y 2 = e–x sin x

and

y1′ = – e–x (sin x + cos x)

y2′ = e–x (cos x – sin x)

W = y1 y2′ − y2 y1′ = e–2x Also,

φ(x) = e–x sec3x A′ =

bg

– y2 φ x W −x

B′ = −x

–e

A = – A = –

z

3

B′ =

tan x sec 2 x dx + C1 ,

tan 2 x + C1 , 2

B =

Substituting these values of A and B in Eqn. (1), we get

e

z

−x

sec 2 x dx + C2

B = tan x + C2

F – tan x + C I e cos x + btan x + C g e sin x GH 2 JK e tan x sin x + e tan x sin x e bC cos x + C sin x g – 2 2

y =

W

cos x ⋅ e − x sec 3 x e– 2 x 2 B′ = sec x

sin x ⋅ e sec x e– 2 x A′ = – tan x sec2x, A′ =

bg

– y1 φ x

−x

1

–x

2

–x

=

–x

–x

1

2

Thus,

b

g

e – x tan x sin x 2 This is complete solution of the given equation. –x y = e C1 cos x + C2 sin x +

EXERCISE 6.1 Solve the following equations by the method of variation of parameters: 1.

d2y + y = tan2 x. dx 2

2.

d2y + y = x sin x. dx 2

3.

d2y dy – 5⋅ + 6 y = e4 x. 2 dx dx

4.

d2y + 4 y = 4 sec2 2x. dx 2

5. (D2 + D) y = x cos x.

b

g

Ans. y = C1 cos x + C2 sin x + sin x log sec x + tan x – 2

LMAns. y = C cos x + C sin x + 1 x sin x – 1 x cos xOP 2 4 N Q LMAns. y = C e + C e + 1 e OP 2 N Q Ans. y = C cos 2 x + C sin 2 x + sin 2 x log bsec 2 x + tan 2 x g − 1 LMAns. y = C + C e + 1 x bsin x − cos xg + cos x + 2 sin xOP 2 N Q 2

1

2

2x

1

1

2

–x

1

2

3x

2

4x

292


LMAns. y = C cos ax + C sin ax – 1 x cos ax + 1 sin ax log sin ax OP a a N Q

6. (D2 + a2) y = cosec ax.

1

2

Ans. y = C1 cos x + C2 sin x + x cos x − sin x + sin x log sec x

7. (D2 + 1) y = sec x tan x.

LMAns. y = bC + C xge N

8. (D2 + 2D + 1) y = e–x log x.

9. (D2 – 3D + 2) y =

2

1

–x

2

1 . 1 + e– x

+

g OPQ

b

1 2 −x 2 log x − 3 x e 4

d

i

d

Ans. y = C1e x + C2 e 2 x − xe x + e x log 1 + e x + e 2 x log 1 + e – x 10. (D2 – 6D + 9) y =

e3x . x2

LMAns. y = bC N

13. (D2 + 1) y =

1 . 1+ sin x

1

1

ex . x

15. (D2 + 6D + 9) y =

e –3 x . x5

+ C2 x e 3 x +

2

3x

2

1

14. (D2 – 2D + 1) y =

6.2

g

Ans. y = C1 + C2 x e 3 x – e 3 x log x

g 18 d2 x − 4 x + 3i e OPQ Ans. y = C cos x + C sin x + sin x log bsec x + tan xg + logcos x − 1 Ans. y = C cos x + C sin x – x cos x + sin x log b1 + sin x g – 1

11. (D2 – 2D + 1) y = x2 e3x. 12. (D2 + 1) y = log cos x.

b

i

2

b

g

Ans. y = C1 + C2 x e x + xe x log x

LMAns. y = bC N

1

g

+ C2 x e –3 x +

1 –3 –3 x x e 12

OP Q

SOLUTION OF CAUCHY’S HOMOGENEOUS LINEAR EQUATION AND LENGENDRE’S LINEAR EQUATION

A linear differential equation of the form

xn

n −1 n −2 dny y y dy n −1 d n− 2 d a x a x + ⋅ + + ⋅⋅⋅ + an −1 x ⋅ + an y = φ( x ) 2 1 dx dx n dx n −1 dx n − 2

...(1)

Where a1, a2, a3 ...an are constants and φ(x) is a function of x is called a homogeneous linear differential equation of order n. The equation can be transformed into an equation with constant coefficients by changing the independent variable x to z by using the substitution x = ez or z = log x Now

z = log x ⇒

dz 1 = dx x

293


dy dy dz = ⋅ = dx dz dx

Consider ∴

x

1 dy ⋅ x dz

dy dy = Dy = dx dz

d . dz Differentiating w.r.t. ‘x’ we get,

where D =

x

d 2 y dy + ⋅1 = dx 2 dx d2y dx 2

x

i.e.,

d 2 y dz ⋅ dz 2 dx

=

d 2 y 1 dy ⋅ – dz 2 x dx

=

1 d 2 y 1 dy ⋅ – ⋅ x dz 2 x dz

i.e.,

x2

d 2 y dy d2y − = dz 2 dz dx 2

i.e.,

x2

d2y = (D2 – D) y = D (D – 1) y dx 2

Similarly,

x3

d 3y = D (D – 1) (D – 2) y dx 3 ............................................................. .............................................................

xn

dny = D (D – 1) ... (D – n + 1) y dx n

n dy 2 d 2 y n d y ,x x ⋅⋅⋅⋅⋅ ⋅⋅⋅⋅ in Eqn. (1), it reduces to a linear dx dx 2 dx n differential equation with constant coefficient can be solved by the method used earlier. Also, an equation of the form,

Substituting these values of x

bax + bg

n

⋅

b

dny + a1 ax + b dx n

g

n −1

⋅

d n −1 y + ... any = ( x ) dx n −1

...(2)

where a1, a2 .....an are constants and φ (x) is a function of x is called a homogeneous linear differential equation of order n. It is also called “Legendre’s linear differential equation”. This equation can be reduced to a linear differential equation with constant coefficients by using the substitution. ax + b = ez or z = log (ax + b) As above we can prove that

bax + bg ⋅ dydx

= a Dy

294


bax + bg

2

bax + bg

n

d2y = a2 D (D – 1) y dx 2 ............................................................. ............................................................. ⋅

dny = an D (D – 1)(D – 2) ..... (D – n + 1) y dx n The reduced equation can be solved by using the methods of the previous section. ⋅

WORKED OUT EXAMPLES dy d2y − 2x − 4y = x4. 2 dx dx Solution. The given equation is

1. Solve x 2

d2y dy − 2x − 4 y = x4 2 dx dx Substitute x = ez or z = log x x2

So that

x

dy = Dy, dx

x2

...(1)

d2y = D (D – 1) y dx 2

The given equation reduces to D (D – 1) y – 2Dy – 4y = (ez)4 [D (D – 1) – 2D – 4] y = e4z i.e.,

(D2 – 3D – 4) y = e4z

...(2)

which is an equation with constant coefficients A.E. is i.e., ∴ C.F. is

m2 – 3m – 4 = 0 (m – 4) (m + 1) = 0 m = 4, –1 C.F. = C1e4z + C2e–z P.I. = =

1 e4 z 3 4 D − D− 2

b4 g

1 2

bg

–3 4 −4

=

1 ze 4 z 2D − 3

=

1 ze 4 z ( 2) 4 − 3

=

1 4z ze 5

bg

e4 z

D→ 4 Dr = 0 D→ 4

295


∴ The general solution of (2) is y = C.F. + P.I. y = C1 e4z + C2 e–z +

1 4z ze 5

Substituting ez = x or z = log x, we get –1 4 y = C1 x + C2 x +

d i

1 log x x 4 5

C2 x 4 log x + 5 x

4 y = C1 x +

is the general solution of the Eqn. (1).

d2y dy − 3x + 4y = (x + 1)2. dx dx 2 Solution. The given equation is 2 2. Solve x

d2y dy − 3x + 4 y = (x + 1)2 dx dx 2 Substituting x = ez or z = log x x2

x

Then

dy = Dy, dx

x2

...(1)

d2y = D (D – 1) y dx 2

∴ Eqn. (1) reduces to D (D – 1) y – 3 Dy + 4y = (ez + 1)2 i.e.,

(D2 – 4D + 4) y = e2z + 2ez + 1

which is a linear equation with constant coefficients. A.E. is i.e., ∴

m2 – 4m + 4 = 0 (m – 2)2 = 0 m = 2, 2 C.F. = (C1 + C2z) e2z P.I. =

=

b

1 D−2

e gd 2

e2 z

+

2z

i

+ 2e z + 1 2e z

+

...(2)

e0z

b D – 2g b D – 2 g b D – 2 g 2

2

2

= P.I.1 + P.I.2 + P.I.3 P.I.1 = = =

e2 z

b D – 2g

(D → 2)

2

e2 z

b2 – 2 g

ze 2 z 2 D–2

b

(Dr = 0)

2

g

(D → 2)

296


= P.I.1 = P.I.2 = =

ze 2 z 2 2−2

b

g

(Dr = 0)

z2e2z 2 2e z

b D − 2g

2

(D → 1)

2

(D → 0)

2e z

b− 1g

2

P.I.2 = 2ez P.I.3 = =

e0 z

b D − 2g e0z 1 = 4 4

1 z 2 2z e + 2e z + 2 4 The general solution of Eqn. (2) is y = C.F. + P.I. P.I. =

y =

bC + C z g e

y =

bC + C log xg x

1 z 2 e2 z + 2e z + 2 4 ez = x or z = log x, we get

Substituting

1

1

2z

2

2

+

2

+

b g

x 2 log x 2

2

+ 2x +

1 4

is the general solution of the equation (1).

d2y dy + 2x − 12y = x2 log x. 2 dx dx Solution. The given Eqn. is

2 3. Solve x

d2y dy + 2x − 12 y = x2 log x 2 dx dx or Substituting x = ez x2

x

dy = Dy, dx

...(1) z = log x, so that

2 and x

d2y = D (D – 1) y dx 2

Then Eqn. (1) reduces to D (D – 1) y + 2 Dy – 12y = e2zz i.e.,

(D2 + D – 12) y = ze2z

which is the Linear differential equation with constant coefficients. A.E. is

m2 + m – 12 = 0

...(2)

297


i.e.,

(m + 4) (m – 3) = 0

∴

m = – 4, 3 C.F. = C1e–4z + C2e3z P.I. =

1 ze 2 z D + D – 12 2

2z = e

= e2z

z

b D + 2g + b D + 2g − 12 LM z OP N D + 5D − 6 Q

(D → D + 2)

2

2

1 5 − z− 6 36 – 6 + 5D + D2

z z–

5 6

5 6 5 6 0

LM N

2z P.I. = e −

OP Q

LM N

5 – e2z z 5 − z+ = 6 36 6 6

∴ General solution of Eqn. (2) is y = C.F. + P.I.

FG H

5 e2 z z+ 6 6 z e = x or z = log x, we get − 4z + C2 e 3z – y = C1e

Substituting

–4 3 y = C1x + C2 x –

FG H

IJ K

5 x2 log x + 6 6

FG H

5 C1 x2 log x + + C2 x 3 – 4 6 6 x which is the general solution of Eqn. (1). y =

OP Q

IJ K

IJ K

d2y dy –x + y = 2 log x. 2 dx dx Solution. The given equation is 4. Solve x 2

x2

d2y dy –x +y 2 dx dx

= 2 log x

...(1)

298


x = ez

Substitute

x

Then

or

z = log x, in Eqn. (1) 2

dy = Dy, dx

and

x2

d y = D (D – 1) y dx 2

Eqn. (1) reduces to, D (D – 1) y – Dy + y = 2z (D2 – 2D + 1) y = 2z

i.e.,

...(2)

which is an equation with constant coefficients m2 – 2m + 1 = 0

A.E. is

(m – 1)2 = 0

i.e., ∴

m = 1, 1 C.F. = (C1 + C2z) ez P.I. =

1 ⋅ 2z D − 2D + 1 2

2z + 4 1 – 2D + D2

2z 2z – 4 4 4 0

P.I. = 2z + 4 ∴ The general solution of Eqn. (2) is y = (C1 + C2z) ez + 2z + 4 Hence the general solution of Eqn. (1) is y = (C1 + C2 log x) · x + 2 log x + 4.

d2y dy – 4x + 6y = cos (2 log x). 2 dx dx Solution. The given equation is

5. Solve x 2

d2 y dy – 4x + 6y = cos (2 log x) 2 dx dx Substituting x = ez or z = log x, we have x2

x

...(1)

2 dy 2 d y = Dy, and x = D (D – 1) y dx dx 2

∴ Eqn. (1) reduces to D(D – 1)y – 4Dy + 6y = cos 2z i.e., A.E. is

(D2 – 5D + 6) y = cos 2z 2

m – 5m + 6 = 0

...(2)

299


i.e.,

(m – 2) (m – 3) = 0

∴

m = 2, 3 C.F. = C1e2z + C2e3z P.I. =

1 cos 2 z D − 5D + 6

D2 → – 22

2

=

1 cos 2 z – 2 − 5D + 6

=

1 cos 2 z – 5D + 2

=

2 + 5D 1 cos 2 z × 2 – 5D 2 + 5D

=

2 cos 2 z − 10 sin 2 z 4 − 25D2

=

2 cos 2 z − 10 sin 2 z 104

2

b

1 cos 2z – 5 sin 2 z 52 The general solution of Eqn. (2) is y = C.F. + P.I. P.I. =

2z 3z = C1 e + C 2 e +

D2 → – 22

g b

1 cos 2 z − 5 sin 2 z 52

g

Hence the general solution is y = C1 x 2 + C2 x 3 +

b

b

g

b

g

1 cos 2 log x – 5 sin 2 log x . 52

g

d2y dy +x + 2y = x cos log x . 2 dx dx Solution. The given equation is 6. Solve x 2

d2y dy +x + 2 y = x cos (log x) 2 dx dx Substitute x = ez or z = log x, x2

x

Then we have

...(1)

dy d2y = Dy and x 2 = D (D – 1) y dx dx 2

∴ Eqn. (1) reduces to D (D – 1) y + Dy + 2y = ez cos z i.e., A.E. is i.e.,

(D2 + 2) y = ez cos z 2

m + 2 = 0 m 2 = – 2;

...(2)

300


⇒

m 2 = 2i2

i.e.,

m = ± 2i

C.F. = C1 cos 2 z + C2 sin 2 z P.I. =

1 e z cos z D +2

z = e

= ez

1

b D + 1g

2

+2

cos z

1 cos z D + 2D + 3

(D2 → –12)

2

LM cos z OP N –1 + 2 D + 3Q L cos z OP e M N 2D + 2 Q e L cos z D − 1O M × P 2 N D + 1 D − 1Q e L – sin z − cos z O M D − 1 PQ 2 N e L – sin z − cos z O M – 2 PQ 2 N e bsin z + cos zg 4

= ez

=

(D → D + 1)

2

2

z

z

=

z

=

(D2 → –12)

2

z

=

z

=

∴ General solution of Eqn. (2) is y = C.F. + P.I. = C1 cos 2 z + C2 sin 2 z +

b

ez sin z + cos z 4

g

∴ The general solution of Eqn. (1), as y = C1 cos 2 log x + C2 sin 2 log x +

b

g

b g

b

g

x sin log x + cos log x . 4

sin log x d2y dy y = 5– · – +3 2 x2 dx x dx Solution. Multiplying throughout the equation by x, we get 7. Solve 2x

b

g

sin log x d2y dy + 3x − y = 5x – 2 dx x dx z Substitute x = e or z = log x, in Eqn. (1) 2x2

...(1)

301


Then we obtain, {2D (D – 1) + 3D – 1} y = 5ez – e–z sin z i.e., A.E. is i.e., ∴

(2D2 + D – 1) y = 5ez – e–z sin z

...(2)

2

2m + m – 1 = 0 (m + 1) (2m – 1) = 0 1 2

m = – 1,

FG 1 IJ z

C.F. = C1e −2 + C2 e H 2 K P.I. = =

d

1 5e z − e – z sin z 2D2 + D − 1

i

1 1 5e z – e – z sin z 2 2D + D − 1 2D + D – 1 2

= P.I.1 – P.I.2

5e z P.I.1 = 2 D2 + D – 1 = P.I.2 =

(D → 1)

5e z 2 1 e – z sin z 2D + D − 1

−z = e

= e−z

= e−z = e−z

(D → D – 1)

2

1

b

g b g LM sin z OP N 2 D − 3D Q LM sin z OP MN 2 d–1 i − 3D PQ 2

2 D −1 + D −1 −1 2

sin z (D2 → – 12)

2

sin z – 2 − 3D

−z = −e

= – e−z −z = –e

LM sin z OP N 3D + 2 Q LM sin z × 3D − 2 OP N 3D + 2 3D − 2 Q LM 3 cos z − 2 sin z OP N 9D − 4 Q 2

2

(D2 → –12)

302


LM 3 cos z − 2 sin z OP N – 13 Q e b3cos z − 2 sin zg 13

= – e−z −z

=

b

5 z 1 −z e + e 3 cos z − 2 sin z 2 13 Complete solution of Eqn. (2) is P.I. =

g

y = C.F. + P.I.

FG 1 IJ z

y = C1e − z + C2 e H 2 K +

b

g

b g

b g

5 z 1 –z e + e 3 cos z − 2 sin z 2 13

∴ The general solution of Eqn. (1) is y = 3 8. Solve x

C1 + C2 x

x+

FG H

5 1 3 cos log x – 2 sin log x . x+ 2 13x

IJ K

d3y d2y 1 + 2x 2 + 2y = 10 x + · 3 2 x dx dx

Solution. The given equation

FG H

IJ K

d 3y d2y 1 + 2x2 + 2 y = 10 x + 2 3 x dx dx Substitute x = ez or z = log x x3

x

Hence,

dy = Dy, dx

x2

...(1)

d2y = D (D – 1) y dx 2

d 3y = D (D – 1)(D – 2) y dx 3 Eqn. (1) reduces to linear differential equation as x3

FG H

z [D (D – 1)(D – 2) + 2D (D –1) + 2] y = 10 e +

i.e.,

d

z −z (D3 – D2 + 2) y = 10 e + e

i

1 ez

IJ K

which is linear differential equation with constant coefficients A.E. is m3 – m2 + 2 = 0 i.e., (m + 1) (m2 – 2m + 2) = 0 Hence Therefore,

m = – 1, 1 ± i C.F. = C1e–z + ez (C2 cos z + C3 sin z) P.I. =

d

1 10 e z + e – z D3 – D2 + 2

i

303


= 10

RS TD

1 1 ez + 3 ⋅ e– z 2 2 D − D +2 − D +2

3

= 10 [P.I.1 + P.I.2] P.I.1 = = P.I.2 = =

=

=

P.I.2 =

1 ez D3 − D2 + 2

UV W

D→1

ez 2 1 e– z 2 D −D +2

D → –1

3

1

b –1g − b– 1g 3

2

+2

e−z

Dr = 0

1 ze − z 3D − 2 D

D→– 1

2

ze – z

b g

2

b g

3 –1 − 2 –1 ze – z 5

RS e + ze UV T2 5 W z

P.I. = 10

–z

Hence the complete solution is y = C.F. + P.I.

b

g

Substituting

ez = x or z = log x

We get

y =

m

b g

LM e + ze OP N2 5 Q z

= C1e – z + e z C2 cos z + C3 sin z + 10

−z

b gr

2 log x C1 + x C2 cos log x + C3 sin log x + 5x + ⋅ x x

b

g

2 d3y dy 3 d y 2x – x2 + + xy = sin log x . 3 2 dx dx dx Solution. Dividing throughout the equation by ‘x’, we get

9. Solve x 4

b g

2 sin log x d 3y dy 2 d y 2 –x x + +y = 2 3 x dx dx dx Now substitute x = ez and z = log x

x3

x

2 dy 2 d y = Dy, x = D (D – 1) y, dx dx 2

...(1)

304


x3

d 3y = D (D – 1)(D – 2) y dx 3

Eqn. (1) reduces to sin z ez

[D (D – 1)(D – 2) + 2D (D –1) – D + 1] y = i.e.,

(D3 – D2 – D + 1) y = e–z sin z m3 – m2 – m + 1 = 0

A.E. is

⇒ m2 (m – 1) – 1 (m – 1) = 0 ⇒

(m2 – 1) (m – 1) = 0

⇒

m2 – 1 = 0,

⇒

m–1=0

m = ± 1, 1

⇒

m = – 1, + 1, + 1

C.F. = (C1 + C2 Z.) ez + C3 e–z and P.I. = –z

Taking e

1

b D − 1g b D + 1g 2

e – z sin z (D → D – 1)

outside and replacing

1

–z = e

b D − 2g D

–z = e

dD

–z = e

2

1 2

sin z

i

+ 4 – 4D D

sin z

sin z D + 4D – 4D2

(D2 → –12)

3

sin z –z = e − D + 4D + 4 –z = e

4D – 3 sin z × 4D + 3 4D − 3

–z = e

LM 4 cos z – 3sin z OP N 16D − 9 Q

–z = e

4 cos z – 3 sin z – 25

=

(D2 → –12)

2

b

– 1 –z e 4 cos z − 3 sin z 25

g

∴ The complete solution is y = C.F. + P.I. =

bC + C z g e 1

2

z

+ C3e – z +

b

1 –z e 4 sin z − 3 cos z 25

g

305


x = ez

z = log x

bC

=

b

g

1

b

g

+ C2 log x ⋅ x +

b

g

b

g

1 C3 4 sin log x – 3 cos log x . + 25x x

g

dy d2y 2 + 3 3x + 2 − 36y = 3x + 4x + 1. 2 dx dx Solution. The given equation is 10. Solve 3x + 2

b3x + 2g Substitute

2

⋅

2

⋅

b

g

dy d2y + 3 3x + 2 − 36 y = 3x2 + 4x + 1 2 dx dx 3x + 2 = ez or z = log (3x + 2) (3 x + 2)

So that

b3x + 2g

2

...(1)

dy = 3Dy dx

d2y = 32 D (D – 1) y dx 2

ez – 2 3 Substituting these values in Eqn. (1), we get

Also,

x =

F e – 2I 3G H 3 JK z

3 D (D – 1) y + 3.3 Dy – 36y = 2

9 (D2 – 4) y =

2

F e – 2I + 1 GH 3 JK z

+4

e2z − 1 3

d

i

1 2z e −1 27 Which is a linear differential equation with constant coefficients

i.e.,

Now the A.E.

(D2 – 4) y =

...(2)

m2 – 4 = 0 Whose roots are m2 = 4 m = ±2 C.F. = C1e2z + C2e–

2z

and

d

i

1 1 2z P.I. = D 2 − 4 27 e − 1

LM e O 1 − e P MN b D − 2gbD + 2g D − 4 PQ 2z

=

1 27

=

1 P.I.1 − P.I.2 27

P.I.1 =

b

oz

2

e2z D−2 D+2

gb

g

(D → 2) (Dr = 0)

306


=

ze 2 z 2D

=

ze 2 z 4

(D → 2)

ze oz P.I.2 = D2 − 4 = P.I. = P.I. =

(D → 0)

1 −4

LM 1 ze + 1 OP 4Q N4 1 d ze + 1i 108 1 27

2z

2z

General solution is y = C.F. + P.I. –2 z 2z = C1e + C2 e +

b

y = C1 3x + 2

g

2

d

i

1 ze 2 z + 1 108

b

+ C2 3 x + 2

g

–2

+

b

1 3x + 2 108

g

2

b

g

log 3x + 2 + 1 .

EXERCISE 6.2 Solve the following equations: 2 1. x

LMAns. = L cos F 3 log I + sin F 3 log I O + OP MN y x MMNC GH 2 xJK C GH 2 xJK PPQ x PQ LMAns. y = x bC + C log xg + x OP d y dy + 5x + 4y = x . 36 Q dx dx N LMAns. y = C x + C x – L x + x + 1 OOP d y dy – 20 y = (x + 1) . + 2x MN 14 9 20 PQPQ dx dx MN d2y + y = 3x2. dx 2

2

1

2

2

2. x

2

1

2

2 4. x

4

–5

2

1

2

2

d2y dy –x + 2 y = x sin (log x). 2 dx dx

LMAns. y = x C cos blog xg + C sin blog xg – x log x cos blog xg OP 2 N Q LMAns. y = x C + C log x + x + x blog xg OP d y dy 1 + 3x + y = x+ . dx x dx 4 2 MN PQ 1

2

2

2 5. x

2

2

2

2 3. x

4

−2

4

2

−1

–1

1

2

2

307


2 6. x

dy d2y + 4x + 2 y = ex. 2 dx dx

Ans. y = C1 x –1 + C2 x −2 e x

dy d2y –x + 4 y = cos (log x) + x sin (log x). 2 dx dx 1 x 3 cos log x − 2 sin log x + sin log x Ans. y = x C1 cos 3 log x + C2 sin 3 log x + 13 2

2 7. x

LM N

e

2 8. x

j

e

b g

j

dy d2y – 3x + 5 y = x2 sin (log x). 2 dx dx

LMAns. y = x N

2

b g

b

2 11. x

d2y dy + 5x + 4 y = x log x. 2 dx dx

3 12. x

2 d3y dy 2 d y 3 x + +x + 8 y = 65 cos (log x). 2 3 dx dx dx –2

1

15.

–1

1

b

g

+ x C2 cos 3 log x + C2 sin

b2 x + 1g

b g

2

e

b2 x – 1g

2

b gOQP

j

b g

b g

x2 log x cos log x 2

3 log x + 8 cos log x – sin log x

b g

C1 cos log x + C2 sin log x –

b gOP Q

d2y dy + 1+ x + y = 4 cos log (1 + x). 2 dx dx Ans. y = C1 cos log 1 + x + C2 sin log 1 + x + 2 log 1 + x sin log 1 + x 2

⋅

b g

b

g

b g

3

b g

d2y dy – 2 2x + 1 – 12 y = 16x. dx dx 2

LMAns. y = C b2 x + 1g N

3

1

16.

–1

3

1

LMAns. y = x N

2

2

–2

d2y dy – 3x + 5 y = x2 sin (log x). 2 dx dx

b1 + xg

g

LMAns. y = x bC + C log xg + C x + 1 x log x OP 4 N Q LMAns. y = x bC + C log xg + x FG log x – 2 IJ OP 9H 3K Q N

2 d 3y dy 3 d y 2 – x2 x + + xy = 1. 2 3 dx dx dx

14.

b gOP Q

x2 log x cos log x 2

Ans. y = C1 + C2 log x x + log x + 2

4 10. x

2 13. x

b g

C1 cos log x + C2 sin log x –

9. x2y″ – xy′ + y = log x.

LMAns. y = C x N

b gOPQ

b g

b

g

b

g

+ C2 2 x − 1

–1

−

b g

b

g OPQ

b

g

3 1 2x + 1 + 16 4

d 3y dy + 2x – 1 + 2 y = 0. 3 dx dx

LMAns. y = 2 x − 1 LC + C 2 x − 1 b g MM b g MN N 1

2

3 2

+ C2 2 x − 1

– 3 2

OPOP PQPQ

308


6.3 SOLUTION OF INITIAL AND BOUNDARY VALUE PROBLEMS The differential equation in which the conditions are specified at a single value of the independent variable say x = x0 is called an Initial Value Problem (IVP). If y = y(x), the initial conditions usually will be of the form. y(x0) = x0, y′(x0) = y1, ... y(n –1) (x0) = yn–1 The differential equation in which the conditions are specified for a given set of n values of the independent variables is called a Boundary Value Problem (BVP). If

y = y(x) the n boundary conditions will be y (x1) = y1 ,

y(x2) = y2 ,

y(x3) = y3 ... y(xn) = yn .

We can also have problems involving a system of d.e. (simultaneous de.s) with these type of conditions.

WORKED OUT EXAMPLES 1. Solve the initial value problem

d 2x dx +5 + 6x = 0, given that 2 dt dt

bg

dx 0 = 15. dt Solution. We have (D2 + 5D + 6) y = 0 x (0) = 0,

A.E..

m2 + 5m + 6 = 0 (m + 2) (m + 3) = 0

⇒

m = – 2, – 3

Therefore general solution is x = x (t) = C1e–2t + C2e–3t

...(1)

This is the general solution of the given equation Now, consider,

x (0) = 0

Eqn. (1), becomes

x (0) = C1(1) + C2(1) C1 + C2 = 0

i.e.,

...(2)

Also we have from Eqn. (1), dx = – 2C1e–2t– 3C2e–3t dt

Applying the conditions,

bg

dx 0 = 15 dt

We obtain – 2C1 – 3C2 = 15 Solving equations (2) and (3), we get C1 = 15, C2 = – 15 thus

x (t) = 15 (e–2t – e–3t).

...(3)

309


d2y dy +4 + 3y = e–x subject to the conditions y(0) = y′(0). 2 dx dx Solution. We have (D2 + 4D + 3) y = e–x 2. Solve

A.E. or

m2 + 4m + 3 = 0 (m + 1) (m + 3) = 0 m = – 1, – 3 C.F. = C1e–x + C2e–3x P.I. =

=

=

e–x D2 + 4 D + 3

D→– 1

e– x

b –1g + 4b –1g + 3

Dr = 0

2

x e– x 2D + 4

D→– 1

x e– x 2 y = C.F. + P.I. =

y = C1 e – x + C2 e –3 x +

x e–x 2

...(1)

d

dy 1 = – C1 e – x – 3 C2 e –3 x + – x e– x + e− x dx 2 Consider the conditions y(0) = 1 and y′(0) = 1

y′ =

i

...(2)

Eqn. (1) and (2) become, 1 = C1 + C2 and 1 = – C1 – 3 C2 +

1 . 2

By solving these equations we get, C1 = Thus

y =

7 –3 and C2 = 4 4

7 – x 3 –3 x x e − x e − e + is the particular solution. 4 4 2

d2x dx +6 + 25x = 0. If the particle 2 dt dt is started at x = 0 with an initial velocity of 12 ft/sec to the left, determine x in terms of t. 3. A particle moves along the x-axis according to the law

Solution. We have (D2 + 6D + 25) x = 0 From the given data, the initial conditions x = 0 when t = 0 and

dx = – 12 when t = 0 dt

310


A.E.

m2 + 6m + 25 = 0 m = – 3 ± 4i

∴

x = x(t) = e–3t (C1 cos 4t + C2 sin 4t) –3t

Now

x′(t) = + 3e

(– C1 sin 4t · 4 + 4 C2 cos 4t)

–3t

= – 12 e Consider

...(1)

(C1 cos 4t – C2 sin 4t)

...(2)

x (0) = 0 and x′(0) = – 12

Eqns. (1) and (2) become, 0 = C1 and – 12 = 4C2 – 3C1 ∴

C = 0 and C2 = – 3 x (t) = – 3e–3t sin 4t.

EXERCISE 6.3 Solve the following initial value problems: 1.

d2y dy 2 +6 + 9 y = 12e–3x = y (0) =0, = y ′ (0) 2 dx dx

Ans. y = e –3x ⋅ x 4

LMAns. y = 1 bcos h x + cos x gOP 2 N Q LMAns. y = 1 d– 5 + e + 4 cos t – 2 sin t iOP 10 N Q

d4y – y = 0; y (0) = 1 and y′(0) = 0 = y″(0) = y ′′′ ( 0). 2. dx 4

bg

3. y ′′′( t ) + y ′ t = e2t, y(0) = 0 = y′(0) = y ′′′ ( 0 ).

2t

ADDITIONAL PROBLEMS (From Previous Years VTU Exams.) 1. Using the method of variation of parameters solve: Solution. We have (D2 + 1) y =

d2y 1 +y = 2 1 + sin x dx

1 1 + sin x

m2 + 1 = 0 and hence m = ± i C.F. = C1 cos x + C2 sin x y = A(x) cos x + B(x) sin x be the complete solution of the given d.e. where A(x) and B(x) are to be found. y2 = sin x We have y 1 = cos x A.E. is ∴

y1′ = – sin x W = y1 y2′ − y2 y1′ = 1

y2′ = cos x Also

φ(x) =

1 1 + sin x

...(1)

311


bg

Now,

A′ =

– y2 φ x W

i.e.,

A′ =

– sin x 1 + sin x

consider

A′ = A =

b

and and

g = –1 +

– 1 + sin x − 1

z LMN

1 + sin x –1 +

= – x+ = – x+

B′ =

OP Q

B′ =

bg

y1 φ x W

cos x 1 + sin x

1 sin x

1 dx + k1 1 + sin x

z zd

1 − sin x dx + k1 cos2 x

i

sec 2 x − sec x tan x dx + k1

A = – x + tan x – sec x + k1 Also,

B′ = B = =

b

...(2)

g

cos x 1 + sin x cos x 1 − sin x = = 2 1 + sin x cos x cos x

z zb

1 − sin x dx + k 2 cos x

g

sec x − tan x dx + k 2

= log (sec x + tan x) + log (cos x) + k2 = log

FG 1 + sin x IJ + log bcos x g + k H cos x K

2

= log (1 + sin x) – log (cos x) + log (cos x) + k2 B = log (1 + sin x) + k2 ...(3) Using Equations (2) and (3) in (1), we have y = [– x + tan x – sec x + k1] cos x + [log (1 + sin x) + k2] sin x i.e., y = k1 cos x + k2 sin x – x cos x + sin x – 1 + sin x log (1 + sin x) The term sin x can be neglected in view of them k2 sin x present in the solution Thus, y = k1 cos x + k2 sin x – (x cos x + 1) + sin x log (1 + sin x). 2. Solve by the method of variation of parameters

d2y + 4y = 4 tan 2x. dx 2

Solution. Refer page no. 282. Example 2. 3. Solve

d2y dy −2 + 2y = ex tan x using the method of variation of parameters 2 dx dx

Solution. Refer page no. 287, Example 7.

312


d2y + a 2 y = tan ax. 4. Solve 2 dx Solution. We have (D2 + a2) y = tan ax A.E. is m2 + a2 C.F. : be the complete solution We have y1

= 0 ⇒ m = ± ia = C1 cos ax + C2 sin ax of the d.e. where A and B are functions of x to be found. = cos ax y2 = sin ax

y1′ = – a sin ax

y2′ = a cos ax

W = y1 y2′ − y2 y1′ = a

φ(x) = tan ax

Also

bg

– y2 φ x W

B′ =

y1 φ x W

A′ =

– 1 sin 2 ax a cos ax

B′ =

cos ax tan ax a

A′ =

2 – 1 1 − cos ax a cos ax

B′ =

sin ax a

d

i

zb

A =

y = Thus

RS 1 Ta

2

z

g

1 B = cos ax − sec ax dx + k1 , a 1 ⇒ A = 2 sin ax − log sec ax + tan ax + k1 a cosax + k2 B = – a2 Substituting these values in Eqn. (1), we get ⇒

bg

A′ =

b

g

sin ax dx + k 2 a

UV W

RS T

sin ax − log sec ax + tan ax + k1 cos ax + –

y = k1 cos ax + k 2 sin ax −

b

g

UV W

cos ax + k 2 sin ax a2

1 log sec ax + tan ax cos ax . a2

5. Using the method of variation of parameters find the solution of

ex x A.E. is m2 – 2m + 1 = 0 ⇒ (m – 1)2 = 0 m = 1, 1 are the roots of A.E. ∴ C.F. = (C1 + C2 x) ex y = (A + Bx) ex where A = A(x), B = B(x) be the complete solution of the d.e. and we shall find A, B, we have y2 = x ex y1 = ex,

d2y dy ex −2 +y= · 2 dx x dx

Solution. We have (D2 + 2D + 1) y =

...(1)

313


y1′ = ex

∴

y2′ = (x + 1) ex

2x W = y1 y2′ − y2 y1′ = e Also

Further, we have

A′ =

bg

– y2 φ x W – xe x ⋅

A′ =

and

⇒

A =

e x

bg

y1 φ x W ex ⋅

B′ =

e2 x

z

B′ =

ex x

x

ex x

e2 x

1 B′ = x

A′ = – 1

i.e.,

φ(x) =

z

1 dx + k 2 x B = log x + k2.

–1⋅ dx + k1

B =

i.e., A = – x + k1 Using these values in Eqn. (1), we have y = (– x + k1) ex + (log x + k2) x ex i.e., y = (k1 + k2 x) ex + (log x – 1) x ex x The term – xe can be neglected in view of the term k2 xex present in the solution. Thus y = (k1 + k2 x) ex + x log x ex. 3 6. Solve x

d3y d2y dy + 3x 2 +x + 8 y = 65 cos (log x). 2 3 dx dx dx

Solution. Put Thus, we have

t = log x or x = et

xy′ = Dy, n 2 y ′′ = D (D – 1) y x 3 y ′′′ = D (D – 1) (D – 2) y

where D =

Hence, the given d.e. becomes [D (D – 1) (D – 2) + 3D (D – 1) + D + 8] y = 65 cos t i.e., = (D3 – 3D2 + 2D + 3D2 – 3D + D + 8) y = 65 cos t or (D3 + 8) y = 65 cos t ⇒ m3 – 23 = 0 A.E. : m3 + 8 = 0 (m + 2) (m2 – 2m + 4) = 0 m = – 2 and m2 – 2m + 4 = 0 By solving m2 – 2m + 4 = 0, we have m =

2 ± 4 − 16 2 ± 2i 3 = = 1± i 3 2 2

o

t –2 t C.F. = C1e + e C2 cos 3t + C3 sin 3t

t

d dt

314


Also

P.I. =

65 cos t D3 + 8

P.I. =

65 cos t 65 8 + D cos t , = –D + 8 64 − D2

=

D2 → –12 = –1

b

g

b

65 8 cos t − sin t

g

65 P.I. = 8 cos t – sin t Complete solution : y = C.F. + P.I. with

=

{

C1 + x C2 cos x

D2 → –1

e

t = log x, et = x

j

3 log x + C3 sin

e

3 log x

j}

+ 8 cos (log x) – sin (log x). 2 7. Solve x

d3y d 2 y dy 3x + + = x2 log x. dx 3 dx 2 dx

Solution. Multiplying the equation by x, we have

x3 put then

d 3y d2y dy 3 x + +x = x3 log x 3 2 dx dx dx t = log x or

...(1) x = et

x 2 y ′′ = D (D – 1) y

xy′ = Dy,

x 3 y ′′′ = D (D – 1) (D – 2) y Hence Eqn. (1) becomes [D (D – 1) (D – 2) + 3D (D – 1) + D] e3t . t i.e., D3 y = 0 A.E. : m 3 = 0 and hence m = 0, 0, 0 ∴ C.F. = (C1 + C2 t + C3 t2) eot C.F. = C1 + C2 t + C3 t2 P.I. =

e 3t t t , = e 3t 3 2 D D−3

b

3t = e

t D 3 + 9 D 2 + 27 D + 27

P.I.3 is found by division

27 + 27D + 9D2 + D3

g

t 1 − 27 27 t t +1

–1 –1 0

D→D+3

315


3t P.I. = e ⋅

bt − 1g

27 The complete solution : y = C.F. + P.I. with t = log x and Thus

{C + C log x + C blog xg } + 27x blog x − 1g .

b

g

2

b

1

2

3

g

dy d2y − 2x + 3 − 12y = 6x. 2 dx dx

Solution. Put

t = log (2x + 3)

Hence

x =

Also, we have and

3

2

y =

8. Solve 2x + 3

x = et

b2 x + 3g y′ b2 x + 3g y ′′ 2

d

or et = 2x + 3

i

1 t e −3 2

= 2Dy = 22 D (D – 1) y

Hence, the given d.e. becomes

d

i

1 t e −3 2 i.e., 2 (2D2 – 2D – D – 6) y = 3 (et – 3) [4D (D – 1) – 2D – 12] y = 6 ⋅

i.e., A.E. is

d

i

3 t e −3 2 2m2 – 3m – 6 = 0

(2D2 – 3D – 6) y =

m =

∴ ∴

Also

3 ± 9 + 48 3 ± 57 = 4 4

C.F. = C1 e C.F. =

P.I. =

3t e4

d

e3+ 57 j t 4

LM MNC e 1

+ C2 e

57 t 4

e3+ 57 j t

+ C2 e

3e t

2 2 D 2 − 3D − 6

4

−

57 t 4

OP PQ

−

i 2 d2 D

3e t 9e ot − = 2 2 − 3− 6 2 0 − 0− 6

b

3e t 3 + 14 4 y = C.F. + P.I. t = log (2x + 3), et = 2x + 3

P.I. = − Complete solution with

g b

9e ot 2

g

i

− 3D − 6

316

ENGINEERING MATHEMATICS—II 3t

R| S| T

y = e 4 C1 e

Thus,

57 t 4

+ C2 e

– 57 t 4

where t = log (2x + 3) and et = 2x + 3. 9. Solve the initial value problem:

U| 3 V| − 14 e W

t

+

3 4

dy d2y dy = 1 at x = 0 . +4 + 5y + 2cos hx = 0, given y = 0, 2 dx dx dx

Solution. We have (D2 + 4D + 5) y = – 2 cos hx i.e., (D2 + 4D + 5) y = – (ex + e–x) A.E. is m2 + 4m + 5 = 0 – 4 ± 16 − 20 = –2±i 2 C.F. = e–2x (C1 cos x + C2 sin x)

m =

∴

P.I. =

– ex e– x − 2 D + 4D + 5 D + 4D + 5 2

=

– ex e–x − 1+ 4 + 5 1+ 4 + 5

=

– e x e– x − 10 2

LM e + e OP N10 2 Q x

P.I. = –

–x

Complete solution: y = C.F. + P.I.

g FGH

b

e x e– x –2 x + C1 cos x + C2 sin x − y = e 10 2 Now, we apply the given initial conditions, y = 0,

I JK

...(i)

dy = 1 at x = 0 dx

From Eqn. (i), we get

b

g

b

dy = e –2 x – C1 sin x + C2 cos x – 2e –2 x C1 cos x + C2 sin x dx Using

y = 0 at x = 0, Eqn. (i) becomes 0 = C1 –

Using

FG 1 + 1 IJ H 10 2 K

or

C1 =

3 5

dy = 1 at x = 0, Eqn. (ii) becomes dx 1 = C2 − 2C1 −

1 1 + 10 2

or

C2 − 2 C1 =

3 5

g

−

e x e– x + 10 2 ...(ii)

317


3 9 , we get C2 = 5 5 Thus, the required particular solution from Eqn. (i) is given by Using

C1 =

y =

g FGH

b

I JK

3 –2 x e x e– x · e cos x + 3 sin x − + 5 10 2

dy d2y dy = 2, y = 1 at x = 0. –4 + 5y = 0 . Subject to the conditions 2 dx dx dx Solution. We have (D2 – 4D + 5) y = 0 A.E. : m2 – 4m + 5 = 0

10. Solve

m =

4 ± 16 − 20 4 ± 2i = = 2 ±i 2 2

∴ ∴

C.F. = e2x (C1 cos x + C2 sin x) y = e2x (C1 cos x + C2 sin x)

Also

dy dx

Consider

= e2x (– C1 sin x + C2 cos x) + 2e2x (C1 cos x + C2 sin x) y = 1 1 = 1 (C1 + 0)

Also by the condition

Using Thus

dy dx

...(1) ...(2)

at x = 0, Eqn. (1) becomes ∴ C1 = 0

= 2 at x = 0, Eqn. (2) becomes

2 = C2 + 2C1 C1 = 1, we get C2 = 0 y = e2x (cos x) is the particular solution.

11. Solve the initial value problem

bg

d2y + y = sin (x + a) satisfying the condition y = (0) = 0, dx 2

y ′ 0 = 0. Solution. We have (D2 + 1) y = sin (x + a) A.E. : m2 + 1 = 0 and hence m = ± i ∴ C.F. = C1 cos x + C2 sin x P.I. =

b

sin x + a

D2 + 1 The denominator becomes zero P.I. = x = P.I. =

g

D2 → – 12 = – 1

b

g

b

g

sin x + a D × 2D D

x cos x + a 2 2 D2

b

– x cos x + a 2

D2 → – 12 = – 1

g

318


∴ The complete solution is y = C.F. + P.I.

b

– x cos x + a 2

y = C1 cos x + C2 sin x

b

Using

g

...(1)

b

cos x + a x sin x + a − 2 2

y ′ = – C1 sin x + C2 cos x

Now,

g g

...(2)

y (0) = 0, y ′ (0) = 0 in Eqns. (1) and (2) respectively, we have

cos a 2 Thus by using these values in Eqn. (1), we get the particular solution, C1 = 0 and C2 =

y =

b

x cos x + a cos a sin x – 2 2

b

g

1 cos a sin x − x cos x + a 2 12. Solve the initial value problem

=

g

bg

dx d2x dx 0 = 15 . +4 + 29x = 0, given x(0) = 0, 2 dt dt dt Solution. We have (D2 + 4D + 29) y = 0 A.E. : m2 + 4m + 29 = 0 – 4 ± 16 − 116 – 4 ± 10i = = – 2 ± 5i 2 2 x(t) = e–2t (C1 cos 5t + C2 sin 5t)

m =

∴ Now,

...(1)

dx = x′(t) = e–2t (– 5C1 sin 5t + 5C2 cos 5t) – 2e–2t (C1 cos 5t + C2 sin 5t) ...(2) dt us consider x(0) = 0 and x′(0) = 15

Let Equations (1) and (2) respectively becomes 0 = C1 and 15 = 5C2 – 2C1 ∴ C1 = 0 and C2 = 3 Thus,

x(t) = 3e–2t sin 5t is the required particular solution.

OBJECTIVE QUESTIONS 1. Match the following and find the correct alternative I. Cauchy’s equation

II. Bernoulli’s equation

(i)

b x + 2g

2 (ii) x ⋅

2

b

g

d2y dy + x+2 + y=5 2 dx dx

d 3y d2y – x 2 = ex 3 dx dx

319


III. Method of variation of parameters

(iii)

dy + xy = x 2 dx

dy + xy = x 2 y 2 dx (v) y dx2 – x dy2 = 0 (vi) (D2 + a) y = tan x

(iv)

(vii)

dy y − x = dx y + x

(a) I (i), II (iii), III (vii)

(b) I (ii), II (iii), III (v)

(c) I (i), II (iv), III (vi)

(d) I (ii), II (iv), III (vi)

Ans. d

2. The homogeneous linear differential equation whose auxillary equation has roots 1, 1 and – 2 is (b) (D3 + 3D – 2) y = 0 (a) (D3 + D2 + 2D + 2) y = 0 (c) (D3 + 3D + 2) y = 0 3. The general solution of (a) y = C1 + C2 ex

(x2

(d) (D + 1)2 (D – 2) y = 0. D2

(c) y = C1 + C2 x2

Ans. c

– xD), y = 0 is (b) y = C1 + C2 x (d) y = C1 x + C2 x2.

Ans. c

4. Every solution of y ′′ + ay ′ + by = 0, where a and b are constants approaches to zero as x → α provided. (a) a > 0, b > 0 (b) a > 0, b < 0 (c) c < 0, b < 0

(d) a < 0, b > 0.

Ans. a

5. By the method of variation of parameters y ′′ + a 2 y = sec ax, the value of A is (a)

b

– log sec ax a

2

g+k

1

(b)

– log sec ax + k1 a

log sec ax + k1 (d) None. a3 6. By the method of variation of parameters, the value of W is called (a) The Demorgan’s function (b) Euler’s function (c)

(c) Wronskian of the function

(d) Robert’s function.

Ans. a

Ans. c

7. The method of variation of parameters, the formular for A′ is

bg

(a)

y1φ x W

(c)

– y2 φ x W

bg

(b)

bg

y2 φ x W

(d) None.

Ans. c

320


b

g

8. The equation a0 (ax + b)2 y ′′ + a1 ax + b y ′ + a2 y = φ(x) is called (a) Legendre’s linear equation

(b) Method of undetermined coefficients

(c) Cauchy’s linear equation

(d) Simultaneous equation

Ans. a

b)2

9. The D.E. (ax + y″ is 3 (a) a D (D – 1) (D – 2) y (c) D · Dy

(b) a2 D (D – 1) y (d) None.

Ans. b

10. The equation a0 y″ + a1 xy′ + a2y = φ(x) is called (a) Legendre’s linear equation (b) Cauchy’s linear equation (c) Simultaneous equation (d) Method of undetermined coefficients. x2

Ans. b 11. If t = log x, the value of x is (a) et (c) xet 12. The initial value problem, (a) C1 – C2 = 0 (c) C1 = 0 13.

(b) ex (d) ext.

Ans. a

d 2x dx +5 + 6 x = 0 x(0) = 0 is 2 dt dt (b) C1 + C2 = 0 (d) C2 = 0

Ans. b

d2y dy +4 + 3 y = e – x. Subject to the condition y (0) = 1 is dx dx 2 (b) C1 + C2 = 0 (a) C1 + C2 = 1 (c) C1 = 0

(d) C2 = – 1

Ans. a GGG

UNIT

811

Laplace Transforms 7.1I INTRODUCTION The Laplace transform method is used for solving the differential equations with initial and boundary conditions. The advantage of this method is that it solves the differential equations with initial conditions directly without the necessity of first finding the general solution and then evaluating the arbitrary constants using the initial conditions. In particular, this method is used in problems where the driving force (mechanical or electrical) has discontinuities for a short time or is periodic. In this unit we study the basic concepts of Laplace transforms and its applications to solve the differential equations arising in mechanics, electrical circuits and bending of beams.

7.2 DEFINITION Let f (t) be a real valued function defined for all t ≥ 0. Then the Laplace transform of f (t) denoted by L{ f (t)} is defined by

z

∞

L { f (t)} =

bg

e – st f t dt

...(1)

0

where s is a real or a complex number. In the integral on the right hand side (1) exists, it is a function of s and is usually denoted by F (s). Here s is called the parameter. Thus,

z

∞

F (s) = L { f (t)} =

bg

e – st f t dt

0

7.3 PROPERTIES OF LAPLACE TRANSFORMS If f (t) and g (t) are two functions defined for all positive values of t and k is a constant then (1) L {k f (t)} = k L { f (t)} (2) L { f (t) + g (t)}= L { f (t)} + L {g (t)} 321

322


z

e – st f t dt

z

e – st k f t dt

∞

Proof: (1) we have L{ f (t)} =

bg

0

Therefore,

∞

L {k f (t)} =

bg

0

z

∞

= k

bg

e – st f t dt

0

= k L { f (t)} (2) Consider

z z ∞

L { f (t) + g (t)} =

e – st

m f bt g + g bt gr dt

0

∞

=

bg

e – st f t dt +

0

z ∞

bg

e – st g t dt

0

= L { f (t)} + L {g (t)}.

7.3.1 Laplace Transforms of Some Standard Functions 1. Laplace transform of a constant Let f (t) = a, where a is constant. Then from the definition of Laplace transform, we get

z

∞

L (a) =

0

e – st a dt

z

∞

= a

e – st dt

0

L e OP aM N–sQ – st

= =

∞

0

– a –∞ e – e0 , s

since

e–∞ = 0 e0 = 1

a s a L (a) = s =

Hence In particular cases,

L (1) =

1 . s

...(2)

323

LAPLACE TRANSFORMS

2. Laplace transform of eat Substituting f (t) = eat in the definition of Laplace transform, we get

z z ∞

L (eat) =

e – st e at dt

0

∞

=

eb

g

– s+a t

dt

0

LM e b g OP MN – bs − ag PQ – s−a t

=

∴

∞

0

=

– 1 –∞ e – e0 s−a

=

1 if s > a > 0 s−a

L (eat) =

1 , s>a>0 s−a

...(3)

1 , s > −a. s+a

...(4)

Replacing a by – a, we get L (e–at) =

3. Laplace transform of sin h at We have

sin h at =

e at − e – at 2

f (t) = sin h at =

Substituting

e at – e – at 2

RS e – e UV = 1 L de i − L de i T 2 W 2 1 R 1 1 U − S V By using, Eqns. (3) and (4) 2 Ts − a s + aW at

– at

L (sin h at) = L

=

Hence L (sin h at)

=

a , if s > a s2 − a 2

=

1 ,s>a s − a2 2

at

– at

...(5)

324


4. Laplace transform of cos h at We have

e at + e – at cos h at = 2

RS e + e UV T 2 W 1 {L de i + L de i} 2 1R 1 1 U + V S 2 Ts − q s + a W at

Then

– at

L (cos h at) = L

= =

at

– at

s s − a2 s , s > a. Thus, L (cos h at) = 2 s − a2 5. Laplace transform of sin at and cos at We know by Euler’s formula that, eiat = cos at + i sin at ∴ L (cos at + i sin at) = L {eiat} =

i.e., L (cos at) + i L (sin at) =

2

1 , s − ia s + ia s − ia s + ia

gb

using eqns. (3) and (4) if s > a ...(6)

Replacing a by ia in (3)

=

b

g

=

s + ia s2 + a 2

=

s a +i 2 2 s +a s + a2 2

On equating the real and imaginary parts, we obtain L (cos at) =

s s + a2

L (sin at) =

a · s2 + a 2

2

...(7)

6. Laplace transform of t n Let f (t) = tn, where n is a non-negative real number or n is a negative non-integers. Then from the definition,

z

∞

n

L (t ) =

0

e – st t n dt

325

LAPLACE TRANSFORMS

dx and s

Substitute st = x, so that dt =

x s t = 0, t = ∞,

t = When

z

∞

∴

n

L (t ) =

e–x

0

=

= L (tn) =

Thus,

1 s

n+1

FG x IJ H sK

z ∞

x = 0 and x= ∞ n

dx , s

e − x ⋅ x n dx

0

b g Γ bn + 1g 1

s

n+1

Γ n +1

...(8)

sn + 1 In particular if n is a non-negative integers, we have

Γ (n + 1) = n! n! Hence, L (tn) = n + 1 s where n is a non-negative integer.

...(9)

Laplace transforms of some Standard Functions f (t) 1.

a

2.

eat

3.

e–at

4.

sin h at

L { f (t)} = F (s) a

,s>0

f (t) 5.

cos h at

1 ,s>0 s−a

6.

sin at

1 ,s>–a s+a

7.

cos at

8.

tn n = 1, 2, 3.......

s

a 2

s –a

2

,s>a

WORKED OUT EXAMPLES 1. Find the Laplace transform of the following functions: (1) 2t2 + 3t + 4

(2) 2e–3t – 4e5t

(3) sin2 at

(4) cos3at

(5) sin 5t cos 3t

(6) cos h2at

(7) at

(8) cos t cos 2t cos 3t.

L { f (t)} = F (s)

s 2

s − a2 a 2

s + a2 s s2 + a 2 n! n+1

s

,s > a ,s>0 ,s>0

,s>0

326


Solution (1) Now

L (2t2 + 3t – 4) = 2L (t2) + 3L (t) + 4L (1) = 2⋅

2! 1 1 +3 2 +4 3 s s s

4 + 3s + 4 s 2 s3 – 4e5t ) = 2L (e–3t ) – 4L (e+5t ) =

(2)

L (2e–3t

= 2 = (3)

L (sin2 at) =

1 1 –4 s+3 s −5

b

g

– 2 s + 11

bs + 3gbs − 5g F 1 − cos 2at IJ LG H 2 K b

1 L 1 − cos 2at 2

=

1 L 1 − L cos 2at 2

=

1 1 s − 2 2 s s + 2a

bg b

LM OP MN b g PQ 1 LM s + 4a − s OP 2 M s d s + 4a i P N Q 2

=

= (4) we know that ∴ Thus,

g

=

2

d

s s 2 + 4a 2

2

2

2

2a 2

g

2

i

cos 3at = 4 cos3 at – 3 cos at cos3at =

L (cos3 at) = =

=

b g L1 O L M bcos 3at + 3 cos at gP N4 Q 1 L bcos 3at g + 3 L bcos at g 4 1 L s 3s O M P + 4 M s + b3a g s + a PQ N s d s + 7a i . d s + 9a id s + a i 1 cos 3at + 3 cos at 4

2

2

2

=

2

2

2

2

2

2

2

327

LAPLACE TRANSFORMS

(5) Since,

sin 5t cos 3t = =

b

g

b

1 sin 5t + 3t + sin 5t − 3t 2

g

1 sin 8t + sin 2t 2

RS 1 sin 8t + sin 2t UV T2 W 1 mL bsin 8t g + L bsin 2t gr 2 1L 8 2 O + M P 2 Ns + 8 s +2 Q 5 d s + 16i ds + 64id s + 4i

Therefore, L (sin 5t cos 3t) = L = =

2

2

2

2

2

=

2

2

(6) We have,

cos h 2at = 2 cos h2 at – 1

So that,

cos h2 at =

Hence,

L [cos h2 at] =

=

= (7) we have Hence,

b

1 1 + cosh 2at 2

g

m bg 1 R| 1 S + s 2 | s s − b 2a g T

1 L 1 + L cos h 2at 2

2

s 2 − 2a 2

d

s s 2 − 4a 2

2

U| V| W

r

i·

at = et log a, a > 0

o

t

t log a = L (at ) = L e

(8) We have cos t cos 2t cos 3t = = = = =

1 · s − log a

b

1 cos t cos 5t + cos t 2

g

1 cos 5t cos t + cos2 t 2

LM b g b N 1 bcos 6t + cos 4t g + 14 b1 + cos 2t g 4

1 1 1 cos 6t + cos 4t + 1 + cos 2 t 2 2 2

1 1 + cos 2t + cos 4t + cos 6t 4

gOPQ

328


Therefore, L (cos t cos 2t cos 3t) = =

bg b

g b

LM N

1 1 s s + 2 + 2 + 2 4 s s +2 s + 42 s2

2. Find the Laplace transforms of the functions, 1 (1) t (2) t Solution (1) We have

∴

L (tn) =

F I L e t j = L Gt J H K 1 2

b g

sn + 1

(2) Now

=

1

s2 + 1

FG IJ HK

1 1 Γ 2 2 3 s2

π

=

3

2s 2

Γ (n + 1) = n Γ (n) and Γ

F 1I F I L G J = L Gt J H tK H K 1 − 2

Γ =

FG –1 + 1IJ H2 K s

Γ =

(3)

FG 1 + 1IJ H2 K

Γ

F I L et t j = L G t J H K 3 2

=

1 +1 2

FG 1 IJ H 2K

1 s2

Γ =

−

FG 3 + 1IJ H2 K 3 +1 2 s

FG IJ HK

3 1 1 ⋅ Γ 2 2 2 5

s2 =

=

3 π 4⋅

5 s2

·

π s

2

(3) t t .

Γ n +1

=

Since

g b s O P. +6 Q

1 L 1 + L cos 2t + L cos 4t + L cos 6t 4

FG 1 IJ H 2K

=

π.

g

329

LAPLACE TRANSFORMS

3. If f (t) =

RS2, Tt,

03

, find L { f (t)}.

z z

∞

Solution. Now,

L { f (t)} =

bg

e – st f t dt

0

e – st ⋅ 2 dt +

0

– st

4. If

R| 1, f (t) = S t, |T t ,

04

2

e – st ⋅ t dt

3

L e OP 2M N –s Q

=

z

∞

3

=

OP − 1⋅ b– sg PQ L 3e − e OP − 1 + 0 – M– N s s Q

3

0

Le + Mt MN – s

– st

2

3

–2 s

=

–2 –3s e s

=

2 s + 1 – 3s + 2 ⋅e . s s

∞

e – st

–3s 2

, find L { f (t)}.

Solution. We have,

z z ∞

L { f (t)} =

bg

e – st f t dt

0

z

2

=

z

∞

4

e

– st

⋅ 1 ⋅ dt +

0

e

– st

⋅ t dt +

2

e – st ⋅ t 2 ⋅ dt

4

LM e OP + LM t e − e OP + LMt e − 2t e + 2 e OP = –s N Q MN – s b– sg PQ MN – s b– sg b– sg PQ –1 F 4 1 I F 2 1 I F 16 8 + 2 IJ e − 1i – e G + J + e G + J + e G + d = Hs s K Hs s K H s s s K s 1 F1 1 I F 12 7 + 2 IJ e . = +G + J e +G + Hs s sK s Hs s K 1 F sI L { f (t)} = F (s). Prove that L { f (at)} = F G J . a H aK − st

2

– st

4

– st

– st

– st

2

0

2

4

–4 s

–2 s

–4 s

2

2

–2 s

–4 s

2

2

z

∞

Solution. Now,

L{ f (at)} =

b g

e – st f at dt

0

Substitute

3

2

–2 s

5. If

∞

– st

2

at =

u or t =

u a

3

2

3

330


Hence, When,

du a t = 0, u = 0 and t = ∞, u = ∞

dt =

z

∞

∴

L{ f (at)} =

– su a

e

0

z FG zH ∞

1 = a

bg

⋅f u ⋅

– su e a

du a

bg

⋅ f u du

0

∞

1 = a

e

0

IJ K

–s u a

bg

⋅ f u du

FG IJ H K

1 s F ⋅ a a

=

EXERCISE 7.1 Find the Laplace transforms of the following functions:

LMAns. 3s + 4 OP s N Q LMAns. 8 – 12s + 9s OP s N Q 2

2

1. 2t + 3

3

2

3. (2t – 3)2

3

LM Ans. N

5. 5t.

OP Q LM s + 2a OP Ans. MN s d s + 4a i PQ LM OP 2a Ans. MN s d s − 4a i PQ 1 s − log 5

2

2

7. cos at.

2

2

2

2

2

9. sin h at.

2

2

LMAns. 8 – 5s + 6s OP 2. 4t – 5t + 6. s N Q L 1 48 − 24s + 6s − s OP 4. (2t –1) . M Ans. N s Q LM 2 d2s − 4s + 1i OP MN Ans. s b s − 1gb s − 2g PQ 6. (1 + e ) . LM OP 6a Ans. 8. sin at. MN ds + a id s + 9a i PQ LM OP s + 15 Ans. 10. sin 3t cos 2t. MN ds + 25id s + 1i PQ LM s d s + 29 i O Ans. MN d s + 49i ds + 9i PPQ LM OP 48s Ans. MN ds + 4id s + 100i PQ 2

2

3

3

2

3

4

2

t 2

3

3

2

2

2

2

2

2

2

2

11. cos 5t cos 2t.

12. sin 6t sin 4t.

2

2

2

2

331

LAPLACE TRANSFORMS

LM OP 2 d s − 60i Ans. MN ds + 100id s + 36i PQ LMAns. s cosb – a sin b OP s +a N Q 1L 1 3 2 OO − + PP M 2 N s + 4 s + 36 s + 16 Q PQ 2

13. cos 8t sin 2t.

2

14. cos (at + b).

2

2

LM Ans. MN

15. sin t sin 2t sin 3t.

2

Find L{f (t)}, in each of the following functions:

R 5, 0 < t < 3 . 1. f (t) = S T 0, t > 3 R 1, 0 < t < 2 2. f (t) = S T t, t > 2 R| 1, 0 < t < 2 3. f (t) = S 2, 2 < t < 4 . |T 3, t > 4 R e , 0 1 R| 0, 0 < t < 1 t, 1 < t < 2 5. f (t) = S |T 0, t > 2 . R| cos at , 0 ≤ t ≤ π 6. f (t) = S |T 0, t>π R| cos FG t − 2π IJ , t > 2π H 3K 3 . 7. f (t) = S 2π || 0, 0
2

2

LM 2 d1 – e MNAns. s LM Ans. 1 + 1 b1 + sg e N s s

–3s

.

LM Ans. 1 d1 + e N s LMAns. N

iOPQ − 1O e P 1− s Q

–2 s

.

+ e –4 s

1− s

LM Ans. 1 obs + 1g e – b2s + 1g e N s –s

2

LM Ans. MN

b

i OP PQ OP Q

–2 s

2

t

.

2

–2 s

tOPQ

g OP PQ OP se s + 1 PP Q

s + – s cos a π + a sin a π e – πs s +a 2

2

LM MM Ans. N

– 2 πs 3

2

7.3.2 Laplace Transforms of the form eat f (t) If the Laplace transform of f (t) is known, then the Laplace transform of eat f (t) where a is a constant can be determined by using the shifting property.

332


Shifting property: If

L{ f (t)} = F (s) then L (eat f (t)} = F (s – a)

z z z

∞

Proof: We have L { f (t)} =

bg

bg

e – st f t dt = F s

0

∞

Therefore,

L {eat f (t)} =

bg

e – st e at f t dt

0

∞

=

e

b

g

– s−a t

bg

b

g

f t dt = F s − a .

0

Replacing a by – a, We get,

L {e–at f (t)} = F (s + a)

In view of the shifting property we can find the Laplace transform of the standard functions discussed in the preceeding section multiplied by eat or e–at 1.

L (sin bt) =

b , s + b2

L (eat sin bt) =

2.

L (cos bt) =

s , s + b2

L (eat cos bt) =

3. L (sin h bt) =

b , s − b2

L (eat sin h bt) =

4. L (cos h bt) =

s , s – b2

L (eat cos h bt) =

5. L (tn)

=

2

2

2

2

b g,

Γ n+1 s

L (eat tn) =

n+1

b

bs − ag

2π 3 , find L { f (t)}. 2π 0

Solution ⇒ We have,

L (cos t) =

s s +1 2

+ b2

s−a

bs − ag

2

+ b2

b

bs − ag

2

– b2

s−a

bs − ag

2

− b2

.

.

b g , for n = 0. bs − ag Γ n +1

WORKED OUT EXAMPLES

R| cos FG t − 2π IJ , H 3K 1. If f (t) = S || 0, T

2

n+1

333

LAPLACE TRANSFORMS

By using shifting Rule

a =

L { f (t)} =

2π 3 –2 πs e 3

–2 πs

s se 3 ⋅ 2 = 2 · s +1 s +1

2. Find the Laplace transform of the following functions (1) t2 e2t

(2) e–3t sin 2t

(3) e4t cos h 3t

(4) e–t cos23t Solution

(5) e3t sin3 2t

(6)

t et .

2 s3 By using the shifting property, we get

(1) We have

L (t2) =

L (t2 e2t) = (2) Since,

L (sin 2t) = L (e–3t sin 2t) =

= (3)

L (cos h 3t) = L (e4t cos h 3t) =

=

(4) Consider

2

bs − 2g

(s → s – 2)

3

2 s2 + 22

2

bs + 3g

2

(s → s + 3)

+4

2 s2 + 6s + 13 s s − 32 2

s−4

b s − 4g

2

(s → s – 4)

−9

s−4 . s − 8s + 7 2

FG 1 + cos 6t IJ H 2 K 1 L b1g + L bcos 6t g 2 1 L1 s O + M P 2 Ns s + 6 Q

L (cos2 3t) = L = =

=

2

s 2 + 18

d

2

i

s s 2 + 36

334


bs + 1g + 18 bs + 1g bs + 1g + 36 2

∴

L (e–t cos2 3t) =

=

sin3A =

(5) We have

sin3 2t =

Hence

(s → s + 1)

2

s2 + 2 s + 19

bs + 1g d s + 2s + 37i . 1 b3 sin A – sin 3 Ag 4 2

b

1 3 sin 2t – sin 6t 4

b

g

g b

L (sin3 2t) =

1 3L sin 2t − L sin 6t 4

=

1 2 6 3⋅ 2 − 2 2 4 s +2 s + 62

=

LM N

ds

48

2

id

g OP Q

i

+ 4 s2 + 36

By using shifting Rule, we get, s → s – 3 L {e3t sin3 2t} =

=

L

(6) Now

L

Hence

e

ds Γ

e tj

=

j

=

t et

bs − 3g

48 2

bs − 3g

+4

2

+ 36

48 2

id

i

− 6s + 13 s 2 − 6s + 45

FG 1 + 1IJ 1 Γ 1 H2 K = 2 2 = 3 s2

1 +1 s2

π

b g

2 s−1

3 2

π 3

2s 2 (s → s + 1)

·

3. If L {f (t)} = F (s), Prove that (1)

L {cosh at f (t)} =

1 F s−a + F s+a 2

b

g b

g

(2)

L {sin h at f (t)} =

1 F s−a − F s+a 2

b g b

g

Solution (1) Consider

L {cosh at f (t)} =

L

|RS LM e |T N

at

+ e − at 2

OP f (t )U|V Q |W

335

LAPLACE TRANSFORMS

b gt + L oe

o

=

1 L e at f t 2

=

1 F s−a + F s+a . 2

− at

b gt

f t

By using shifting property

b g b g R| L e – e O f (t )U| LSM P V T| N 2 Q W| 1 L oe f bt gt – L oe f bt gt 2 1 F b s − ag − F bs + a g . 2 − at

at

(2) Consider,

L {sin h at f (t)} =

at

= =

– at


LM 1 Ans. MN bs + 2g

1. te–2t.

3. e sin 3t.

5. e–2t sin h 3t.

t

2

7. e sin t.

LMAns. N s LMAns. N s 2

2t

OP PQ OP 3 − 4 s + 13 Q 2

OP Q LM OP 2 Ans. MN bs − 1g ds − 2s + 5i PQ 2

3 + 4s – 5

2

LM 2 d2s – 12s + 19i OP MNAns. bs − 3g PQ 2

2

2. (t2 + 4) e3t.

LMAns. N s LMAns. MN s

4. e4t cos 4t.

6. e5t cos h 2t.

2

2

OP Q OP s–5 − 10s + 21 PQ s–4 − 8s + 32

LM OP s + 8 s + 18 Ans. cos t. MN bs + 4g d s + 8s + 20i PQ LM OP 6 Ans. MN ds − 4s + 5id s – 4s + 13i PQ OP LM 2 e s – 4 s – 17j Ans. MN e s – 4 s + 53j e s – 4 s + 13j PQ LM OP s − a g + 2b b cos bt. MM Ans. bs − ag bs − ag + b PP N Q 2

–4t

8. e

9. e2t sin3 t.

2

2

2

2

2

2t

10. e cos 5t sin 2t.

11.

et t

.

2

LM π MM Ans. s − 1 N b g

1 2

OP PP Q

2

2

at

12. e

2

2

2

2

336


7.3.3 Laplace Transforms of the form t n f (t) where n is a positive integer In this section we shall find the Laplace transforms of the functions of the form tn f (t) where n is a positive integer if the Laplace transform of f (t) is known. Theorem: If L { f (t)} = F (s) then

b– 1g

m b gr

dn F s dsn Proof: We shall prove the theorem for n = 1 L {tn f (t)} =

i.e.,

We have,

L {t f (t)} =

–

n

m b gr

d F s ds

m b gr =

F (s) = L f t

z ∞

bg

e – st f t dt

0

Differentiating w.r.t. ‘s’, we get

m b gr

d F s ds

z

∞

=

0

b gt dt

o

d e – st f t ds

In the R.H.S., we shall apply Leibnitz rule for differentiation under the integral sign,

z

∞

=

0

b g bg

e – st – t f t dt

z ∞

= −

m b gr dt .

e – st t f t

0

= – L {t f (t)}

m b gr = –dsd L m f bt gr

–d F S ds

∴

L {t f (t)} =

Further,

L {t2 f (t)} = L t t f t

b g s = –dsd L mt f bt gr – d L– d O L m f bt grP ds MN ds Q b– 1g dsd L m f bt gr n

=

=

2

2

2

=

b– 1g

2

L {tn f (t)} =

b– 1g

n

m b gr

d2 F s ds2 By repeated this process of the above theorem, we get ⋅

m b gr

dn F s . ds n

337

LAPLACE TRANSFORMS

bg

f t t

7.3.4 Laplace Transforms of

bg

f t

If L f (t) is known then we can find the Laplace transform of Theorem: If

lim t→0

L { f (t)} = F (s) and

R f bt g UV LS T t W

z

∞

=

bg

f t t

t

by using the following.

exists then

bg

F s ds

s

z

...(1)

∞

Proof: We have, F (s) = L { f (t)} =

bg

e – st f t dt

0

On integrating both sides w.r.t. s from s to ∞, we get

z

∞

s

s

∞

=

LM MN LM MN

zz zz ∞

bg

F s ds =

0

∞

...(2)

e – st

...(3)

0

∞

s

(By changing the order of integration.)

z

∞

Now

e – st ds =

s

LM – e OP N –t Q – st

∞

= s

∴ Eqn. (3) gives,

z

∞

z

∞

∴

–1 –∞ + e – st e – e – st = t t

z

∞

bg

F s ds =

s

0

bg

F s ds

= L

s

O b g PP Q O dsP f bt g dt , PQ

e – st f t dt ds

bg

e – st f t dt = L t

RS f bt g UV T t W

(e–∞ = 0)

RS f bt g UV T t W

This completes the proof.

WORKED OUT EXAMPLES 1. Find the Laplace transforms of the following functions: (1) t sin at

(2) t cos at

(3) t sin h at

2

(6) t et sin h t

(4) t cos h at

(5) t cos at

(7) te–2t cos 2t

(8) t3 sin t.

Solution (1)

L {t sin at} =

b

–d L sin at ds

g

338


=

=

(2)

L {t cos at} = =

=

(3)

L {t sin h at} = =

=

(4)

L {t cos h at} = =

RS Ts

–d ds

ds

2as + a2

2

b –d R S ds T s

ds

+ a2

2

g

s + a2

2

i

2

UV W

⋅

2

s2 − a 2

UV W

·

b

g UV W

–d L sin h at ds

RS Ts

–d ds

ds

− a2

2

a – a2

2

2 as

i

2

·

b

–d L cos h at ds –d ds

RS Ts

s – a2

2

UV W

g

ds + a i . = ds − a i d cos at) = b – 1g ⋅ m L bcos at gr ds d R s U = +1 S V ds T s + a W – 2 s d 3a − s i = ds + a i – 2s d3a − s i . cos at} = s + a d i 2

2 2

2

L (t2

i

–d L cos at ds

2

(5)

a + a2

2

2

2

2

2

2

2

2

2

2 3

2

2

2

L {t

2

2

2

2 3

339

LAPLACE TRANSFORMS

(6)

L {t sin h t} =

mL bsin h t gr –d R 1 U S V ds T s – 1 W

–d ds

=

2

2s

d s – 1i 2 b s − 1g bs − 1g − 1 2 b s − 1g ⋅ s b s − 2g

=

2

2

L (et t sin h t) =

2

=

s→s–1

2

2

2

(7) From Eqn. (2) above with a = 2, we have L (t cos 2t) =

s2 − 4

ds

i

+4

2

2

Hence by using shifting Rule, we get s → s + 2

b s + 2g – 4 b s + 2g + 4 s b s + 4g d s + 4s + 8i 2

–2t

L {e

t cos 2t} =

2

=

(8) we have

L (sin t) =

∴

2

·

2

2

1 s +1

L (t3 sin t) =

2

b– 1g ⋅ dsd RS s 1+ 1UV T W 24 s d s − 1i · ds + 1i 3

3

3

2

2

=

z z

∞

2. Prove that

e –3t t sint dt =

0

Solution

2

4

3 . 50

∞

⇒ We have

e

– st

t sin t dt = L (t sin t) = F (s)

0

Now,

F (s) = L (t sin t) =

m b gr

–d L sin t ds

340


=

z

2s

d s + 1i

e – 3t t sin t dt =

F 3 =

2

z

z

2.3

d3 + 1i 2

2

=

6 3 = . 100 50

12 . 169

t e –2 t sin 3t dt =

0

Solution:

2

bg

0

∞

3. Prove that

2

= ∞

∴

RS 1 UV T s + 1W

–d ds

∞

⇒ We have

e

– st

t sin 3t dt = L (t sin 3t) = F (s)

0

Now,

F (s) = L (t sin 3t) = = =

m b gr –d R 3 U S V ds T s + 9 W –d L sin 3t ds 2

ds

6s 2

i

+9

2

Given the integral, F (2) =

e2

6.2 2

j

+9

2

=

12 . 169

4. Find the Laplace transforms of the following functions:

sin at (1) t

1 − e at (2) t

cos at − cos bt t Solution

(5)

(4)

(1) Now

lim

t→0

e –at – e –bt t

LM N

FG H

IJ K

sin at sin at = a 3 lim ⋅a = a t →0 t at

L (sin at) =

R sin at UV LS T t W

a s + a2 2

z

∞

=

s

=

a ds s + a2 2

LMtan FG s IJ OP N H aKQ

∞

–1

s

OP Q

(3)

1 − cos at t

(6)

sin h at · t

and

341

LAPLACE TRANSFORMS

–1 –1 = tan ∞ – tan

=

FG IJ H K

π s − tan –1 2 a

= cot –1

1 − e at t→0 t lim

(2) Now

FG s IJ H aK

= lim

t→0

FG s IJ . H aK

– ae at =–a 1

(By using L’ Hospital Rule)

L (1 – eat) = L (1) – L (eat) 1 1 – = s s−a

Also,

RS1 − e UV T t W

z FGH ∞

at

L

=

s

=

IJ K

1 1 ds − s s−a

b g LMlog F s I OP MN GH s − a JK PQ LM F I OP G 1 MMlog G F a I JJ PP MN GGH 1 − GH s JK JJK PQ LM OP 1 log 1 – log M MM 1 − FG a IJ PPP N H sK Q log s – log s − a

∞ s

∞

=

s

∞

=

s

=

= – log = log

s s−a

FG s − a IJ · H s K

1 − cos at a sin at =0 = lim t→0 1 t L (1 – cos at) = L (1) – L (cos at)

(3) Consider lim

t→0

We have

=

L

RS1 − cos at UV T t W

s 1 − s s2 + a 2

z FGH ∞

=

s

I JK

1 s ds − 2 s s + a2

(By using L’ Hospital Rule)

342


=

LMlog s − 1 log ds + a iOP 2 N Q OP 1 s log 2 s +a Q O 1 1 P P log 2 a P 1+ s PQ R| U| 1| 1 | log 1 – log S V 2| a | 1+ |T s |W 2

s

∞

2

=

∞

2

2

2

s

∞

=

2

2

=

s

2

2

–1 s2 1 s2 + a 2 log = log . = 2 s2 + a 2 2 s2 (4) Let

f (t) = cos at – cos bt

lim

Now,

t→0

bg

f t t

= lim

cos at – cos bt t

= lim

– a sin at + b sin bt 1

t→0

t→0

(Using L’ Hospital Rule) = 0 Now

L {cos at – cos bt} =

L

RS cos at − cos bt UV T t W

s s – 2 2 s +a s + b2 2

z

∞

=

s

=

LM Ns

2

OP Q

s s – 2 ds 2 s + b2 +a

LM 1 log ds N2

2

i

+ a2 –

OP Q OP PP PQ

1 s2 + a 2 log 2 = 2 s + b2 a2 1+ 2 1 s log = 2 b2 1+ 2 s

∞

s

∞

s

d

1 log s 2 + b 2 2

iOPQ

∞ s

343

LAPLACE TRANSFORMS

a2 1 1 s2 log 1 – log = 2 2 b2 1+ 2 s 1+

= –

=

s2 + a 2 1 log 2 2 s + b2

1 s2 + b2 log 2 . 2 s + a2

f (t) = e–at – e–bt

(5) Let

lim

Now,

t→0

bg

f t t

= lim

t→0

– ae – at + be – bt e – at – e – bt = lim (Using L’ Hospital Rule) t→0 1 t

= b – a, which is finite. L { f (t)} = L {e–at – e–bt}

Now

m b gr RST e

L f t

=L

– at

– e – bt t

UV W

=

1 1 − s+a s+b

=

z FGH

IJ K log b s + a g – log b s + bg s + aO log P s + bQ aO 1+ P s log bP 1+ P sQ ∞

s

=

1 1 ds – s+a s+b

∞

=

s

∞

=

s

a s = log 1 – log b 1+ s 1+

= – log = log

s+a s+b

s+b · s+a

∞ s

344


(6) Let

f (t) = sin h at

lim

Now,

t→0

bg

f t t

= tlim →0

sin h at t

= lim

e at – e – at 2t

t→0

ae at + ae – at t→0 2 = a, which is finite.

= lim

We have,

L (sin h at) =

R sin h at UV LS T t W

a s − a2 2

z

∞

=

s

a ds s − a2 2

a s−a log = 2 s+a

=

a 2

=

a 2

OP Q

∞

s

LMlog 1 − log F s − a I OP GH s + a JK PQ MN LM– log F s − a I OP MN GH s + a JK PQ

s+a a = 2 log s − a .

z

∞

5. If L{ f (t)} = F (s) then prove that

z

∞

Hence prove that

0

π sin t dt = . t 2

z

∞

i.e.,

0

e – st

R f bt g UV LS T t W RS f bt g UV dt T t W

∞

s

z z

∞

=

z

b g dt = F bsg ds provided both the integrals exist. t

f t

0

Solution. We have,

bg

F s ds

s

∞

=

bg

F s ds

s

Taking the limits on both sides as s → 0+, we get

z

∞

0

b g dt

f t t

(Using L’ Hospital Rule)

z

∞

=

This completes the proof of the example.

s

bg

F s ds

345

LAPLACE TRANSFORMS

b g

f (t) = sin t, then F (s) = L sin t =

Let

z

∞

∴

0

z

∞

sin t dt = t

0

OP PQ

1 ds = tan –1 s 2 s +1

= tan ∞ – tan 0 –1

–1

1 s +1 2

∞

0

π · 2

=

EXERCISE 7.3 Find the Laplace transform of the following functions:

LM s + 2s + 8 OP MMAns. s ds + 4i PP Q N LM 24s d s + 5i OP MMAns. ds + 1i ds + 9i PP Q N 4

1. t cos2 t.

2

2

2

2

2

3. t sin3 t.

2

2

2

2

LM 2 d3s + 4i OP 2. t sin t. MMAns. s ds + 4i PP Q N LM 1 L s − 9 3 ds − 1i O OP M PP P + 4. t cos t. M Ans. 4 M MN MN ds + 9i ds + 1i PQ PQ LM 2s d3s + 48s + 288i OP MMAns. ds + 16i ds + 4i PP Q N LM 2s ds + 3a i OP 7. t cos h at. MMAns. ds − a i PP Q N LM s – 2s – 3 OP MMAns. ds – 2s + 5i PP 9. t e cos 2t Q N 2

2

2

2

2

3

2

2

2

4

5. t sin 3t cos t.

2

LM 2a ds – a i OP MMAns. ds + a i PP Q N O LM 6 b s + 2g P Ans. MM ds + 4s + 13i PP Q N 2

6.

t2

8.

te–2t

sin at.

sin 3t.

Evaluate the following:

z z z

∞

1.

e –2 t ⋅ t sin 4t dt .

0

2

2

2 2

2

2

LMAns. 1 OP N 25 Q

0

∞

5.

0

t 3 e – t cos t dt .

Ans. 1

LM N

–3 Ans. 2

OP Q

2

2

2

2

2

2

2 3

2

t

z z z

∞

2.

te −3t cos 2 t dt .

0

∞

t 2 e t cos t dt .

2

2

2

∞

3.

2

2

4.

t 2 e – t sin 2 t dt .

0

∞

6.

0

t 3 e 2 t sin t dt .

2

2

LMAns. 5 OP N 169 Q LMAns. – 12 OP N 25 Q LMAns. – 24 OP N 625 Q

346


Find the Laplace transforms of the following functions: 1.

LMAns. log s – 1OP s Q N Ans. cos b s – 1g

1− et . t

et sin t . 3. t

LMAns. tan N

s s – tan –1 2 8

–1

OP Q

OP Q LMAns. 1 log s + 4 OP N 4 s Q LMAns. 1 log s − b OP N 2 s −a Q –1

s 2

2

sin 2 t . 4. t

–1

2 sin 3t cos 5t . 5. t

LMAns. cos N

sin 2t . t

2.

2

cos h at – cos ht . 6. t

2

2

2

2

Prove the following:

z z z

∞

1.

0

∞

2.

0

∞

3.

0

cos 6t − cos 4t 3 dt = log . t 2 e – t − e –3t dt = log 3. t e – t sin t π dt = . 4 t

7.4 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS A function f (t) is said to be periodic function with period α > 0, if f (t + α) = f (t). Theorem: If f (t) is a periodic function of period α > 0, then 1 L { f (t)} = 1 − e – sα

z z

∞

Proof: We have,

L { f (t)} =

z α

bg

e – st f t dt

bg

e – st f t dt

0

α

=

...(1)

0

z

2α

bg

e – st f t dt +

bg

e – st f t dt +

α

0

Substitute, t = u + α in the second integral Then dt = du When t = α, u = 0, and t = 2α, u = α.

z

2α

Hence

α

Since

e

– st

bg

f t dt =

z α

e

b

–s u + α

0

f (u + α) = f (u)

g

b

g

f u + α du

z

3α

2α

bg

e – st f t dt + ⋅⋅ ⋅

...(2)

347

LAPLACE TRANSFORMS

= e

z z α

– sα

bg

e – su f u du

0

= e

α

– sα

bg

e – st f t dt

(replacing u by t)

0

Similarly by substituting t = u + 2α in the 3rd integral, we get,

z

3α

bg

e – st f t dt = e –2 sα

2α

z α

bg

e – st f t dt

0

and so on. ∴ Equation (2) reduces to

z α

L { f (t)} =

e

– st

bg

f t dt + e

– sα

0

=

=

Since,

1 + r + r2 + ... to ∞ =

LM1 + e N

z α

0

– sα

d i

+ e – sα

z

2

α

1 1 − e – sα

e

OP Q

+ ⋅⋅⋅

bg

– st

z

f t dt + e

−2 sα

z α

e − st f ( t ) dt + ⋅⋅ ⋅

0

α

bg

e – st f t dt

0

bg

e – st f t dt

0

1 and 1– r

– sα < 1. |r| = e

1 L { f (t)} = 1 – e – sα

Thus,

z α

bg

e – st f t dt

0

This completes the proof of the theorem.

WORKED OUT EXAMPLES 1. If f (x) =

RS 3t, T 6,

0
Solution. Since f (t) is a periodic function with period α = 4 from (1), we get L { f (t)} =

z 4

Now,

e

– st

1 1 − e –4 s

z

z

bg

4

e – st f t dt

0

2

bg

f t dt =

4

e – st ⋅ 3t dt +

0

0

L e 3 Mt ⋅ MN – s

– st

=

z

...(1)

e – st ⋅ 6 dt

2

OP L e O − 1⋅ + 6 b – sg PQ MN – s PQ 2

e – st

– st

4

2

0

2

348


LM N

OP L Q MN

3 –2

=

3 3e –2 s 6e –4 s − 2 − s s2 s

=

3 1 – e –2 s − 2 s e –4 s 2 s

d

Therefore from (1), we get

d

3 1 – e –2 s – 2 s e –4 s L { f (t)} =

OP d Q

e –2 s e –2 s 1 6 − 2 – 3 0 – 2 – e –4 s – e –2 s s s s s

=

d

s 2 1 – e –4 s

i

i

i

i·

2. If A periodic function f (t) of period 2a is defined by f (t) =

RS a T–a

for 0 ≤ t < a for a ≤ t ≤ 2a

FG IJ H K

a as tan h . s 2 Solution. Since f (t) is a periodic function with period 2a, we have from Eqn. (1), Show that L { f (t)} =

L { f (t)} =

z

Now

e

– st

z

bg

e – st f t dt

0

a

bg

2a

z

2a

1 1 − e –2 as

f t dt =

e

⋅ a dt +

LM ae OP N –s Q –a oe s – st

=

=

b g

e – st – a dt

a

0

0

z

2a – st

a

0

– as

L −ae OP +M N –s Q a − 1t + e s – st

o

a 1 – 2e – as + e –2 as s

=

a 1 − e – as s

d

i

t

2

Therefore from (i), we obtain,

=

d d

i

– as a 1− e ⋅ s 1 − e –2 as

d

2

i i

2

1 − e – as a s 1 − e – as 1 + e – as

d

2a

a

–2 as

=

L { f (t)} =

...(1)

id

i

− e – as

349

LAPLACE TRANSFORMS

=

a 1 – e – as ⋅ s 1 + e – as

=

a e2 −e 2 as – as s e2 +e 2

R| S| T

– as

U| V| W

as

Multiplying numerator and denominator by =

as

e2

FG IJ H K

a as tan h ⋅ s 2

3. If f (t) = t2, 0 < t < 2 and f (t + 2) = f (t) for t > 2, find L {f (t)}. Solution. Here f (t) is a periodic function with period 2. Therefore from eqn. (1), we have L { f (t)} =

z 2

Consider,

e

– st

2

t dt =

0

On Integrating by parts,

1 1 − e −2 s

LMt MN

2

z 2

e – st ⋅ t 2 dt

...(1)

0

OP b g PQ

e – st e – st e – st 2 − 2t + 2 3 –s –s –s

b g

=

RS – 4 e Ts

=

2 e –2 s − 3 4 s2 + 4s + 2 3 s s

=

2 1 − 2 s2 + 2 s + 1 e –2 s 3 s

–2 s

−

2

0

UV RS UV W T W

4e –2 s 2 – 2 s 2 – 3e – – 3 2 s s s

d

d

i

i

Therefore from (i), we get L { f (t)} =

2π is defined by ω π E sin ωt, 0≤t< ω π 2π 0, ≤t < ω ω

4. A periodic function of period

f (t) =

R| S| T

where E and ω are constants. Show that, L { f (t)} =

d

2

s 1− e 3

Eω

. F I ds + ω i GH1 − e JK 2

2

– πs ω

–2 s

i d

i

1 − 2 s 2 + 2 s + 1 e –2 s

350


Solution. We have a periodic function f (t)

z α

1 L { f (t)} = 1 − e – sα

Hence

α =

bg

e – st f t dt

0

2π ω

z

2π ω

1

=

1–

– 2 πs e ω

0

LM MM MN

1

=

1–

– 2 πs e ω

1–

z

z

2π ω

e – st E sin ωt dt +

e – st

π ω

0

z

z

2π ω

OP ⋅ 0 dt P PP Q

π ω

1

=

bg

e – st f t dt

– 2 πs e ω 0

e – st E sin ωt dt

...(1)

π ω

Consider

=

e – st E sin ωt dt

0

z

π ω

= E

e – st sin ωt dt

0

=

=

=

=

=

=

L e EM MN b– sg

O – s sin ωt – ω cos ωt gP b PQ +ω e b + s sin ωt + ω cos ωt g LMe bs sin π + ω cos πg – e bs sin 0 + ω cos 0gOP MN PQ LMe c0 + ω b –1gh – ωOP NM QP LM–ωe − ωOP MN PQ F1 + e I GH JK π ω

– st

2

–E s + ω2 2

–E s2 + ω 2 –E s2 + ω 2 –E s2 + ω 2 Eω 2 s + ω2

2

0

πω

− st

0

– sπ ω

0

– sπ ω

– sπ ω

– sπ ω

351

LAPLACE TRANSFORMS

Hence from Eqn. (1), we get =

F1 + e G s +ω H 1– e F I E ω G1 + e J H K F IF d s + ω i GH1 − e JK GH1 + e 1

⋅

– 2 πs ω

– sπ ω

Eω

2

2

I JK

– πs ω

L { f (t)} =

2

L { f (t)} =

– πs ω

2

Eω

F I ds + ω i GH1 − e JK 2

– πs ω

2

– πs ω

I JK

Hence proved.

EXERCISE 7.3 Find the Laplace transforms of the following periodic functions:

LM 1 2e MNAns. s – s d1 − e

OP i PQ O 2 – d9 s + 6s + 2 i e P PQ i LM as − 1 + e OP MNAns. s d1 − e i PQ LM 1 – e b g OP Ans. MN b s + 1g d1 − e i PQ LMAns. 1 tan h FG s IJ OP N s H 2K Q OP LM 1 MNAns. s d1 + e i PQ LMAns. 1 tan h FG as IJ OP N s H 2 KQ –2 s

1. f (t) = t, 0 < t < 2 and f (t + 2) = f (t).

2. f (t) =

t2,

0 < t < 3 and f (t + 3) = f (t).

2

LM 1 MNAns. s d1 – e 3

–2 s

–3s

2

–3s

– as

3. f (t) = a – t, 0 < t < a and f (t + a) = f (t).

2

– as

– s +1

4. f (t) =

e– t.

0 < t < 1 and f (t + 1) = f (t).

RS 1, 0 < t ≤ 1 and f (t + 2) = f (t). T– 1, 1 < t ≤ 2 R|1, 0 < t ≤ 1 6. f (t) = S |T0, 1 ≤ t < 2 and f (t + 2a) = f (t). 0≤t
–s

–s

2

352


8. f (t) =

RS t , T1,

Rt, 9. f (t) = S T0, 7.5

LM 1 − e – s e OP MNAns. s d1 – e i PQ LM 1 – b s + 1g e OP MNAns. s d1 − e i PQ

0 < t < 1 and f (t + 2) = f (t). 1 ≤ t < 2

–s

2

–2 s

–2 s

–s

0< t <1 and f (t + 2) = f (t). 1< t < 2

2

–2 s

LAPLACE TRANSFORMS OF UNIT STEP FUNCTION AND UNIT IMPULSE FUNCTION

Unit Step Function (Heaviside function) The unit step function or Heaviside function u (t – a) is defined as follows u (t – a) =

RS 0, T 1,

when t ≤ a when t > a

where a ≥ 0 The graph of this function is as shown in the below. u (t – a) 1

a

O

Fig. 7.1

where a is the +ve constant.

7.5.1 Properties Associated with the Unit Step Function e – as s (ii) L { f (t – a) u (t – a)} = e–as F (s) = e–as L {f (t)}. Proof: (i) Using the definition of Laplace transform, we have (i) L u (t – a) =

z z z

∞

L {u (t – a)} =

b g

e – st u t − a dt

0

a

=

e

– st

u(t − a ) dt +

0

0

z

e – st u ( t − a ) dt

a

∞

a

=

z

∞

e – st ⋅ 0 dt +

a

e – st ⋅ 1 dt

...(1)

353

LAPLACE TRANSFORMS

=

=

e – st –s

0+

OP Q

∞

a

– 1 –∞ e – as e – e – as = s s

e – as Thus, L {u (t – a)} = s (ii) Heaviside Shifting Theorem: If

L { f (t)} = F (s) then L { f (t – a) u (t – a)} = e–as F (s) = e–as L {f (t)}

where u (t – a) is the unit step function. Proof: By definition, we have

z z z z

∞

L { f (t – a) u (t – a)} =

b g b g

e – st f t − a u t − a dt

0

=

e – st f t − a u t − a dt +

0

a

=

e

– st

b gb g

f t − a 0 dt +

z

∞

b g b g

e – st f t − a u t − a dt

a

b g

e – st f t − a ⋅ 1 dt

a

0

∞

=

z ∞

b gb g

a

b g

e – st f t − a dt

a

Substitute t – a = x so that dt = dx, When

t = a,

x = 0 and

t = ∞,

x=∞, t=a+x

Hence,

z

∞

L { f (t – a) u (t – a)} =

e

b

–s a + x

0

z z

gf

∞

=

e – sa

b x g dx bg

e – sx f x dx

0

∞

=

e – as

bg

e – st f t dt change x to t

0

= e–as L { f (t)} = e–as F (s). Hence proved.

354


7.5.2 Laplace Transform of the Unit Impulse Function Unit Impulse Function Definition: The unit impulse function denoted by δ (t – a) is defined as follows δ (t – a) =

δε (t – a) =

Where

b g

lim δ ε t − a , a ≥ 0

ε→0

R| 0, S| 1ε , T 0,

...(1)

if

t
if

a
if

t >a+ε

...(2)

The graph of the function δε(t – a) is as shown below: @A (t – a) 1 — A

a A

O

a+A

t

Fig. 7.2

Laplace transform of the unit impulse function

z z ∞

Consider

L {δε (t – a)} =

b g

e – st δ ε t − a dt

0

a

=

0

=

1 ε

e

– st

z

b0g dt +

z

a+ε

e

a

a+ε

e

– st

a

1 dt = ε

z

∞

– st

LM e OP N–sQ – st

bg

1 dt + e – st 0 dt ε a+ε a +ε

a

1 s a e – b + ε g − e – as εs

=

−

=

e – as

LM1 − e OP N εs Q – εs

Taking the limits on both sides as ε → 0, we get,

m b gr

lim L δ ε ⋅ t − a

ε→0

i.e., If

=

e – as lim

L {δ (t – a)} = e–as

ε→0

LM1 – e OP N εs Q –ε s

a = 0 then L {δ (t)} = 1

(Using L’ Hospital Rule)

355

LAPLACE TRANSFORMS

1. Find the Laplace transforms of the following functions: (1) (2t – 1) u (t – 2)

(2) t2 u (t – 3)

(3) (t2 + t + 1) u (t – 1)

(4) e3t u (t – 2)

(5)

b sin t + cos t g u FGH t − 2π IJK .

Solution (1) Now

2t – 1 = 2 (t – 2) + 3

∴ Using Heaviside shift theorem, we get L {(2t – 1) u (t – 2)} = L {[2 (t – 2) + 3] u (t – 2)} = e–2s L {2t + 3} –2s

= e

–2 s = e

Replacing t – 2 by t

{2 L (t) + L (3)}

RS 2 + 3 UV · Ts sW 2

t2 = [(t – 3) + 3]2

(2) Now,

= (t – 3)2 + 6 (t – 3) + 9 Then

L {t2 u (t – 3)} = L {[(t – 3)2 + 6 (t – 3) + 9] u (t – 3)}

Replacing t – 3 by t = e–3s L{t2 + 6t + 9} Using Heaviside shift theorem = e–3s {L (t2) + 6 L (t) + 9 L (1)} –3s = e

(3) Now

RS 2 + 6 + 9 UV ⋅ Ts s sW 3

2

t2 + t + 1 = (t – 1)2 + 3t

= (t – 1)2 + 3 (t – 1) + 3 Then L {(t2 + t + 1) u (t – 1)}= L {[(t – 1)2 + 3 (t – 1) + 3] u (t – 3)} Replacing t – 1 by t = e–s L {t2 + 3t + 3} –s = e

(4) Now, Hence

RS 2 + 3 + 3 UV Ts s sW 3

2

e3t = e3(t – 2) + 6 = e6 . e3 (t – 2)

L {e3t u (t – 2)} = L {e6 e3(t – 2) u (t – 2)}

Replacing t – 2 by t = e6 . e–2s L {e3t} e6 ⋅ e –2 s e –2b −3g . = = s−3 s−3 s

356


FG π − t IJ + sin FG π − t IJ H2 K H2 K F πI F πI = cos GH t – JK – sin GH t − JK 2 2 R F π I U RL F π I F π I O F π I U L Sbsin t + cos t g u G t – J V = L S Mcos G t − J − sin G t − J P u G t − J V H 2 K W TN H 2 K H 2 K Q H 2 K W T sin t + cos t = cos

(5) Now

∴

Replacing t −

π by t. 2 e

−

FG π IJ s H 2 K L {cos t – sin t}

Using Heaviside function = =

FG – π IJ eH 2 K

m Lbcos t g – L bsin t gr FG IJ R s 1 U eH K ⋅ S − T s + 1 s + 1VW FG IJ e H K ⋅ b s − 1g . –π s 2

2

−

=

s

2

π s 2

s2 + 1

2. Express the following functions in terms of the Heaviside’s unit step function and hence find their Laplace transforms.

Rt , 0 < t < 2 (1) f (t) = S T 4t, t > 2 Re , 0 < t < 3 . (3) f (t) = S T 0, t > 3 2

(2) f (t) =

RS3t, T 5,

0 4

–t

Solution. (1) Given function f (t) can be expressed in terms of the Heaviside’s unit step function as f (t) = t2 + (4t – t2) u (t – 2) Taking Laplace transform on both sides, we get, L { f (t)} = L (t2) + L {(4t – t2) u (t – 2)} = = =

{d i b g} 2 – L { bt – 2g − 4 u b t – 2g} s 2 – L t 2 – 4t u t – 2 s3 2

3

o

t

2 – e –2 s – L t 2 – 4 s3

(Using Heaviside shifting theorem.)

357

LAPLACE TRANSFORMS

{d i RS 2 − 4 UV. Ts s W

b g}

=

2 – e –2 s – L t 2 – 4 L 1 s3

=

2 – e –2 s s3

3

(2) In terms of the Heaviside step function We have f (t) = 3t + (5 – 3t) u (t – 4) ∴

L [ f (t)] = 3 L (t) + L {(5 – 3t) u (t – 4)}

o b g

b gt

=

3 + L –3 t–4 –7u t –4 s2

=

3 + e –4 s L 2 s

=

=

mb– 3t – 7gr 3 R – 3 − 7 UV +e S s T s sW 3 R 3 7U –S + Ve . s Ts sW –4 s

2

2

–4 s

2

2

f (t) = e–t +[0 – e–t] u (t – 3)

(3) Now, ∴

= e–t – e–t u (t – 3) = e–t – e– (t – 3) u (t – 3) e–3 L { f (t)} = L (e–t) – e–3 L {e– (t – 3) u (t – 3)}

d i

=

1 – e –3 e –3s L e – t s +1

=

1 1 –3 s + 1 –e b g⋅ s +1 s+1

1– e b = s+1

g

–3 s + 1

·

3. Express f (t) in terms of the Heavisides unit step function and find its Laplace transform:

R| t , f (t) = S 4t, |T 8, 2

04

Solution. We get f (t) = t2 + (4t – t2) u (t – 2) + (8 – 4t) u (t – 4) f (t) = t2 + [4 – (t – 2)2] u (t – 2) + [– 4 (t – 4) – 8] u (t – 4)

358


Taking Laplace transform on both sides, we get, L { f (t)} = L (t2) + L {[4 –(t – 2)2 ] u (t – 2)} + L {[– 4 (t – 4) – 8] u (t – 4)} =

2 s

3

d

b

i

g

+ e –2 s L 4 – t 2 + e –4 s L – 4t – 8

Using Heaviside shift theorem.

FG H

IJ K

FG H

IJ K

=

2 4 2 –4 8 – + e –2 s − + e –4 s s s3 s s3 s2

=

2 2 1 1 2 + 2e –2 s − 3 – 4 e –4 s 2 + . 3 s s s s s

FG H

IJ K

FG H

IJ K

4. Find L 2δ(t − 1) + 3δ(t − 2) + 4δ(t + 3) . Solution. We have = 2L δ (t – 1) + 3L δ (t – 2) + 4L δ (t + 3) Since L δ (t – a) = e–as

= 2e–s + 3e–2s + 4e3s. 5. Find L [cos h 3t δ (t – 2)]. Solution cos h 3t δ (t – 2) = L [cos h 3t δ (t – 2)] =

o

t b g

{

b g

1 3t e + e –3t δ t – 2 2

b g}

1 L e 3t δ t − 2 + L e −3t δ t − 2 2

s–3→s s+3→s

= shifting

=

=

=

{ b g + L δ bt – 2 g 1R UV + de i e i d S 2T W 1 {e b g + e b g } 2 1 L δ t–2 2 –2 s

–2 s

s→ s – 3

–2 s − 3

o

e –2 s 6 e + e –6 2 L [cosh 3t δ (t – 2)] = cosh 6 e– 2s =

s→s– 3

–2 s + 3

t

s→s+3

s→ s +3

}

359

LAPLACE TRANSFORMS


LMAns. e FG 2 + 5IJ OP H s sK Q N LMAns. e FG 2 + 4 + 4 IJ OP H s s sK Q N LMAns. e FG 2 + 8 + 14 IJ OP H s s s KQ N LMAns. e b g OP MN s − 3 PQ LMAns. e OP MN s b s + 1g PQ –s

1. (2t + 3) u (t –1).

2

– 2s

2. t2 u (t – 2).

3

2

3

2

–2 s

3. (t2 + 2t – 1) u (t – 3).

– s−3

4.

e3t

u (t – 1).

–2 s

5. (1 –

e2t)

u (t – 2).

Express the following functions in terms of the Heaviside’s unit step function and hence find their Laplace transformation.

RS T R 3, 2. f (t) = S T t,

1. f (t) =

3. f (t) = 4. f (t) =

2, 3,

|RS 4 , T| t , 2

RS 4t , Tt , 2

Re , f (t) = S T 2, 2t

5.

02 03

–2 s

2

–3s

.

02 01

R| sin t , 0 < t < π 2. 6. f (t) = S π |T cos t , t > 2 R| 1, 0 < t < 1 7. f (t) = S t , 1 < t < 3 . |Tt , t > 3 2

LMAns. 2 + e OP s Q N LMAns. 3 + e FG 1 − 1IJ OP N s H s sK Q LMAns. 4 + e FG 2 + 6 + 5IJ OP N s H s s sK Q LMAns. 4 + e FG 2 – 4 IJ OP H s sKQ N s LMAns. 1 + e F 2 – e I OP MN s − 2 GH s s − 1JK PQ FG IJ R 1 + s UO LM 1 PP – eH K S Ans. V 1 s + MN s + 1 T WQ –s

01

3

2

–2 s

2

3

2

–s

–π s 2

2

LMAns. 1 + e N s s

–s 2

2

+ e −3 s

FG 2 + 5 + 6 IJ OP H s s sKQ 3

2

360


R| cos t , 8. f (t) = Scos 2t , |Tcos 3t , R|cos t , 9. f (t) = S 1, |Tsin t ,

R|t , 10. f (t) = S 4, |T 0, 2

0 2π

LM R s +e S Ans. MN s + 1 T s − πs

2

UV W d

5se –2 πs s s + + 2 + 4 s2 + 1 s2 + 4 s2 + 9

LMAns. s + R 1 + s U e MN s + 1 ST s s + 1VW

0 2π

2

– πs

2

–

id

2 < t < 4. t >4

3

3

–2 πs

OP PQ

4 –4 s e s

OP PQ

RS 1 – 1 UV e T s s + 1W 2

LMAns. 2 – R 2 + 4 U e MN s ST s s VW

0
OP i PQ

2

–2 s

–

ADDITIONAL PROBLEMS (From Previous Years VTU Exams.) 1. Find the Laplace transforms of: cos 2t − cos 3t + t sin t t Solution. The given function be denoted by f (t) and let

2t +

f (t) = F (t) + G (t) + H (t) where ∴ Now

F (t) = 2t, G (t) =

cos 2t − cos 3t , t

H (t) = t sin t L [f (t)] = L [F (t)] + L [G (t)] + L [H (t)] L [F (t)] = L [2t] = L [elog2 . t ] = L [G (t)] = L =

z z

=

LM cos 2t − cos 3t OP N t Q L bcos 2t − cos 3t g ds FG s − s IJ ds H s + 4 s + 9K

∞

s

=

∞

s

2

2

LM 1 log d s + 4i − 1 log d s 2 N2 LM s + 4 OP MNlog s + 9 PQ 2

2

=

1 s − log 2

∞

2

s

2

+9

iOPQ

∞ s

...(1)

361

LAPLACE TRANSFORMS

=

LM MNlog

1 + 4 s2 1 + 9 s2

OP PQ

s2 + 4 s2 + 9

− log s=∞

s2 + 9 s2 + 4

= log 1 − log s2 + 9 s2 + 4 H (t) = t sin t

L [G (t)] = log Further

L [H (t)] = Hence

L [H (t)] =

LM N

b g

−d −d 1 L sin t = dt ds s 2 + 1

OP Q

2s

ds + 1i 2

2

Thus the required L f (t) is given by =

1 2s s2 + 9 + log 2 + 2 s − log 2 s +4 s2 + 1

d

2. Find the Laplace transforms of

t2

Solution. We shall first find L

(t2

we have

e–3t

=

d2 L (sin 2t) ds 2

LM N

OP Q O L d M −4 s P ds M s + 4 P MN d i PQ ds + 4i b−4g + 4s ⋅ 2ds + 4i ⋅ 2s d s + 4i 4 d3s − 4 i ds + 4i 4 3b s + 3g − 4 bs + 3g + 4 d d 2 2 ds ds s + 4

2

2

2

2

=

sin 2t.

sin 2t)

L [t2 (sin 2t )] = (–1)2 L (t2 sin 2t) =

2

2

2

L

(t2

sin 2t) =

2

3

2

Thus

L (e–3t t2 sin 2t) =

i

2

3

4

362


3. (i) Evaluate: L {t (sin3 t – cos3 t)} Solution sin3 t – cos3 t = L (sin3 t – cos3 t) =

1 1 (3 sin t – sin 3t) – (3 cos t + cos 3t) 4 4

RS T

L [t

t –

cos3

bg

UV W

−d f s , we have ds

Using the property: L [t f (t)] =

(sin3

UV RS W T

1 3 3 1 3s s − + 2 − 2 2 2 4 s +1 s +9 4 s +1 s +9

O L LM −2s s + 9 − 2s 2 s P 1 M 3d s + 1i − 2 s MM ds + 1i + ds + 9i PP + 4 MM ds + 1i + ds + 9i Q N N OP 1 LM 1 − s 9 − s OP L 1 3s M 1 − + 3⋅ + P 2 M d s + 1i 4 M s +1 s + 9 d i d i ds + 9i PPQ PQ MN NM 2

−3 t)] = 4

2

2

2

2

2

2

2

2

2

2

=

2

2

2

2

(ii) Using Laplace transforms:

z

Evaluate

∞

e − t t sin 2 3t dt .

0

Solution. We shall first find L (t sin2 3t) sin2 3t = L (sin2 3t) = L (t sin2 3t) =

1 (1 – cos 6t) 2

LM N

s 1 1 − 2 2 s s + 36

LM N

OP Q

1 −d 1 s ⋅ − 2 ds s s 2 + 36

LM MM N

d

i

OP Q

OP i PPQ

2 2 −1 −1 s + 36 − 2 s − = 2 2 s2 s2 + 36

∴

L (t

sin2

LM MM N

d

d

Using the basic definition in LHS, we have

z

∞

0

e

− st

OP i PPQ

1 1 36 − s2 + 3t) = 2 2 2 s s 2 + 36

LM MM N

OP i PPQ

1 1 36 − s2 t sin 3t dt = + 2 2 s2 s 2 + 36 2

d

2

2

2

2

2

2

2

OP PP Q

363

LAPLACE TRANSFORMS

Thus by putting s = 1, we get

z

∞

e − t t sin 2 3t dt =

0

LM N

1 35 1+ 2 2 37

OP Q

702 1369

=

4. Find the Laplace transform of: 1 − cos 3t t

(i) e2t cos2 t

(ii)

Solution. (i) Let

f (t) = e2t cos2 t = e2t · L [f (t)] = = = =

Thus

L (e2t cos2 t) =

(ii) Let

1 [L (e2t) + L (e2t cos 2t)] 2

LM m b gr N 1L 1 R s UV +S M 2 N s − 2 Ts + 4W s−2 O 1L 1 + M 2 N s − 2 s − 4 s + 8 PQ s−2 O 1L 1 + M 2 N s − 2 s − 4 s + 8 PQ 1 1 + L cos 2 t 2 s−2 2

s→ s − 2

s→ s − 2

OP Q

OP Q

2

2

f (t) = 1 – cos 3t —

∴

f (s) = L [f (t)] =

We have the property: L i.e.,

1 (1 + cos 2t) 2

L

LM1 − cos 3t OP N t Q

bg

f t t

=

z LMN ∞

= =

s

z

∞

s 1 − s s2 + 9

bg

f s ds

s

OP Q

s 1 ds − 2 s s +9

LMlog s − 1 log ds + 9iOP N 2 Q LMlog s OP NM s + 9 QP LM OP s log MN s 1 + 9 s PQ − log ∞

2

s

∞

=

2

s

0

=

2

s=∞

= log 1 − log

L

LM1 − cos 3t OP N t Q

= log

LM MN

s2 + 9 s

s s2 + 9

OP PQ

= log

s s +9 2

s2 + 9 s

364


5. Find the Laplace transform of the full wave rectifier f (t) = E sin wt, 0 < t < π/w having period π/w. Solution. Refer page No. 349. Example 3. 6. Express the followiong functions interms of Heaviside’s unit step function and hence find its Laplace transform where

RSt , 0 < t ≤ 2 T4t, t > 2 2

f (t) =

Solution. (1) Refer page No. 356. Example 2. 7. Express the function. f (t) =

RSπ − t, 0 < t ≤ π Tsin t, t > π

in terms of unit step function and hance find its laplace transform. Solution. f (t) = (π – t) + [sin t – (π – t)] u (t – π) by standard property. i.e., f (t) = (π – t) + [sin t – π + t] u (t – π) L [f (t)] = L (π – t) + L {[sin t – π + t] u (t – π)} Taking F (t – π) = sin t – π + t, we have F (t) = sin (t + π) – π + (t + π) i.e., F (t) = – sin t + t ∴

—

F (s) = L [F (t)] = —

...(1)

−1 1 + s2 + 1 s2

Also L [F (t – π) u (t – π)] = e–πs F (s) ∴ L [(sin t – π + t) u (t – π)] = e–πs

FG 1 − 1 IJ H s s + 1K 2

...(2)

2

Thus by using (2) in (1) with L (π – t) =

π 1 − s s2

L [ f (t)] =

FG π − 1 IJ + e FG 1 − 1 IJ H s s K H s s + 1K

we get

− πs

2

2

2

8. Find (i) L [t2 u (t – 3)], (ii) L [e3t u (t – 2)]. Solution. (2) & (4) Refer page No. 355. Example 1. 9. Define Heaviside unit step function. Using unit step function find the laplace transform of:

R|sin t, 0 ≤ t < π f (t) = Ssin 2t, π ≤ t < 2π |Tsin 3t, t > 2π

365

LAPLACE TRANSFORMS

Solution. The given f (t) can be written in the following form by a standard property. f (t) = sin t + [sin 2t – sin t] u (t – π) + [sin 3t – sin 2t] u (t – 2π) Now L [f (t)] = L (sin t) + L {[sin 2t – sin t] u (t – π)} + L {[sin 3t – sin 2t] u (t – 2π)} ...(1) Consider L [sin 2t – sin t] u (t – π) Let F (t – π) = sin 2t – sin t ⇒ F (t) = sin 2 (t + π) – sin (t + π) F (t) = sin (2π + 2t) – sin (π + t) or F (t) = sin 2t + sin t 2 1 + 2 s + 4 s +1

—

∴

F (s) = L [F (t)] =

2

—

But L [F (t – π) u (t – π)] = e–πs F (s) − πs i.e., L {[sin 2t – sin t] u (t – π)} = e

Also i.e., i.e.,

2

...(2)

2

G (t – 2π) = sin 3t – sin 2t G (t) = sin 3 (t + 2π) – sin 2 (t + 2π) G (t) = sin 3t – sin 2t —

∴

L [G (t)] = G (s) =

3 2 − 2 s +9 s +4 2

—

L [G (t – 2π) u (t – 2π)] = e–2πs G (s)

But i.e.,

LM 2 + 1 OP N s + 4 s + 1Q

L {(sin 3t – sin 2t] u (t – 2π)} = e–2πs

LM 3 − 2 OP Ns + 9 s + 4Q 2

...(3)

2

Thus (1) as a result of (2) and (3) becomes L [f (t)] =

LM N

OP Q

LM N

OP Q

1 2 1 3 2 + e–2πs 2 + e −πs 2 + 2 − 2 · s +1 s + 4 s +1 s +9 s +4 2

OBJECTIVE QUESTIONS 1. Laplace transform of tn e+at is: (a)

(c)

n!

b s + ag

n

(b)

n +1

(d)

n!

b s − ag

bn + 1g! b s + ag bn + 1g! b s + ag

n +1

n +1

[Ans. c]

366


2. Laplace transform of f (t) = te at sin (at), t > 0. (a)

b g bs − ag + a 2a s − a 2

(c)

2

2

2

b g bs − ag + a bs − ag bs − ag + a a s−a 2

2

2

s−a

b s − ag

(b)

(d)

− a2

2

2

[Ans. a]

3. The Laplace transform of te–t cos h 2t is: (a)

(c)

4. L

s2 + 2s + 5

ds ds

2

i

2

(d)

+ 2s − 3 4s + 4

2

2

(b)

i

+ 2s − 3

LM sin t OP = Nt Q

1 s +1 (c) cot–1 (s – 1) (a)

2

s2 − 2 s + 5

ds ds

2

i

+ 2s − 3 4s − 4

2

i

+ 2s − 3

2

2

[Ans. a]

(b) cot–1 s (d) tan–1 s

[Ans. b]

5. The relation between unit step function and unit impulse function is: (a) L [u (t – a)] = L [δ (t)]

(b) L [u′ (t – a)] = L [δ (t – a)]

(c) L [u (t)] = L [δ′ (t – a)]

(d) None of these

[Ans. b]

6. The laplace transform of sin2 3t is: (a)

(c)

3 s + 36

(b)

2

d

18

i

s s + 36 2

(d)

d

6

i

s s + 36 2

18 s + 36

[Ans. c]

2

7. L [t2 et] = (a)

(c)

2

b s − 2g

2

(b)

3

(d)

1

bs − 2g

2

b s − 2g

3

1

b s − 1g

[Ans. b]

3

8. L [e–t sin h t] = (a)

(c)

1

b s + 1g

2

1 s s+2

+1

b g

(b)

(d)

1

b s − 1g

2

+1

s−1

b s − 1g

2

+1

[Ans. c]

367

LAPLACE TRANSFORMS

9. L (e–3t cos 3t) = s−3 s − 6 s − 18 s+3 (c) 2 s − 6s + 18 10. L [(t2 + 1) u (t – 1)] = (a) 2e–s (1 + s + s2)/s3 (c) 2es (1 + s + s2)/s3 11. Laplace transform of (t sin t) is:

(a)

(a)

(c)

2

2s

ds + 1i 2

s+3 s + 6s + 18 s−3 (d) 2 s + 6s − 18 (b)

(b) e–s (1 + s + s2)/s3 (d) None of these

(b)

2

2s s +1

(d)

2

12. Laplace transform of f (t): t ≥ 0 is defined as: (a)

z z

1

bg

e − st f t dt

0

(c)

2

bg

f t dt

0

2

(b)

[Ans. b]

[Ans. a]

s s+1 2s

ds − 1i 2

z

∞

[Ans. a]

2

bg

e − st f t dt

0

(d) None of these

[Ans. b]

13. A unit step function is defined as:

RS0, t < a a ≥ 0 T1, t ≥ a R0, t > a (c) u (t – a) = S T1, t ≥ a (a) u (t – a) =

(b) t – a = 0 (d) None of these

[Ans. a]

14. L (e2t sin t) is: (a) (c)

1 s +5 1

(b)

2

s2 − 4s + 5 15. L (et) = (a)

1 s − log 2

1 s+2 16. L (e–t t k ) is:

(c)

(a) (c)

k!

b s + 1g

k +1

k!

b s − 1g

k +1

1 s − 4s 2

(d) None of these

(b)

1 s + log 2

(d) None of these

(b)

[Ans. c]

[Ans. a]

s!

b s + 1g

k +1

(d) None of these

[Ans. a]

368


17. L (cos3 4t) = (a) (c)

FG H 1 F1 G − 2 Hs s

IJ K s I J + 16 K

s 1 1 + 2 3 s s − 16 2

(b)

FG H

s 1 1 + 2 s s 2 − 16

IJ K

(d) None of these

18. Laplace transform of the unit impulse function δ (t – a) is: (b) eas (a) e–as (c) e–t (d) None of these 19. L [cos (2t + 3)] = (a)

s cos 3 − 2 sin 3 s2 + 4

s cos 3 + 2 sin 3 s2 − 4 20. L [e at] is: (c)

(a)

1 s+a

1 s−a L [ f ′ (t)] is: (a) S L [f (t)] – f (0) (c) S L [f (t)] L f (t) u (t – a) is: (a) eas Lf (t) (c) f (t + a) The Laplace transform of t3 δ (t – 4) is: (a) 43 e–4s (c) e4s 32 L [eiat] is: (c)

21.

22.

23.

24.

(a)

1 s + ia

1 s 25. The Laplace transform of (1 + cos 2t) is: (c)

(b)

[Ans. a]

s2 cos 3 − 2 sin 3 s2 − 4

(d) None of these

(b)

[Ans. b]

[Ans. a]

1 s

(d) None of these

[Ans. c]

(b) F (s) (d) None of these

[Ans. a]

(b) e–as L [f (t + a)] (d) None of these

[Ans. b]

(b) 34 e3s (d) None of these

[Ans. a]

(b)

1 s − ia

(d) None of these

(a)

1 s + 2 s s +4

(b)

(c)

1 1 + s2 s

(d) None of these

[Ans. b]

1 s − 2 s s −4 [Ans. a] GGG

UNIT

8111

Inverse Laplace Transforms 8.1 INTRODUCTION If L { f (t)} = F (s), then f (t) is called the Inverse Laplace Transform of F (s) and symbolically, we write f (t) = L–1 { F (s)}. Here L–1 is called the inverse Laplace transform operator. For example, (i)

L (e at ) =

(ii)

L (t) =

RS T

1 , 1 s > a, L–1 s–a s–a 1 , s2

L–1

RS 1 UV = t Ts W

UV = e W

at

2

8.2 INVERSE LAPLACE TRANSFORMS OF SOME STANDARD FUNCTIONS 1. Since

L (1) = L (eat ) =

2.

1 , s

L–1

RS 1UV T sW

RS 1 UV = e Ts + a W

3.

a , s2 + a 2

L–1

4.

RS Ts

2

1 + a2

UV W

=

L (cos at) =

RS T

1 , 1 –1 s > a, L s–a s–a

Replacing a by –a, we get L–1 L (sin at) =

= 1.

– at

s , s + a2 2

RS Ts

L–1

RS Ts

369

at

.

L–1

sin at ⋅ a

UV = e . W

2

a + a2

UV = sin at. W

2

s + a2

UV = cos at. W

370


5.

L (sin h at) =

RS Ts

1 – a2

UV W

a , s – a2 2

=

sin h at ⋅ a

6.

L (cos h at) =

s , s – a2

7.

L (t n) =

L–1

L–1

L–1

2

RS n ! UV Ts W RS 1 UV Ts W

2

L–1

RS Ts

L–1

RS Ts

UV W

a = sin h at. – a2

2

s – a2

2

UV W

= cos h at.

n! where n is a positive integer, we get sn + 1

n+1

= tn

n+1

=

tn · n!

RS 1 UV = t T s W bn – 1g ! n –1

Replacing n by n – 1, we get L–1

RS 1UV T sW RS 1 UV Ts W

L–1

In particular,

L–1

= 1 =

t2 –1 = t ( 2 – 1) !

1 = s3

t3–1 t2 = 3–1 ! 2

2

L–1

L (t n) =

Since,

n

b g Γ bn + 1g , s

n+1

L–1

RS 1 UV Ts W n+1

tn , n > – 1. Γ n +1

b g

=

The following table gives list of the Inverse Laplace Transform of some standard functions. S.No.

F (s)

f (t) = L– 1 {F (s)}

S.No.

1.

1 s–a

e at

6.

2.

1 s+a

e– at

7.

sin at a

8.

cos at

9.

sin h at a

10.

3. 4.

5.

1 s2 + a 2 s 2

s +a

2

1 s2 – a 2

F (s)

s s2 – a 2

1 s 1 s2 1 s

n

s

f (t) = L– 1 {F (s)} cos h at 1 t tn –1 n –1 !

, n = 1, 2, ...

b g

1

tn Γ n+1

,n> –1 n +1

b g

371

INVERSE LAPLACE TRANSFORMS

Following properties immediately follow from the corresponding properties of the Laplace transform.

Properties of inverse Laplace transform 1. Linearity Property If a and b are two constants then L–1 {a F (s) + b G (s)} = a L–1{F (s)} + b L–1 {G (s)} This result can be extended to more than two functions. This shows that like L, L–1 is also a linear operator.

RS 2 – 3s + 4 UV T s – 3 s + 16 s – 9 W R 1 UV – 3 L RS s UV + 4 L RS = 2L S T s – 3W Ts + 4 W Ts

Example: L–1

2

2

–1

–1

–1

2

= 2 e3t – 3 cos 4t + 4

2

2

1 – 32

UV W

sin h 3t · 3

2. Shifting Property L–1{F (s)} = f (t), then

If

L–1 {F (s – a)} = e at f (t) = eat L–1 {F (s)} This follows immediately from the result. If

then L {e at f (t)} = F (s – a)

L { f (t)} = F (s),

Replacing a by – a, we get L–1 {F (s + a)} = e– at f (t) = e– at L–1 {F (s)}. Examples: (i)

L–1

RS Ts

2

1 – 2s + 5

UV W

t = e

t = e

= (ii)

L–1

RS Ts

2

s–3 – 6s + 13

UV W

R| 1 U| S| bs – 1g + 2 V| T W R 1 UV L S Ts + 2 W

–1 = L

2

2

–1

2

2

sin 2t 2

1 t e sin 2t . 2

–1 = L

3t = e

R| s – 3 U| S| bs – 3g + 2 V| T W R s UV L S Ts + 2 W 2

2

–1

= e3t cos 2t.

2

2

372


WORKED OUT EXAMPLES 1. Find the inverse Laplace transforms of the following functions: (i)

1 2s – 3

(ii)

s (v) 2 9s + 4 8 – 6s (ix) 16s2 + 9

1 4s + 5 1

(vi)

(i)

L–1

RS 1 UV T 2s – 3W

(ii)

L–1

RS 1 UV T 4s + 5W

L–1

RS 1 UV Ts + 9W 2

=

= =

(v)

(vi)

–1

RS s UV T s – 16 W RS s UV T 9s + 4 W 2

s

s s 2 – 16

(viii)

2s + 18 s 2 + 25

3

6

–1

–1

–

–1

–1

2

2

1 sin 3t. 3

–1 = L

2

2

–1

2

2

–1

b s + 2g

(iv)

R| 1 U| = 1 L R| 1 U| = S| 2 FG s – 3IJ V| 2 S| s – 3 V| T 2W T H 2K W R 1 U| L | S| 4 FG s + 5IJ V| T H 4K W 1 R 1 U| = 1 e FGH IJK . L | S| s + 5 V| 4 4 T 4W R 1 UV L S Ts + 3 W

RS s UV Ts – 4 W = cos h 4t. 1 R s U| L | L = 9 S| s + FG 2 IJ V| T H 3K W 1 F2 I cos G t J · = H3 K 9 R 1 UV = L R| 1 U| L S S|16 FG s – 9 IJ V| T16s – 9 W T H 16K W 1 R 1 U| L | = 16 S| s – FG 3 IJ V| T H 4K W L–1

1 s2 + 9

–1 = L

=

(iv)

(vii)

2

16s – 9 1 (x) 3 2 ⋅ s

Solution

(iii)

(iii)

2

–1

2

2

–1

2

2

5 t 4

3

1 2t e . 2

373


R| bs + 2g U| S| s V| T W

1 16

1 . sin h 3 t 3 4 4

=

1 3 sin h t · 12 4

FG IJ HK

3

(vii)

–1

L

6

–1 = L

L–1

RS 2s + 18 UV T s + 25W 2

FG IJ H K

RS s + 6s + 12s + 8 UV s T W RS 1 + 6 + 12 + 8 UV Ts s s s W RS 1 UV + 6 L RS 1 UV + 12 L RS 1 UV + 8 L RS 1 UV Ts W Ts W Ts W Ts W 3

–1 = L

–1 = L

(viii)

FG IJ H K

=

2

6

3

4

RS 8 – 6s UV T16s + 9 W 2

–1

5

=

t2 t3 t4 t5 + 6 + 12 +8 2! 3! 4! 5!

=

t2 t4 t5 + t3 + + ⋅ 2 2 15

= 2 L–1

RS Ts RS Ts

2

UV W s U V + 18 L +5 W

UV W 1 U V +5 W

s 1 + 18 L–1 2 s + 25 + 25 –1

2

RS T RS Ts

2

2

2

18 sin 5t. 5

–1

2

2

–1

–1

2

=

=

2

–1

–1

2

2

=

6

RS 1 UV – 6 L RS s UV T16s + 9 W T16s + 9 W R 1 U| – 6 L R| s U| 8L | S|16 FG s + 9 IJ V| S|16 FG s + 9 IJ V| H K 16 T W T H 16 K W 8 R 1 U| − 6 L R| s U| L | 16 S| s + FG 3 IJ V| 16 S| s + FG 3IJ V| T H 4K W T H 4K W 2 1 F3 I 3 F3 I sin G t J – cos G t J 3 F 3I GH 4 JK H 4 K 8 H 4 K 8 F3 I 3 F3 I sin t − cos t ⋅ H4 K 8 H4 K 9

–1 = 8L

=

–1

4

= 2 cos 5t +

L–1

6

–1

3

–1 = 2L

(ix)

5

2

2

374


L–1

(x)

1 sn + 1

RS 1 UV Ts W RS 1 UV Ts W

L–1

or

L–1

we have,

n

32

tn = Γ n +1

b g

n –1

= =

t Γ n

bg

t

3 2 −1

Γ

t1 2 t1 2 = = 1 1 1 Γ +1 Γ 2 2 2

FG 3IJ H 2K

FG H

IJ K

FG IJ HK

=

FG where ΓFG 1 IJ H 2K H

2 t

π 2. Find the inverse Laplace transforms of the following functions: 1 s+2 3 (ii) 2 (iii) 2 (i) 4 s + 4s + 20 s + 8s + 25 s–2

b g

(iv)

2s – 3 s 2 − 2s + 5

s 4s 2 + 12s + 5

(v)

s+3 · 4s 2 + 4s + 9

(vi)

Solutions. Using the shifting rule, we have L–1

(i)

R| 3 S| bs – 2g T

4

U| V| W

2 t –1 = 3e L

2t = 3e .

(ii)

L–1

RS Ts

2

1 + 4 s + 20

UV W

–1 = L

–2 t = e

= e– 2t = (iii)

L–1

RS Ts

2

s+2 + 8s + 25

UV W

RS 1 UV Ts W

s→s–2

4

t 3 e2t t3 = · 3! 2

R| 1 U| S| bs + 2g + 4 V| T W R 1 UV L S Ts + 4 W 2

2

–1

2

2

1 sin 4t 4

1 – 2t e sin 4t. 4

–1 = L

– 4t = e

– 4t = e

– 4t = e

R| bs + 4g – 2 U| S| bs + 4g + 3 V| s→s+4 T W R s – 2 UV L S Ts + 3 W LM L R s U – 2 L R 1 UOP ST s + 3 VWP MN ST s + 3 VW Q LMcos 3t – 2 sin 3t OP. 3 N Q 2

2

–1

2

2

–1

–1

2

2

2

2

=

π

IJ K

375


(iv)

L–1

RS 2s – 3 UV T s – 2s + 5 W 2

=

= = (v)

L–1

RS T 4s

s + 12s + 5

2

UV W

=

t

2

2

–1

2

t

2

–1

–1

2

2

2

2

t

–1

2

=

s2 + 3s + 5/4 =

Now,

R| 2bs – 1g – 1 U| (s – 1 → s) S| bs – 1g + 2 V| T W R 2s – 1 UV e L S Ts + 2 W L R s U – L R 1 UOP e M2 L S MN T s + 2 VW ST s + 2 VWPQ 1 L O e M2 cos 2t – sin 2t P. 2 N Q s R U| L | S| 4 FG s + 3s + 5 IJ V| 4K W T H 1 R s U| L | S| s + 3s + 5 V| 4 4W T FG s + 2 ⋅ 3 ⋅ s + 9 IJ + 5 – 9 H 2 4K 4 4 FG s + 3IJ – 1 H 2K R| FG s + 3 IJ – 3 U| 1 | H 2 K 2 |V L S 4 || FG s + 3IJ – 1|| TH 2K W R| s – 3 U| 1 b g 2 e L S 4 – s |T 1V|W R s UV – 3 L RS 1 UVOP 1 b g L e L S M 4 MN T s – 1W 2 T s – 1WPQ 1 b g L 3 e MNcos h t – 2 sin ht OPQ. 4 R s + 3 U| L | S| 4 FG s + s + 9 IJ V| 4K W TH 1 R s + 3 U| L | S| s + s + 9 V| 4 4W T

–1 = L

–1

2

2

2

=

∴

L–1

RS T 4s

s + 12s + 5

2

UV W

=

=

= = (vi)

L–1

RS T 4s

2

s+3 + 4s + 9

UV W

=

–1

2

– 32 t

–1

2

– 32 t

–1

– 32 t

–1

2

=

–1

2

–1

2

2

376


9 s +s+ = 4 2

Now,

LMs + 2 ⋅ s ⋅ 1 + FG 1 IJ OP + 9 – 1 2 H 2 K PQ 4 4 MN FG s + 1 IJ + 2 H 2K R| FG s + 1 IJ + 5 U| 1 | H 2 K 2 |V L S 4 || FG s + 1 IJ + e 2 j || TH 2K W R| s + 5 U| 1 b g 2 e L S 4 |T s + e 2 j V|W L R| s U| 5 R| 1 1 b g M e MM L S| s + b 2 g V| + 2 L S| s + 2 4 W T e j N T O 1 b g L 5 e cos 2 t + sin 2 t P . M 4 2 2 N Q 2

2

2

=

∴

1 –1 L 4

R| U| |S s + 3 |V || FG s + 1 IJ + 2 || TH 2K W 2

=

=

=

=

–1

2

– 12 t

2

–1

2

2

– 12 t

–1

–1

2

2

2

2

U|OP V|P WPQ

– 12 t

8.3 INVERSE LAPLACE TRANSFORMS USING PARTIAL FRACTIONS In this method we first resolve the given rational function of s into partial fractions and then find the inverse Laplace transform of each fraction. 3. Find the inverse Laplace transforms of the following functions: s 2s – 1 (i) 2 (ii) 2s – 1 3s – 1 s – 5s + 6

b

(iii) (v)

2s – 3 s–1 s–2 s–3

b gb gb g 3s – 1

b s – 3g d s

2

+4

(iv)

i

(vi)

gb

4s + 5

bs – 1g bs + 2g 2

s2

ds + 4ids 2

Solution (i) Let

2s – 1 = s 2 – 5s + 6

2s – 1

bs – 2gbs – 3g

=

A B + s–2 s–3

2s – 1 = A (s – 3) + B (s – 2) Put

s = 3,

B= 5

Put

s = 2,

A = –3

∴

2s – 1 3 5 + = – s–2 s–3 s 2 – 5s + 6

g

2

+9

i

.

377


L–1

Then

RS 2s – 1 UV = T s – 5s + 6 W 2

– 3 L–1

RS 1 UV + 5 L RS 1 UV Ts – 2W T s – 3W –1

= – 3e 2t + 5e 3t.

s

b2s – 1gb3s – 1g

(ii) Let

=

A B + 2 s – 1 3s – 1

s = A (3s – 1) + B (2s – 1) Put

s = s

∴

L–1

b2s – 1gb3s – 1g |RS s |UV |Tb2s – 1gb3s – 1g |W

=

1, 2

A = 1,

=

2s – 3 s–1 s–2 s–3

B = –1

RS 1 UV – L RS 1 UV T 2s – 1W T 3s – 1W R 1 U| – L R| 1 U| L | S| 2 FG s – 1 IJ V| S| 3 FG s – 1IJ V| T H 2 K W T H 3K W 1 R 1 U| – 1 L R| 1 U| L | S| s – 1 V| 3 S| s – 1 V| 2 T 2W T 3W –1

–1

–1

–1

–1

1

=

1, 3

1 1 – 2 s – 1 3s – 1

–1 = L

=

(iii) Let

and s =

1

1 2t 1 3t e – e . 2 3 A B C + + s–1 s–2 s–3

b gb gb g b g b g b g =

2s – 3 = A (s – 2) (s – 3) + B (s – 1) (s – 3) + C (s – 1) (s – 2) Put

s = 1,

⇒ A =

–1 2

3 2 s = 2, ⇒ B = – 1 s = 3,

Thus,

L–1

⇒ C =

1 3 1 2s – 3 2 – + 2 = s–1 s–2 s–3 s–1 s–2 s–3

b gb gb g R|S 2s – 3 U|V |Tbs – 1gbs – 2gbs – 3g |W

–

RS T

UV W

RS T

UV W

RS T

=

– 1 –1 1 1 3 1 – L– 1 + L– 1 L 2 2 s–1 s–2 s–3

=

–1 t 3 e – e 2 t + e 3t . 2 2

UV W

378


4s + 5

bs – 1g bs + 2g

(iv) Let

2

=

A B C + + 2 s–1 s–1 s+2

b g

4s + 5 = A (s – 1) (s + 2) + B (s + 2) + C (s – 1)2 s = 1, ⇒B = 3

Put

s = – 2, ⇒ C =

–1 3

To find A, put s = 0, Then

5 = – 2A + 2B + C

This gives

A =

1 3

Thus the partial fraction is

4s + 5

bs – 1g bs + 2g R| 4s + 5 U| S| b s – 1g bs + 2g V| T W 2

∴

L–1

2

1 1 3 – 3 = 3 + s–1 s –1 2 s+2 =

=

(v) Let Then

3s – 1

bs – 3g d s

2

i

+4

b g b g R 1 UV + 3 L R|S 1 1 L S 3 T s – 1W |T bs – 1g 1 L R 1 UO 1 e + 3e M L S VP – e 3 N T s WQ 3 –1

t

–1

t

U| 1 R 1 U V| – 3 L ST s + 2 VW W –1

2

–2 t

–1

2

=

1 t 1 e + 3e t ⋅ t – e – 2 t 3 3

=

1 t 1 e + 3t e t – e – 2 t . 3 3

=

Bs + C A + s – 3 s2 + 4

3s – 1 = A (s2 + 4) + (Bs + C) (s – 3)

8 13 Now, 3s – 1 = (A + B) s2 + (– 3B + C ) s + (4A – 3C ) Comparing the coefficients, we get A + B = 0, – 3B + C = 3, 4A – 3C = – 1 Put

s = 3 A=

A =

∴

3s – 1

bs – 3g d s

2

i

+4

–8 15 8 , , C= B = 13 13 13

8 15 –8 s+ 13 + 13 13 = 2 s–3 s +4

=

8 1 1 8s – 15 ⋅ – 13 s – 3 13 s 2 + 4

379


L–1

∴

R| 3s – 1 U| S| bs – 3g ds + 4i V| T W 2

b

gb

g

RS T

UV W

UV W

8 –1 1 1 –1 8s – 15 – L L 13 13 s–3 s2 + 4

=

8 3t 1 1 s 8 L–1 2 e – – 15 L–1 2 13 13 s +4 s +4

=

8 3t e 13

=

8 3t e 13

=

A B + x+4 x+9

(vi) Let s2 = x, then

x x+4 x+9

RS T

=

LM R U RS UVOP S V MN T W T WPQ R s UV – 15 L RS 1 UVOP 1 L 8L S – M 13 MN Ts + 2 W T s + 2 WPQ 1 L 15 O 8 cos 2t – – sin 2 t P . M 13 N 2 Q –1

–1

2

2

2

x = A (x + 9) + B (x + 4) Put

x

x = – 4, A =

–4 5

x = – 9, B =

9 5

b x + 4gb x + 9g

∴

ds

i.e.,

L–1

R| S| ds T

s

2

−4 9 5 + 5 = 2 s + 4 s2 + 9

i U| s V= + 4 id s + 9 i | W

2

id

−4 9 5 + 5 = x+4 x+9

+ 4 s2 + 9 2

2

2

RS T RS Ts

UV W

RS T

– 4 –1 1 9 1 L + L–1 2 2 5 5 s +4 s +9

UV W

=

– 4 –1 L 5

=

–4 1 9 1 ⋅ sin 2 t + ⋅ sin 3t 5 2 5 3

2

RS T

1 9 1 + L–1 2 5 + 22 s + 32

–2 3 sin 2 t + sin 3t . 5 5 4. Find the inverse Laplace transforms of the following functions:

=

(i)

s 2 + 2s – 4 s 4 + 2s3 + s2

UV W

(ii)

s . s + s2 + 1 4

UV W

2

380


Solution

s 2 + 2s – 4 s2 + 2s – 4 s2 + 2 s – 4 = 4 3 2 = 2 s + 2s + s s2 s 2 + 2 s + 1 s2 s + 1

d

(i) Now

s2 + 2s – 4 s

2

bs + 1g

2

=

b g

i

A B C D + 2 + + s s s+1 s+1 2

b g

2

s + 2s – 4 = A (s) (s + 1) + B (s + 1)2 + C s2 (s + 1) + D s2

Hence,

2

Put

s = 0, B = – 4 s = – 1,

D = –5

2

s + 2s – 4 = (A + C ) s3 + (2A + B + C + D) s2 + (A + 2B) s + B

Now,

Comparing the coefficient, we get A + C = 0, 2A + B + C + D = 1, Hence

A + 2B = 2, B = – 4

A = 10, C = – 10

s + 2s – 4 10 4 10 5 – – – = s s2 s + 1 s + 1 2 s + 2 s3 + s 2 2

∴

b g LM s + 2s – 4 OP = 10 L RS 1 UV – 4 L RS 1 UV – 10 L RS 1 UV – 5 L TsW Ts W T s + 1W N s + 2s + s Q R1U = 10.1 – 4 . t – 10e – 5e L S V Ts W 4

2

L–1

–1

4

3

–1

–1

–1

2

2

–t

–1

–t

2

= 10 – 4t – 10e– t – 5t e– t.

(ii) Since,

s4 + s2 + 1 = (s2 + 1)2 – s2 = (s2 + 1 – s) (s2 + 1 + s) s = s 4 + s2 + 1

We have,

ds

s

2

id

i

+ 1 – s s2 + 1 + s

d s + s + 1i – ds – s + 1i ds + s + 1ids – s + 1i OP 1L 1 1 – = M 2 N s – s + 1 s + s + 1Q 1 1 1 L OP – = M 2 F MM GH s – 21 IJK + 43 FGH s + 12 IJK + 43 PP N Q UV = 1 L L R| 1 U| – L R| 1 2 MM + 1W S| FG s – 1 IJ + 3 V| S| FG s + 1 IJ MN T H 2 K 4 W T H 2 K 1 = 2

2

2

2

2

2

2

2

–1 Hence, L

RS Ts

4

s + s2

2

–1

–1

2

2

U|OP 3V + |P 4 WPQ

R| 1 S| bs + 1g T

2

U| V| W

381


=

=

= 5. Show that

z

∞

0

Solution

LM MM N 1L b Me 2 MN

=

{ }

L e – tx

2

= =

= = Hence,

g

12 t

3

f (t) =

L–1

z

∞

0

RS 1 UV Ts W

e

3 t. 2

sin

2

e – tx dx e – st

0

∞

∞

n

2

0

0

1 1 + x2

z

∞

0

–1

π

2 s π –1 L 2

RS 1 UV T sW –1

FG IJ HK

tn–1 Γ n

bg 1

π t 2 dx = 2 π

OP sQ

x

2

ze ∞

1 dx = s + x2

LM 1 tan N s

OP Q O dt P dx Q

e – tx dx dt

e – st ⋅ e – tx

π t2 = 1 2 Γ 2 =

∞

0

−

– tx 2

b g

− 12 t

0

1

Since,

3 – 12 t 2 t –e b g sin 2 3

sin

−e

z z LMNz z LMNz ∞

L { f (t)} =

Since,

2

0

∞

0

s

j

1 2

dx + x2

R| 1 S| s + FG 3 IJ T H2K 3 O tP 2 PQ

L–1

2

3

eb

1

∞

f (t) =

∴

g

12 t

g

−1 2 t

2

π ⋅ 2

2

e – x dx =

Let

R| 1 U| – eb S| s + FG 3 IJ V| T H2K W

1 (1/ 2 ) t –1 e L 2

2

2

U|OP V|PP WQ

382


π 1 ⋅ 2 t t = 1, We get =

Taking

z

∞

0

2

e – x dx =

π · 2

EXERCISE 8.1 Find the inverse Laplace transforms of the following functions: 1.

2 . s3

s . s +9 s 5. 2 . s –3

3.

2

Ans. t 2

2.

Ans. cos 3t

4.

Ans. cos h 3 t

LMAns. 1 eb g OP N 2 Q LMAns. 2 cos 5t – 1 sin 5t OP 5 N Q

7.

1 . 2s – 1

9.

2s – 1 . s2 + 25

11.

5s + 4 . s3

13.

2s + 3 . 4 s 2 + 20

14.

3s – 4 . 4s2 – 1

15.

1 – 3s . 4 s2 + 9

12 t

Ans. 5t + 2t 2

LMAns. 1 sin 4t OP N 4 Q LMAns. 1 sin h 2t OP N 2 Q

1 . s + 16 2

1 . s –4 1 . 6. s+6

8. 10. 12.

2

Ans. e – 6t

LMAns. 1 sin 3 t OP N 6 2Q LMAns. 4t – 16 t OP 24 – 30 s . s π N Q 8 – 6s LMAns. 2 sin 3t – 3 cos 3t OP . 16s + 9 N 3 4 8 4Q LMAns. 1 cos 5 t + 3 sin 5 t OP 4 5 N 2 Q LMAns. 3 cos h t – 2 sin h t OP 2Q N 4 2 LMAns. 1 sin 3t – 3 cos 3t OP N 6 2 4 2Q 1 . 4 s2 + 9

3

2

Find the inverse Laplace transforms of the following functions: 1.

1

bs – 3g

3

·

2.

s · s + 6s – 7

3.

2s + 3 · s – 4 s + 40

4.

2

2

1

b2s – 1g

2

·

52

4

LMAns. 1 t N 2

OP Q LMAns. e FG cos h 4t – 3 sin h 4t IJ OP H KQ 4 N LMAns. e FG 2 cos 6t + 7 sin 6tIJ OP H KQ 6 N LMAns. 1 t eb g OP N 4 Q 2

e 3t

– 3t

2t

12 t

383


5.

2s – 3 · 4 s + 4 s + 17

6.

s · 3s – 2 s – 5

2

1

−3 t 2

·

2s + 3

10.

−1 2

– 12 t

s +1 8. 2 . s + s +1

9.

– 12 t

13 t

2

7.

LMAns. 1 e b g bcos 2t – sin 2t gOP N 2 Q LMAns. 1 eb g Lcos h 4 t – 1 sin h 4 t OOP N 3 MN 3 4 3 PQQ LMAns. 1 e . e OP MN 2 π PQ LMAns. e b g L 3 cos 3 t + sin 3 t OOP M P 2 2 PQPQ 3 MN MN LMAns. 4t e OP 3 π PQ MN LMAns. e L 7 cos 2 t + 1 sin 2 t OOP MNM 4 PQPP 8 2 MN Q 32

1

b s + 4g

52

.

7s + 4 . 4s + 4s + 9

– 4t

−t 2

2

Find the inverse Laplace transforms of the following functions: 2s + 1 . s –1 s – 3

b gb g

2.

s–1 . s +s–6

3.

1 . s – 4s

4.

s–4 . 2 s + 1 3s – 1

5. 6.

7.

8.

9. 10.

– 3t

4t

2

gb

g

5 . s–1 s – 3 s –5

b gb gb g

2s2 – 4 . s +1 s – 2 s – 3

b gb gb g 3s – 2

bs – 2gbs + 1g

2

.

s2

b s + 2g d s – 4i 2

2s – 1 . s s–4 2

3t

2t

2

b

LMAns. – 3 e + 7 e OP N 2 2 Q LMAns. 1 e + 4 e OP N 5 5 Q LMAns. 1 de – 1iOP N 4 Q LMAns. 1 9 e – 11 e OP N 5 Q LMAns. 1 e – 3 e + 5 e OP N 8 4 8 Q LMAns. – 1 e – 4 e + 7 e OP 3 2 N 6 Q LMAns. 4 e – 4 e + 5t e OP 3 N 9 9 Q LMAns. 3 e – t e + 1 e OP 4 N 4 Q LMAns. 7 e – 7 + 1 e OP N 16 16 4 Q t

1.

bs – 1g ds + 1i 2

.

t 3

t

3t

5t

–t

2t

3t

2t

–t

–t

– 2t

– 2t

2t

4t

b g

3s + 1

.

−t 2

t

Ans. 2e t – 2 cos t + sin t

384


11.

12.

13.

14.

15.

16. 17.

2s – 1

bs + 2g ds

+4

2

i

5s + 3

bs – 1g ds

i

2s 2 – 1

ds + 1ids + 4i 2

bs + 1gbs – 2g

3

t

.

.

2

5s2 – 15s – 11

ds

– 2t

+ 2s + 5

2

id

i

+ 2s + 2 s + 2s + 5 2

ds + 1ids

2

Ans. 2 + e t – e – 2 t

1 2

2t

–t

.

2s2 + 5s – 4 . s3 + s2 – 2s 2

–t

–t

.

s2 + 2s + 3 2

LMAns. – 5 e + 5 cos 2t + 3 sin 2t OP 8 8 N 8 Q LMAns. e – e Lcos 2t – 3 sin 2t OOP MN PQQ 2 N LMAns. – sin t + 3 sin 2t OP 2 N Q LMAns. – 1 e – e L 7 t – 4t – 1 OOP MN 2 3 3 PQ Q N LMAns. 1 e bsin t + sin 2t gOP N 3 Q

·

i

+ 4s + 8

LMAns. – 4 cos t + 7 sin t + 4 e 65 65 65 N

.

– 2t

cos 2t +

1 – 2t sin 2t e 130

OP Q

INVERSE LAPLACE TRANSFORMS OF THE FUNCTIONS OF THE 8.4

bg

F s s

FORM

We have proved that if

L

then

z

L { f (t)} = F (s)

RS f bt g dt UV T W R F bsg UV L S T s W t

0

–1

Hence,

= =

bg

F s s

z

t

0

bg

f t dt

...(1)

WORKED OUT EXAMPLE 1. Evaluate

R|S 1 U|V |T s bs + ag |W R| 1 U| S| s ds + 1i V| T W

(i) L–1

(iv) L–1

3

2

(ii) L–1

(v) L–1

R| 1 U| S| s d s + a i V| T W RS 1 log FG 1 + 1 IJ UV. Ts H s KW 2

2

2

(iii) L–1

R| s + 2 U| S| s bs + 3g V| T W 2

385


Solution

L–1

(i) Consider

RS 1 UV Ts + aW

= e– at

Using the equation (1), we get –1

L

–1 (ii) We have L

R|S 1 U|V |T s bs + ag |W RS 1 UV Ts + a W 2

2

=

z

L e OP = M N–aQ – at

t

e

– at

dt

0

=

t

= 0

d

1 sin at a

∴ Using the equation (1), we get L–1

R| 1 U| S| s d s + a i V| T W 2

2

=

z

t

0

1 = a

1 sin at dt a

LM b– cos at g OP N a Q

t

0

=

1 (1 – cos at). a2

=

A B + s s+3

(iii) By partial fractions s+2 s s+3

b g

(s + 2) = A (s + 3) + B (s) Put

s = 0,

A=

2 3

s = – 3, B =

s+2 s s+3

L–1

b g LM s + 2 OP MN s bs + 3g PQ

1 3

2 1 3+ 3 = s s+3 =

RS T

RS UV TW

2 –1 1 1 1 L + L–1 3 s 3 s+3

2 1 ⋅ 1 + e – 3t 3 3 2 1 – 3t + e = 3 3 =

∴ From (1), we get

L–1

|RS1 ⋅ s + 2 |UV |T s s bs + 3g |W

z FGH t

=

0

IJ K

2 1 – 3t + e dt 3 3

i

1 1 – e – at . a

UV W

386


= =

L–1

(iv)

RS 1 UV T s + 1W 2

LM 2 t – 1 e OP N3 9 Q

t

– 3t

0

2 1 1 t – e – 3t + 3 9 9

= sin t

Hence by equation (1), we get

R| 1 U| S| s d s + 1i V| T W R| 1 U| S| s d s + 1i V| T W R| 1 U| S| s ds + 1i V| T W

L–1

L–1

L–1

and

z zb zb t

2

=

2

2

=

3

2

=

sin t dt = 1 – cos t

0

t

0

g

1 – cos t dt = t – sin t

g

t

t – sin t dt

0

LM t + cos t OP N2 Q 2

=

t

0

2

= (v) Let

RS FG T H

IJ UV KW F 1I log G 1 + J H sK

L–1 log 1 +

1 s2

2

m b gr

d F s ds

Now

t + cos t – 1. 2

= f (t)

= L { f (t)} = F (s)

d

d log s 2 + 1 – log s 2 ds

=

2s 2 – s +1 s 2

LM1 – s OP N s s + 1Q L F 1I F s I OP – 2 ML G J – L G MN H s K H s + 1JK PQ

= –2

L–1

LM d mF bsgrOP N ds Q

i

=

=

2

–1

= – 2 (1 – cos t) i.e.,

– t f (t) = – 2 (1 – cos t)

–1

2

387


or

f (t) =

RS FG 1 IJ UV T H s KW RS1 log FG1 + 1 IJ UV Ts H s KW

L–1 log 1 + L–1

∴

b

2 1 – cos t t

b

2 1 – cos t

2

=

2

= 2

z

g g

t

1 – cos t dt . 0 t t

EXERCISE 8.2 Find the inverse Laplace transforms of the following functions:

3. log

Fs GH s

I JK

+ a2 + b2

2 2

LMAns. 1 – e OP t Q N LMAns. cos at – cos bt OP t N Q at

s–a 1. log . s

12

.

–1 s . 2. cot a

4.

FG H

IJ K

a s+a log . 2 s–a

Find the inverse Laplace transforms of the following functions: 1.

3.

5.

1

d

s s2 – a 2

i

1

d

s2 s2 + a 2

d

1 cot s Evaluate:

i

1. L

.

z

2

4.

1

2

b g

s s–a

b g

LMAns. 1 de – at – 1iOP N a Q LMAns. e – 1 – t – t OP 2Q N LMAns. 1 – e dt OP t N Q at

2

2

1

s s +1

.

3

.

1 s–a log . 6. s s

t

z

at

t

0

0

2

2 2

2

–1

2

2.

t

–1

2

3. L–1

2

R| s – a U| S| s + a V| given that L Td i W R| s + a U| S| s – a V| given that L Td i W R| s + 2 U| S| s + 4s + 5 V| given that iW Td 2

2. L–1

.

s . a

–1

2

–1

i

2

2

1

s3 s2 – 1

7.

LMAns. 1 bcos h at – 1gOP N a Q LMAns. – 1 FG sin at – t IJ OP N a H a KQ LMAns. cos h t – t – 1OP 2 N Q LMAns. sin t dt OP t N Q

.

LMAns. sin at OP t Q N LMAns. sin h at OP t N Q

2 2

2

RS Ts RS Ts

2

2

L–1

s + a2 s – a2

UV = cos at. W UV = cos h at. W

RS 1 UV = sin t. T s + 1W 2

Ans. t cos at

Ans. t cos h at

LMAns. 1 t e N 2

– 2t

sin t

OP Q

388


8.5 CONVOLUTION THEOREM L– 1 {F (s)} = f (t) and

If

L– 1{F (s) G (s)} =

then

z

L– 1{G (s)} = g(t)

bg b g

t

f u g t – u du

0

L– 1 F (s) = f (t) and

Proof. Since we have

F (s) = L { f (t)} =

and

G (s) = L {g(t)} =

L– 1 {G (s)} = g (t)

z z

∞

z z

RS f bug gbt – ug duUV T W Consider R U L S f bug gbt – ug duV T W

e – st

0

To prove (1), it is sufficient to prove that

bg g bt g dt

e – st f t dt

0

∞

L

...(1)

t

0

= F (s) G (s)

0

z z z z ∞

t

= =

u

...(2)

e – st

0

RS f bug gbt – ug duUV dt T W e f bug g bt – ug du dt t

0

∞

t

t=0

u=0

– st

u u=¥

u=t

t=u

t=0 t=¥

t=¥

u=0

O

...(3)

O

t

Fig. 8.1

u=0

t

Fig. 8.2

The domain of integration for the above double integral is from u = 0 to u = t and t = 0 to t = ∞ which is as shown in Fig. 8.1. The double integral given in the R.H.S. of equation (3) indicates that we integrate first parallel to u-axis and then parallel to t-axis. We shall now change the order of integration parallel to t-axis the limits being t = u to t = ∞ and parallel to u-axis the limits being u = 0 to u = ∞. ∴ From equation (3), we get

L

z

RS f bug g bt – ug duUV T W t

0

= =

z z

∞

0 ∞

0

z

b g RST e g bt – ug dt UVW du RS e b g g bt – ug dt UV du f bu g e T W

f u

∞

u

– su

– st

z

∞

u

–s t –u

389


Substitute t – u = v so that dt = dv when t = u, v = 0, and when t = ∞, v = ∞

L

z

RS f bug g bt – ug duUV T W t

0

= = =

z z

z

b g RST e g bvg dvUVW du f bu g e G b sg du G b sg e f bug du ∞

f u e – su

0

∞

– sv

0

∞

– su

z

0

∞

– su

0

= G (s) F (s) ∴

L– 1 {F (s) G (s)} =

z

bg b g

t

f u g t – u du

0

This completes the proof of the theorem.

WORKED OUT EXAMPLES 1. Using Convolution theorem find the inverse Laplace transforms of the following functions: (i)

(iv)

(vii)

1

s

2

ds ds

bs + 1g s

2

+ a2

i

2

(ii)

2

(v)

s2 2

+ a2

ids

2

+ b2

i

1

bs + 1g ds + 1i 2

ds

s 2

+4

i

(iii)

(vi)

2

.

Solution (i) Let

F (s) =

L– 1 {F (s)} =

Then

L– 1 {G (s)} =

1

, G (s) = 1 s2

bs + 1g R| 1 U| L S |T bs + 1g V|W = t e = f (t) (say) R1U L S V = t = g (t) say Ts W 2

–1

–t

2

–1

2

Then by Convolution theorem, we have L– 1 {F (s) G (s)} = L–1

R| 1 S| s bs + 1g T 2

2

U| V| = W

z bg b z b zd i

f u g t – u du

0

t

0

t

=

g

t

0

g

u e – u t – u du ut – u 2 e – u du

s

bs + 1g ds + 1i 2

ds

s2 2

+ a2

2

i

2

390


By integration by parts =

dut – u id– e i – bt – 2ug de i + b– 2g d– e i –u

2

–t

–t

= te

+ 2e

–u

–u

+t– 2

Verification: L {t e–t + 2 e–t + t – 2} = L {t e– t} + 2 L (e– t) + L (t) – 2 L (1) =

= = (ii) Here

F (s) G (s) =

Let

F (s) =

1

bs + 1g

2 1 2 – + s + 1 s2 s

+

b g b g s b s + 1g

b g

2

s2 + 2s2 s + 1 + s + 1 – 2s s + 1 2

2

1

b g

s2 s + 1

2

2

.

1

bs + 1g d s + 1i 2

1 , s2 + 1

–1 L– 1 {F (s) = L

L– 1 {G (s)} = L–1 Let

2

t = u,

G (s) =

1 s +1

RS 1 UV = sin t = f (t) T s + 1W RS 1 UV = e = g (t ) T s + 1W 2

−t

f (u) = sin u and g (u) = e– u Then by Convolution theorem, we obtain L–1

R| 1 U| S| bs + 1g d s + 1i V| T W 2

= L– 1 {F (s) G (s)} =

=

z

sin u ⋅ e – b

t

0

–t = e

z

t

t–u

= e

Using the formula

z

bg b g

t

f u g t – u du

0

g du

e u sin u du

0

LM e bsin u – cos ugOP N2 Q u

–t

z

t

0

ax

e ax sin bx dx = =

b

e (a sin bx – b cos bx) + c a + b2 2

g

1 1 sin t – cos t + e – t . 2 2

t 0

391


(iii) Here

F (s) G (s) =

Let

F (s) =

s

s 1 ⋅ s +1 s +1 2

=

bs + 1g ds + 1i 2

2

s , G (s) = s +1

1

bs + 1g

2

–1 L– 1 {F (s)} = L

L– 1 {G (s)} = L–1

b g

2

2

RS s UV = cos t = f (t) T s + 1W R| 1 U| S| bs + 1g V| = t e = g (t) T W 2

–t

2

t = u, g (u) = u e– u, f (u) = cos u Hence by Convolution theorem, –1

L

R| s U| S| ds + 1i bs + 1g V| = T W 2

2

=

Since,

z

z z

t

0 t

0

–t = e

e u cos u du =

bg b g

f u g t – u du = L– 1 {F (s) G (s)}

b g g du bt – ug cos u e du

cos u t – u e – b

z

t

0

t–u

u

eu (cos u + sin u) 2

Integrating by parts, we get

LMR e O SMTbt – ug 2 bsin u + cos ugUVW – e2 bsin u + cos ugb– 1g duPP N Q LM0 – 1 t + 1 e bsin u + cos ug duOP N 2 2 Q LM– 1 t + 1 e sin uOP N 2 2 Q LM– 1 t + 1 e sin t OP N 2 2 Q

= e

–t = e

z

t

u

–t

z

0

0

t

u

0

t

–t = e

u

0

–t = e

t

1 (t e– t + sin t). 2 1 s s + 2 ⋅ 2 F (s) G (s) = 2 2 s +a s + a2 s2 + a 2

=

(iv) Here

Let

d

F (s) =

i

s , s + a2 2

d

id

G (s) =

1 s + a2 2

t

i

u

392


–1 L– 1 {F (s)} = L

Then

–1 L– 1 {G (s)} = L

RS Ts RS Ts

2

s + a2

2

1 + a2

UV = cos at = f (t) W UV = 1 sin at = g bt g W a

t = u, f (u) = cos au, g (u) =

1 sin au. a

Using Convolution theorem, we have

z z

L– 1 {F (s) G (s)} =

i.e.,

L–1

R| S| s Td

f u g t – u du

0

U| V= +a i | W s

2

bg b g

t

2 2

t

cos au

0

z

b g

1 sin a t – u du a

b

g

1 t cos au sin a t – au du a 0 1 t sin at + sin at – 2au du = 2a 0 =

z

b

LM N

g

b

1 1 u sin at – cos at – 2 au = 2a – 2a

gOPQ

t

0

1 t sin at. 2a (v) Substituting a = 2, in (iv) we get the required inverse Laplace transform.

=

(vi) Here,

F (s) G (s) =

Let Hence,

ds

s2 2

+ a2

s s + a2

F (s) =

2

–1 L– 1 {F (s)} = L

–1 L– 1 {G (s)} = L

i.e.,

RS Ts RS Ts

i

2

=

and

2

s + a2

2

s + a2

s s . 2 2 s + a s + a2 2

G (s) =

s s + a2 2

UV = cos at = f (t) W UV = cos at = g (t) W

f (t) = g (t) = cos at f (u) = g (u) = cos au Using Convolution theorem, we have

z

L– 1 {F (s) G (s)} = i.e.,

L–1

ds

s2 2

+a

t

0

i

2 2

b g cos bat – aug cos au du

cos au ⋅ cos a t – u du =

z

t

0


=

1 1 u cos at − sin at − 2au 2 2a

t

cos at + cos at – 2au du

0

LM b N 1L 1 O t cos at + sin at P · 2 MN a Q s s + a2

and

2

–1 L– 1 F (s) = L

–1 L– 1 G (s) = L

and

RS Ts RS Ts

2

s + a2

2

s + b2

Therefore using Convolution theorem, we get L– 1 {F (s) G (s)} = i.e.,

L–1

R| S| d s T

U| V= + a id s + b i | W s2

2

2

2

g

1 2

F (s) =

Then

b

=

= (vii) Let

z

393

2

=

z z

G (s) =

gOPQ

t

0

s s + b2 2

UV = cos at = f (t) W UV = cos bt = g (t) W

bg b g

t

f u g t – u du

0

b g

t

0

1 2

cos au cos b t – u du

zo t

0

b

g

b

g

cos a – b u + bt + cos a + b u – bt

R| b S| T

b

g

g

1 sin a – b u + bt sin a + b u – bt + = a –b a +b 2 a sin at – b sin bt , = a 2 – b2

2. Using Convolution theorem, evaluate L–1 Solution Let

F (s) =

Since,

–1

L

R| – 2s S| s + 4 Td i 2

2

U| V| W

= –t⋅

U| ⋅ V | + 4i W s

2

3

and G (s) =

2

–1 L– 1 {F (s)} = L

Now,

Hence

1 s +4

R| S| s Td

RS 1 UV = Ts + 4 W 2

1 sin 2t 2

L– 1 {F ′ (s)} = – t f (t)

a ≠ b.

1 sin 2t 2

ds

s 2

+4

i

2

U| V| W

t

0

t du

394


i.e.,

R| S| s Td

L–1

U| V + 4i | W s

2

2

=

1 t sin 2t 4

1 t sin 2t 4 Therefore using Convolution theorem, we have i.e.,

{G (s)} =

L– 1 {F (s) G (s)} =

i.e.,

R| S| s Td

L–1

U| V + 4i | W s

2

3

=

z z

b g bg

t

f t – u g u du

0

0

b g

1 1 sin 2 t – u ⋅ u sin 2u du 2 4

t

z

b

g

1 t u sin 2t – 2u sin 2u du 8 0 1 t u cos 2t – 4u – cos 2t du = 16 0 t 1 t u cos 2t – 4u du – cos 2t u du = 0 16 0 1 = (t sin 2t – 2t2 cos 2t). 64 3. Find f (t) in the following equations: =

(i) f (t) = t + (iii) f′ (t) = Solution

z

t

0

z

z LMz N

b b

g g

b g (ii) f (t) = t + 2 f bug cos b t – ug du, if f (0) = 1. t

e – u f t – u du

0

z

z

t

0

OP Q

b g bg

cos t – u f u du

(i) Taking Laplace transforms on both sides of the given equation, we get

l q RST

L { f (t)} = L t + L Using Convolution theorem, we have

i.e.,

b g LMN

F s 1–

1 s+1

OP Q

=

b g UVW

e – u f t – u du

0

d i m b gr bg

1 s2

s+1 s3 1 1 L { f (t)} = 2 + 3 s s

F (s) =

i.e.,

t

1 + L e– t L f t 2 s 1 1 F s F (s) = 2 + s+1 s

L { f (t)} =

i.e.,

z

395


–1 f (t) = L

Hence,

FG 1 IJ + L FG 1 IJ Hs K Hs K –1

2

3

t2 · 2 (ii) Taking Laplace transforms on both sides, we get f (t) = t +

RS T

bg

L { f (t)} = L t + 2 L

z

b g b g UVW

t

cos t – u f u du

0

Using Convolution theorem, we have 1 = 2 + 2 L (cos t) L { f (t)} s 1 2s F s i.e., F (s) = 2 + 2 s s +1

bg

bg

L { f (t)} = F s =

i.e.,

s2 + 1 s

2

bs – 1g

2

=

b g

s2 s – 1

2

2 1 2 2 + 2 – + s s s –1 s –1 2

b g

–1 f (t) = L

∴

s2 + 1

R| 2 1 2 S| s + s – s – 1 + bs –21g T 2

= 2 + t – 2 et + 2 t et f (t) = 2 + t + 2 (t – 1) et. (iii) From the given equation, we have L { f ′ (t)} = L

z

RS f bug cos bt – ug duUV T W t

0

By using Convolution theorem, we get s L { f (t)} – f (0) = L { f (t)} · L (cos t)

bg

s F (s) – 1 = F s ⋅

b g LM N

F s s–

i.e.,

s s +1 2

OP Q

s s +1 2

= 1

s2 + 1 F (s) = s3

b g s s+ 1 = 1s + s1 L1 1 O L M + P Ns s Q

L { f ′ (t)} = F s = f (t) =

2

3

–1

f (t) = 1 +

3

t2 . 2

3

2

U| V| W

396


EXERCISE 8.3 Verify Convolution theorem: L { f (t)} L {g (t)} = L

z

RS f b ug g b t – ug duUV T W t

0

for the following functions: 1. f (t) = 1, g (t) = cos t.

2. f (t) = t, g (t) = et.

3. f (t) = sin t, g(t) = e– t.

4. f (t) = t, g (t) = t e– t.

5. f (t) = sin 2t, g (t) = cos 2t. Using Convolution theorem find the inverse Laplace transforms of the following functions:

bs – 4gbs + 5g .

2.

1 . s s–2

3.

4.

5.

6.

7.

8.

9.

10.

11.

LMAns. 1 de N 9

iOPQ LMAns. 1 e – 1 t – 1 OP N 4 2 4Q LMAns. 1 FG t – 1 cos 3tIJ OP N 9 H 3 KQ

1

1.

2

b g d

1

i

s s2 + 9 2

1

b g

s s+2 2

.

ds + 4ids + 9i 2

1

bs – 2gbs + 2g

ds + 4i bs + 1g d

i

s s +1

ds

2

2t

2

+a

2

1

ds + 1i 2

3

2

.

ids

2

+ b2

bs – 1g bs + 2g 2

i

.

2

2

2

.

4s + 5

–t

.

s

2

– 2t

–t

2

1

.

.

2

1

3

LMAns. 1 b1 + t g e + 1 bt – 1gOP 4 N t Q LMAns. 1 bcos 2t – cos 3t gOP N 5 Q LMAns. 1 e – b4t + 1g e OP N 16 Q LMAns. e 10 e – b3 sin 2t + 4 cos 2t g OP N 50 Q LMAns. 1 t + cos t – 1OP N 2 Q LMAns. cos bt – cos at OP a –b N Q LMAns. 1 d3 – t i sin t – 3t cos t OP N 8 Q LMAns. 3t e + 1 e – 1 e OP 3 3 N Q – 2t

s

2

– e – 5t

2t

.

2

4t

.

t

t

2t

397


Find f (t) in the following integral equations: 1. f (t) = t +

zb z z

1 6

t

t –u

0

2. f (t) = sin t + 5 t

2 3. f (t) = t –

zb t

4.

0

bg t – ug

bg

6. f ′ t + 5

z

3

bg b g

f u sin t – u du.

b g

2

e u f t – u du.

z

t

b g

2

f t – u cos u du , if f (0) = 4.

0

b g bg

t

3

13

du = t (1 + t).

13

5. f ′ (t) = t +

0

0

f u

t

LMAns. f bt g = 1 bsin t + sin h t gOP 2 N Q LMAns. 1 sin h 2t OP N 2 Q LMAns. t – 1 t OP 3 Q N LMAns. 3 3 t b3t + 2gOP PQ MN 4 π LMAns. f bt g = 4 + 5 t + 1 t OP 2 24 Q N

g f bug du.

4

cos 2 t – u f u du = 10, if f (0) = 2.

0

LMAns. f b t g = 1 b24 + 120t + 30 cos t + 50 sin 3t gOP 27 N Q

8.6 LAPLACE TRANSFORMS OF THE DERIVATIVES

z

If the Laplace transform of f (t) is known then by using the following results we can find the Laplace transforms of the derivatives f ′ (t), f ′′ (t), ... f n (t) and

t

bg

f t dt .

0

Laplace transforms of the derivatives: Functions of exponential order. A continuous function f (t), t > 0 is said to be of exponential order if

lim e– st f (t) = 0

t→∞

Theorem: If f (t) is of exponential order and f ′ (t) is continuous then L { f ′ (t)} = s L { f (t)} – f (0) Proof: By the definition of Laplace transform, we have L f ′ (t) = Apply integration by parts = = =

z

∞

0

bg

e – st f ′ t dt

z

b g – f bt g e b– sg dt 0 – f b 0g + s e f b t g dt – f b 0g + s e f bt g dt e – st f t

∞ 0

z

z

∞

– st

0

∞

– st

0

∞

– st

0

= – f (0) + s L { f (t)}

...(1)

398


L { f ′ (t)} = sL { f (t)} – f (0) Laplace transform of f ′′ (t) L { f ′′ (t)} = s2 L { f (t)} – s f (0) – f ′ (0) f ′ (t) = g (t) so that

Let Consider

...(2)

f ′′ (t) = g′ (t)

L f ′′ (t) = L {g′ (t)} = sL {g (t)} – g (0), using (2) = sL { f ′ (t)} – f ′ (0) = s [sL { f (t)} – f (0)] – f ′ (0) L { f ′′ (t)} = s2L { f (t)} – s f (0) – f ′ (0)

Similarly,

L { f ′′′ (t)} = s3L { f (t)} – s2 f (0) – s f ′ (0) – f ′′ (0) n

n

n–1

L { f (t)} = s L { f (t)} – s

n–2

f (0) – s

f (0) ... f

...(3) n–1

(0)

...(4)

If f (t) = y then (4) can be written in the form L ( yn) = sn L ( y) – sn – 1 y (0) – sn – 2 y′ (0) – ... – yn – 1 (0) where y ′, y ′′, ...... y(n) denoted the successive derivatives.

8.7 SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS One of the important applications of Laplace transforms is to solve linear differential equations with constant coefficients with initial conditions. For example, consider a second order linear differential equation

i.e.,

d2y dy + a0 + a1 y = f (t) 2 dx dx y′′ + a0 y′ + a1 y = f (t)

where a0, a1 are constants with initial conditions y (0) = A and y′ (0) = B. Taking Laplace transforms on both sides of the above equation and using the formulae on Laplace transforms of the derivatives y′ and y′′. We recall the formulae for immediate reference. L ( y′) = s L ( y) – y (0) L ( y′′) = s2 L ( y) – s y (0) – y′(0) L ( y′′′) = s3 L ( y) – s2 y (0) – s y′ (0) – y ′′ (0) and so on.

WORKED OUT EXAMPLES 1. Solve using Laplace transforms.

given that

d2y dy –3 + 2y = e3t 2 dt dt y (0) = 0 and y′ (0) = 0.

399


Solution. Given equation is y″ – 3y′ + 2y = e3t Taking Laplace transforms on both sides, we get L ( y″) – 3 L ( y′) + 2 L( y) = L(e3t) i.e., s2 L ( y) – s y (0) – y′ (0) – 3 [s L ( y) – y (0)] + 2 L ( y) = where y (0) = 0 and

y′ (0) = 0

i.e.,

(s2 – 3s + 2) L ( y) =

i.e.,

L ( y) =

= ∴

y =

=

= y =

1 using the initial conditions s–3

ds

1 2

ib g

– 3s + 2 s – 3

1 s–1 s–2 s–3

b gb gb g L OP 1 L M MN b s – 1gbs – 2gbs – 3g PQ 1 O LM 1 PP using partial fractions, 1 2 2 – + L M N s – 1 s − 2 s − 3Q R 1 UV − L RS 1 UV + 1 L RS 1 UV 1 L S 2 T s − 1W T s − 2 W 2 T s − 3 W –1

–1

–1

−1

1 t 1 e − e 2 t + e 3t 2 2

This is the required solution. 2. Solve using Laplace transforms

d2y dy −3 + 2y = 4e2t 2 dt dt given that y (0) = – 3 and y′ (0) = 5. Solution. Given equation is y′′ – 3y′ + 2y = 4e2t Taking Laplace transforms on both sides, we get L ( y″) – 3 L ( y′) + 2 L ( y) = 4 L (e2t) s2L ( y) – sy (0) – y′ (0) – 3 [sL ( y) – y (0)] + 2L ( y) = where y (0) = – 3 and i.e.,

1 s–3

4 s−2

y′ (0) = 5

s2L ( y) + 3s – 5 – 3 [sL ( y) – 3] + 2L ( y) =

4 s−2

−1

400


i.e.,

(s2 – 3s + 2) L ( y) =

∴

4 − 3s + 14 s−2

=

4 − 3s 2 + 14 s + 6s − 28 s−2

=

− 3s2 + 20s − 24 s−2

L ( y) =

− 3s 2 + 20s − 24

b s − 2g d s

2

− 3s + 2

i

By using partial fraction = L ( y) =

− 3s2 + 20s − 24

bs − 1gb s − 2g

2

−7 4 4 + + s −1 s − 2 s−2

b g

LM − 7 4 4 O + + MN s − 1 s − 2 b s − 2g PPQ F 1 IJ + 4 L LM 1 OP + 4 L −7 L G H s − 1K Ns − 2Q

–1 y = L

Hence,

=

2

–1

y = – 7et + 4e2t + 4e2t t This is the required solution. 3. Solve using Laplace transforms,

d2y +y = t dt 2

Given y (0) = 1 and y′′ (0) = – 2. Solution. Given equation is y′′ + y = t. Taking Laplace transforms on both sides L ( y″) + L ( y) = L (t) i.e., s2L ( y) – s y (0) – y′ (0) + L ( y) =

1 s2

where y (0) = 1 and y′ (0) = – 2 i.e.,

1 s2 1 L ( y) (s2 + 1) – s + 2 = 2 s 1 L ( y) (s2 + 1) = 2 + s − 2 s

s2 L ( y) – s + 2 + L ( y) =

2

−1

−1

LM 1 MN bs − 2g

2

OP PQ

401


1

L ( y) =

i.e.,

s

2

ds + 1i

+

s−2 s2 + 1

1 1 2 s – 2 + 2 − 2 2 s s +1 s +1 s +1

L ( y) =

1 3 s + 2 − 2 2 s s +1 s +1

=

−1 y = L

Hence,

2

RS 1 UV + L RS s UV − 3 L RS 1 UV T s W T s + 1W T s + 1W −1

2

−1

2

2

y = t + cos t – 3 sin t This is the required solution. 4. Solve using Laplace transforms

d2y dy −5 + 6y = sin t , 2 dt dt

1 21 and y′ (0) = · 10 10 Solution. Given equation is

given y (0) =

y″ – 5y + 6 = sin t. Taking Laplace transforms on both sides, we get L ( y″) – 5 L ( y′) + 6 L ( y) = L (sin t) 1 s2 + 1

i.e., s2 L ( y) – s y (0) – y′ (0) – 5 [s L ( y) – y (0)] + 6 L ( y) =

1 21 and y′ (0) = 10 10 1 1 s 21 i.e., s 2 L y − − −5 s L y − +6L y = 2 10 10 10 s +1

where y (0) =

bg

LM b g N

OP Q

(s2 – 5s + 6) L ( y) =

i.e.,

bg

b

g

1 1 + s + 16 s2 + 1 10

1

ds + 1ids

L ( y) =

ds + 1i bs − 3gb s − 2g

2

2

i

+

L ( y) =

− 5s + 6

1

2

1 s + 16 10 s2 − 5s + 6

+

d

i

1 s + 16 ⋅ 10 s − 2 s − 3

b gb g

1 −1 1 1 – 18 19 = 210 + 5 + 10 + + s + 1 s − 2 s − 3 10 s − 2 s − 3

LM N

=

LM N

OP Q

By using partial fractions

−2 s 2 1 1 + + + 2 2 s − 2 s − 3 10 s + 1 s + 1

OP Q

402


Therefore,

|RS − 2 + 2 + 1 LM s + 1 OP|UV |T s − 2 s − 3 10 N s + 1 s + 1Q|W R 1 UV + 2 L RS 1 UV + 1 LM L RS s UV + L RS 1 UVOP −2 L S Ts − 2W T s − 3 W 10 MN T s + 1W T s + 1WPQ

–1 y = L

y =

2

−1

2

−1

−1

−1

2

= – 2e2t + 2e3t +

2

1 (cos t + sin t) 10

which is the required solution. 5. Solve using Laplace transforms y″ + 2y′ + 5y = 8 sin t + 4 cos t given that y (0) = 1 and y (π) = e– π. Solution. Given equation is y″ + 2y′ + 5y = 8 sin t + 4 cos t Taking the Laplace transforms on both sides, we get L ( y″) + 2 L ( y′) + 5 L ( y) = 8 L (sin t) + 4 L (cos t) i.e., s2 L ( y) – s y (0) – y′ (0) + 2 [sL ( y) – y (0)] + 5 L ( y) = Since y (0) = 1 and assuming

y′ (0) = A, we get

(s2 + 2s + 5) L ( y) – s – A – 2 = (s2 + 2s + 5) L ( y) = ∴

L ( y) =

=

=

Therefore,

8 4s + 2 s +1 s +1 2

y =

b g

4 s+2 s2 + 1

b g +s+ A+2

4 s+2 s2 + 1

b g + s+ A+2 ds + 1ids + 2s + 5i s + 2s + 5 L 1 – 1 OP + b s + 1g + A + 1 2M N s + 1 s + 2s + 5Q s + 2s + 5 L 1 OP bs + 1g + A + 1 1 – 2M MN s + 1 bs + 1g + 2 PQ + bs + 1g + 2 L R 1 U R| 1 U|OP 2 ML S MN T s + 1VW − L S|T bs + 1g + 2 V|WPQ R| s + 1 U| R| A + 1 U| +L S +L |T bs + 1g + 2 V|W S|T bs + 1g + 2 V|W sin t L 1 O 2 Msin t − e sin t P + e cos 2 t + b A + 1g e 2 2 N Q 4 s+2

2

2

2

2

2

2

2

2

−1

2

2

−1

2

2

−1

−t

2

−1

2

=

2

−t

2

2

−t

2

403


y = e– t cos 2t + 2 sin t + Since y (π) = e– π, we get

1 (A – 1) e– t sin 2t 2

1 (A – 1) e– π sin 2π = e– π. 2 This shows that for any value of A the given condition y (π) = e– π holds good. Hence this gives the solution of the equation the initial conditions. e– π = e– π cos 2π + 2 sin π +

6. Solve using Laplace transforms, y″ + 2y′ + y = 6t e– t given y (0) = 0, y′ (0) = 0. Solution y″ + 2y′ + y = 6t e– t Taking the Laplace transforms on both sides of the given equation, we get L ( y″) + 2 L ( y′) + L ( y) = 6 L (t e– t) s2 L ( y) – s y (0) – y′ (0) + 2 [s L ( y) – y (0)] + L ( y) = Since i.e.,

6

bs + 1g

y (0) = 0, y′ (0) = 0 s2 L ( y) + 2s L ( y) + L ( y) = (s2 + 2s + 1) L ( y) =

∴

L ( y) =

= L ( y) =

∴

y =

=

6

bs + 1g 6

bs + 1g

2

6

b s + 1g ds 2

2

i

+ 2s + 1

6

b s + 1g b s + 1g 2

6

2

bs + 1g R| 1 U| 6L S V T| b s + 1g W| R1U 6e L S V Ts W 4

−1

4

−t

−t = 6e ⋅

y = t3 e– t This is the required solution.

2

−1

4

t3 3!

2

404


7. Solve using Laplace transforms, y′′′ – 3y″ + 3y′ – y = t2 et given y (0) = 1, y′ (0) = 0, y′′′ (0) = – 2. Solution y″′ – 3y″ + 3y′ – y = t2 et Taking Laplace transforms on both sides, we get L ( y″′) – 3 L ( y″) – 3L ( y′) – L ( y) = L (t2 et ) i.e., [s3 L ( y) – s2 y (0) – sy′ (0) – y″ (0)] – 3 [s2 L ( y) – sy (0) – y′ (0)] + 3 [sL ( y) – y (0)] – L ( y) = Since,

y′ (0) = 0,

(s3 – 3s2 + 3s – 1) L ( y) + 3s – s2 – 1 = (s3 – 3s2 + 3s – 1) L ( y) = i.e.,

i.e.,

(s – 1)3 L ( y) =

L ( y) =

=

2

b s − 1g

3

+ s 2 − 3s + 1

3

+ s 2 − 3s + 1

2

b s − 1g

2

b s − 1g bs − 1g 3

2

b s − 1g

+

3

3

+

2

b s − 1g

3

2

b s − 1g

3

s 2 − 3s + 1

bs − 1g

3

s2 − 2s + 1 − s

bs − 1g b s − 1g − b s − 1g − 1 2 + b s − 1g b s − 1g bs − 1g − b s − 1g − 1 2 + b s − 1g bs − 1g b s − 1g bs − 1g 6

b s − 1g

y (0) = 1, y″ (0) = – 2

s3 L ( y) – s2 + 2 – 3 [s2 L ( y) – s] + 3 [sL ( y) – 1] – L ( y) = i.e.,

2

3

2

=

6

3

2

=

L ( y) =

y =

6

2

3

+

3

1 1 1 − − 2 3 s−1 s−1 s−1

b s − 1g b R| 2 U| +L L S |T b s − 1g V|W 6

−1

−1

6

g b g RS 1 UV − L R|S 1 T s − 1W |T b s − 1g

1 t 5 et + e t − t et − t 2 e t = 60 2 which is the required solution.

3

−1

U| V| − L W

−1

2

R| 1 U| S| b s − 1g V| T W 3

·

405


8. Solve the D.E. y″ + 4y′ + 3y = e–t with y(0) = 1 = y′(0) using Laplace transforms. Solution. Taking Laplace transform on both sides of the given equation, we have L [ y″(t)] + 4L [ y′(t)] + 3L[ y(t)] = L[e–t ] i.e., {s2 L y (t) – s y (0) – y′(0)} + 4 {s L y (t) – y (0)} + 3 L y (t) =

1 s +1

Using the given initial condition, we obtain (s2 + 4s + 3) L y (t) – s – 1 – 4 =

1 s+1

(s2 + 4s + 3) L y (t) = (s + 5) +

i.e.,

(s + 1) (s + 3) L y (t) = ∴

L [ y (t)] =

∴

y (t) =

s 2 + 6s + 6

b s + 1g b s + 3g

Let

2

=

1 s+1

b g

s2 + 6s + 6 s +1 s 2 + 6s + 6

b s + 1g bs + 3g L s + 6s + 6 OP L M MN b s + 1g b s + 3g PQ 2

2

−1

2

A B C + + 2 s +1 s+3 s+1

b g

b g

Multiplying by (s + 1) (s + 3), we get s2 + 6s + 6 = A (s + 1) (s + 3) + B (s + 3) + C (s + 1)2 Put

s = – 1;

⇒

1 = B (2)

B=

...(1)

1 2

3 4 Equating the coefficient of s2 on both sides of (1), we get Put

s = – 3;

⇒

– 3 = C (4)

C= −

FG3 A = 7 IJ H 4K FG 1 IJ + 1 L LM 1 H s + 1K 2 MN b s + 1g

1 = A+C

Hence,

∴

L−1

LM s + 6s + 6 OP MN b s + 1g b s + 3g PQ 2

2

=

y (t) =

7 −1 L 4

−1

7 −t 1 3 e + e − t ⋅ t − e −3t . 4 2 4

2

OP − PQ

FG H

3 −1 1 L 4 s+3

IJ K

406


Solution of Simultaneous Differential Equations dx dy = 2x – 3y, = y – 2x subject dt dt

9. Solve the simultaneous equations using Laplace transforms, to x (0) = 8 and y (0) = 3. Solution. We have, x′ (t) – 2x (t) + 3y (t) = 0 2x (t) + y′ (t) – y (t) = 0 Taking Laplace transforms on both sides of these, we get L [x′ (t)] – 2L [x (t)] + 3L [ y (t)] = 0 2L [x (t)] + L [ y′ (t)] – L [ y (t)] = 0 i.e.,

sL x (t) – x (0) – 2L x (t) + 3L y (t) = 0 2Lx (t) + sLy (t) – y(0) – Ly (t) = 0

Using initial values, Since, x (0) = 8, and y (0) = 3, we get (s – 2) L x (t) + 3L y (t) = 8

...(1)

2L x (t) + (s – 1) L y (t) = 3

...(2)

Solving the Eqns. (1) and (2) Multiplying (s – 1) in the Eqn. (1) and Multiplying 3 by (2) (s – 1) (s – 2) L x (t) + 3 (s – 1) L y (t) = 8 (s – 1) 6L x (t) + 3 (s – 1) L y (t) = 9 Subtracting, we get ∴ ∴ Let,

(s2 – 3s – 4) L x (t) = 8s – 17 L x (t) =

8s − 17 s − 3s − 4 2

−1 x (t) = L

LM 8s − 17 OP MN bs − 4gbs + 1g PQ

8 s − 17 A B + = s−4 s+1 s− 4 s+1

b gb g

8s – 17 = A (s + 1) + B (s – 4) Put

s = 4, A = 3 s = – 1,

∴

−1 x (t) = 3 L

B=5

LM 1 OP + 5 L LM 1 OP Ns − 4Q N s + 1Q −1

x (t) = 3e4t + 5e– t Consider, ∴

dx = 2x – 3y dt 1 dx 2x − y = 3 dt

LM N

...(3)

OP = 1 LM2 x bt g − dx OP dt Q Q 3N

407


dx = 12e4t – 5e– t dt ∴

y (t) =

d

i d

1 2 3e 4 t + 5e − t − 12e 4 t − 5e − t 3

d

1 − 6e 4 t + 15e − t 3 y (t) = 5e– t – 2e4t =

i

i ...(4)

Equations (3) and (4) represent the solution of the given equations. 10. Solve by using Laplace transforms

dx dy − 2y = cos 2t, + 2x = sin 2t, x = 1, y = 0, at t = 0. dt dt Solution. We have a system of equations, x′ (t) – 2y (t) = cos 2t

...(1)

2x (t) + y′ (t) = sin 2t

...(2)

where we have x (0) = 1 and y (0) = 0 Taking Laplace transforms on both sides of Eqns. (1) and (2), we have L [x′ (t)] + 2L [ y (t)] = L (cos 2t) 2 L [x (t)] + L [ y′ (t)] = L (sin 2t) i.e.,

{sL x (t) – x (0)} – 2L y (t) =

s s +4

2L x (t) + {sL y (t) – y (0)} =

2 s +4

2

2

Using the given initial values, we have sL x (t) – 2L y (t) =

s +1 s2 + 4

...(3)

2L x (t) + sL y (t) =

2 s +4

...(4)

2

Let us multiply Eqn. (3) by s and Eqn. (4) by 2

Adding, we get

s2L x (t) – 2sL y (t) =

s2 +s s2 + 4

4L x (t) + 2sL y (t) =

4 s +4 2

4 s2 + 2 +s (s + 4) Lx (t) = 2 s +4 s +4 2

= s+1 i.e.,

2

(s + 4) Lx (t) = s + 1

408


∴

Lx (t) =

s+1 s2 + 4

−1 x(t) = L

∴ Thus

LM Ns

2

x (t) = cos 2t +

OP Q

LM N

1 s + L−1 2 s + 22 + 22

OP Q

1 sin 2t 2

...(5)

To find y(t), let us consider, dx − 2 y = cos 2t dt

∴

y =

∴

y (t) =

where

LM OP N Q 1 L dx O − cos 2t P M 2 N dt Q 1 dx − cos 2t 2 dt

dx = – 2 sin 2t + cos 2t. dt 1 [– 2 sin 2t + cos 2t – cos 2t] 2 y (t) = – sin 2t

∴

y (t) =

∴

...(6)

∴ Eqns. (5) and (6) represent the required solution. 11. Solve the following system of equations using Laplace transforms,

dx − y = et, dt

dy + x = sin t dt

given that x = 1, y = 0 at t = 0. Solution We have

x′ (t) – y (t) = et

...(1)

x (t) + y′ (t) = sin t

...(2)

where we have x (0) = 1, y (0) = 0 Taking Laplace transforms on both sides of Eqns. (1) and (2) L [x′ (t)] – L [ y (t)] = L [et] L [x (t)] + L [ y′ (t)] = L (sin t) i.e.,

sL x (t) – x (0) – L y (t) =

1 s−1

L x (t) + sL y (t) – y (0) =

1 s +1 2

Using the given initial values, we have sL x (t) – 1 – L y (t) =

1 s−1

409


⇒

sL x (t) – L y (t) =

1 +1 s−1

...(3)

⇒

L x (t) + sL y (t) =

1 s2 + 1

...(4)

Let us multiplying 5 by Eqn. (3) and adding Eqn. (4), we get (s2 + 1) L x (t) = s + ∴

L x (t) =

1 s s + + 2 2 2 s +1 s −1 s +1 s +1 2

−1

x (t) = L

s

b s − 1g ds + 1i

Let

2

=

s 1 + s − 1 s2 + 1

b gd LM s OP + L N s + 1Q

i d

−1

2

i

LM s OP MN b s − 1g ds + 1i PQ + L

−1

2

A Bs + C + 2 s−1 s +1

LM 1 OP MM ds + 1i PP Q N 2

2

...(5)

s = A (s2 + 1) + (Bs + C ) (s – 1) s = 1, ∴ A =

Put

1 2

1 2 2 Equating the coefficient of s on both sides, we get

s = 0, C =

0 = A+B ∴B= −

Now,

L−1

LM s OP MN b s − 1g ds + 1i PQ 2

LM N

OP Q

1 2

LM N

OP Q

LM N

=

1 −1 1 1 1 1 s L − L−1 2 + L−1 2 2 2 2 s−1 s +1 s +1

=

1 t (e – cos t + sin t) 2

OP Q ...(6)

Further, we have

L−1

1

ds + a i LM 1 OP MM ds + 1i PP = Q N 2 2

2

∴

L−1

2

2

=

1 (sin at – at cos at) 2a 3 1 (sin t – t cos t) 2

Equation (5) as a consequence of Eqns. (6) and (7) becomes, x (t) = cos t +

1 t 1 (e – cos t + sin t) + (sin t – t cos t) 2 2

...(7)

410


x (t) =

1 t (e + cos t + 2 sin t – t cos t) 2

...(8)

dx − et dt dx 1 t = (e – sin t + 2 cos t + t sin t – cos t) dt 2 1 t (e – sin t + 2 cos t + t sin t – cos t) – et ∴ y (t) = 2 1 y (t) = (t sin t + cos t – sin t – et ) 2 Eqns. (8) and (9) represent the solution of the given equations.

Also from Eqn. (1), y (t) =

...(9)

EXERCISE 8.4 Using Laplace transforms method solve the following differential equations under the given conditions. 1. y′ – 5t = e5t, y (0) = 2. 2. y′ + 3y = 2,

LMAns. y = N

OP Q LMAns. y = 3 e − 1 cos t + 1 sin t OP 2 2 2 N Q LMAns. y = 2e + 1 t − 1 OP 2 4Q N LMAns. y = e + 1 b− cos 2t + 2 sin 2t gOP 5 N Q

y (0) = – 1.

2 5 3t − e 3 3

−t

3. y′ + y = sin t, y (0) = 1. 4. y′ – 2y = t, y (0) =

b g

Ans. y = 2 + t e5t

7 · 4

− 2t

5. y′ – y = cos 2t, y (0) =

4 · 5

6. y″ + y = 0, y (0) = 1,

y′ (0) = – 1.

t

7. y″ + 3y′ + 12y = 0, y (0) = 7,

y′ (0) = – 11.

8. y″ – 7y′ + 12y = 0, y′ (0) = 9, y′ (0) = 2. 9. y″ + 3y′ – 4y = 12e2t, y (0) = 4, y′ (0) = 1. 10. y″ – 3y′ + 2y = 1 – e2t, y (0) = 1, y′ (0) = 0. 11. y″ + 5y′ + 6y = 5e2t, y (0) = 2, y′ (0) = 1.

12. y″ + 9y = cos 2t, y (0) = 1,

y

FG π IJ = –1. H 2K

Ans. y = cos t − sin t Ans. y = 3e − t + 4e − 2 t Ans. y = 4e 3t + 5e 2 t Ans. y = e t + e − 4 t + 2e 2 t

LMAns. y = N

OP Q LMAns. y = 23 e − 4e + 1 e OP 4 4 N Q LMAns. y = 4 cos 3t + 4 sin 3t + 1 cos 2t OP 5 5 5 N Q − 2t

1 1 2t − e − te 2 t 2 2 − 3t

2t

411


LMAns. y = N

13. y″ – 3y′ + 2y = e3t, y (0) = 1, y′ (0) = – 1.

LMAns. y = N

14. y″ + 2y′ + 5y = e– t sin t, y (0) = 0, y′ (0) = 1.

OP Q bsin t + sin 2t gOPQ

7 t 1 e − 3e 2 t + e 3t 2 2 1 −t e 3

Ans. y = 3t + sin 2t

15. y″ + 4y = 9t, y (0) = 0, y′ (0) = 7. –t

16. y″ – 3y′ + 2y = 4t + 12e , y (0) = 6, y′ (0) = 1.

Ans. y = 3e t − 2e 2 t + 2t + 3 + 2e − t 17. y″ – 4y′ + 5y = 12t2, y (0) = 0 = y′ (0) = 0.

b

Ans. y = 25t 2 + 40t + 22 + 2e 2 t 2 sin t − 11 cos t

Ans. y = cos t − 4 sin t + 4t cos t

18. y″ + y = 8 cos t, y (0) = 1, y′(0) = – 1. 19. y″ + 9y = 18t, y (0) = 0,

y

g

FG π IJ = 0. H 2K

Ans. y = 2t + π sin 3t Ans. y = cos t − cos 2t

20. y″ + 2y′ + 5y = 10 sin t, y (0) = 0, y′ (0) = –1.

Using Laplace transforms method solve the following simultaneous equations: 1. x′ = x + y, y′ = 4x – 2y given x (0) = 0,

y (0) = 5.

Ans. x = − e − 3t + e 2 t ; y = 4e − 3t + e 2 t 2. x′ + 2x + y = 0, x + y′ + 2y, given x (0) = 1,

y (0) = 3.

Ans. x = − e − t + 2e − 3t ; y = e − t + 2e − 3t 3. x′ + y′ + 2x + y = 0, y′ + 5x + 3y = 0 given x (0) = 0, y (0) = 4.

Ans. x = 8 sin t ; y = − 12 sin t + 4 cos t 4. x′ + 5x – 2y = t, y′ + 2x + y = 0 given x (0) = y (0) = 0.

LMAns. x = 1 b1 + 3t g − 1 b1 + 6t g e 27 27 N

− 3t

;y =

b

5. x′ – y = et, y′ + x = sin t, given x (0) = 1, y (0) = 0.

LMAns. x = 1 de 2 N

6. x″ + 3x – 2y = 0,

i

g

b

g OPQ

2 2 2 − 3t − 2 + 3t e − 3t 27 27

d

iOPQ OP Q

1 t sin t − e t + cos t − sin t 2 x″ + y″ – 3x + 5y = 0 given x = 0, y = 0, x′ = 3, y′ = 2 when t = 0. 11 1 11 1 Ans. x = sin t + sin 3t ; y = sin t − sin 3t 4 12 4 4 t

+ cos t + 2 sin t − t cos t ; y =

LM N

8.8 APPLICATIONS OF LAPLACE TRANSFORMS In this section we shall consider some applications of the Laplace transforms to solve the problems on vibrations, LRC circuits and bending beams.

412


WORKED OUT EXAMPLES d2x dx +6 + 25x = 0. If the particle is 2 dt dt started at x = 0, with an initial velocity of 12 ft/sec to the left, determine x in terms of t using Laplace transform method. 1. A particle moves along the x-axis according to the law

Solution. The given equation is x″ (t) + 6x′ (t) + 25 x (t) = 0

...(1)

dx = – 12 at t = 0, By data dt i.e., x (0) = 0, x′ (0) = – 12 are the initial conditions.

and x = 0 at t = 0,

Now taking Laplace transform on both sides of (1) we have L [x″ (t)] + 6 L [x′ (t) + 25 L [x (t)] = L (0) i.e., {s2 L x (t) – s x (0) – x′ (0)} + 6 {s L x (t) – x (0)} + 25 L x (t) = 0 Using the initial conditions we obtain, (s2 + 6s + 25) L x (t) = – 12 ⇒

L x (t) =

−12 s + 6s + 25 2

−1 x (t) = −12 L

−1 = −12 L

− t = −12e 3

LM 1 OP N s + 6s + 25 Q LM 1 OP MN bs + 3g + 4 PQ L 1 OP L M Ns + 4 Q 2

2

2

−1

2

2

sin 4t 4 sin 4t.

−3t = −12e

x (t) = – 3e–3t

Thus,

d 2x dx +6 + 25x = 0 . If the dt dt 2 initial position of the particle is at x = 20 and the initial speed is 10, find the displacement of the particle at any time t using Laplace transforms. 2. A particle is moving with damped motion according to the law

Solution. We have x″ (t) + 6x′ (t) + 25x (t) = 0 Initial conditions

x (0) = 20, x′ (0) = 10

Taking Laplace transforms on both sides, we get

L [x″ (t)] + 6L [x′ (t)] + 25L [x (t)] = L (0)

413


i.e., {s2 L x (t) – s x (0) – x′ (0)} + 6{sL x (t) – x (0)} + 25L x (t) = 0 i.e., (s2 + 6s + 25) L x (t) – 20s – 10 – 120 = 0 L x (t) =

20s + 130 s + 6s + 25 2

LM 20s + 130 OP MN bs + 3g + 16 PQ L 20 b s + 3g + 70 OP = L M MN b s + 3g + 4 PQ L 20s + 70 OP L M = e N s +4 Q 70 L O sin 4t P = e M20 cos 4t + 4 N Q 7 F I x (t) = 10 e G 2 cos 4t + sin 4 t J · H K 4

−1 x (t) = L

2

−1

2

− 3t

2

−1

2

2

− 3t

∴

− 3t

3. The current i and charge q in a series circuit containing an inductance L, capacitance C e.m.f dq di q · Express i and q in terms of t given that L, C, E are conE satisfy the D.E. L + = E, i = dt dt C stants and the value of i, q are both zero initially. Solution dq Since i= the D.E. becomes, dt d 2q q L 2 + = E C dt or E d 2q q + = ⇒ 2 L LC dt 2 i.e., q″ (t) + λ q (t) = µ

1 E and µ = . LC L Taking Laplace transforms on both sides, we get L [q″ (t)] + λ2 L [q (t)] = L [µ]

where λ2 =

µ s i = 0, q = 0 at t = 0

{s2 L q (t) – s q (0) – q′ (0)} + λ2 L q (t) = But i.e., ∴

q (0) = 0, (s2 + λ2) L q (t) =

µ s

q′ (0) = 0

414


d

L q (t) =

µ

s s + λ2 2

R| µ U| S| s ds + λ i V| T W 1 1 L1 s O − = M ds + λ i λ N s s + λ PQ By partial fractions L µ OP µ L1 s O = L M MN s ds + λ i PQ = λ L MN s − s + λ PQ −1 q (t) = L

Hence,

s2

i

2

2

2

2

2

2

2

−1

−1

2

∴ where λ =

q (t) =

2

2

µ (1 – cos λt) λ2

1 E and µ = LC L

R|S T|

q (t) = EC 1 − cos

∴

2

2

U|V W|

1 t ⋅ LC

4. A resistance R in series with inductance L is connected with e.m.f. E(t). The current is given by di L + Ri = E (t). If the switch is connected at t = 0 and disconnected at t = a, find the current i in terms dt of ‘t’. Solution. We have i = 0 at t = 0 i.e.,

i (0) = 0

and

E (t) = By date

RS E in 0 < t < a T0 if t ≥ a

Li′ (t) + R i (t) = E (t)

Taking Laplace transforms on both sides L L [i′ (t)] + R L [i (t)] = L [E (t)] L{sL i (t) – i (0)} + RL i (t) = L [E (t)] L i (t) [Ls + R] = L [E (t)]

...(1)

Now to find L [E (t)] We have by definition, L [E (t)] = =

z

∞

0

bg

e − st E t dt =

L e OP = EM N−sQ E d1 − e i s − st

a

0

=

− as

z

a

0

d

e − st ⋅ E dt +

i

− E − as e −1 s

z

∞

a

e − st dt

415


Using R.H.S. of Eqn. (1), we have L i (t) [Ls + R] =

d i E d1 − e i s b Ls + R g E 1 − e − as s − as

L i (t) = =

E E e − as − s Ls + R s Ls + R

b

i (t) = L−1

E s Ls + R

b

Now, let

g

=

g b LM E OP − L MN s b Ls + Rg PQ

−1

g LM E e OP MN s b Ls + Rg PQ − as

A B + s Ls + R

...(2) (Partial fractions)

E = A (Ls + R) + Bs Put

s = 0,

E R

A=

– EL −R , B= R L E 1 EL 1 ⋅ − ⋅ = R s R Ls + R

s = E s Ls + R

b

Thus,

L−1

g LM E OP MN s b Ls + Rg PQ

=

LM OP NQ

E −1 1 E L − L−1 R s R

F GH

E 1− e = R

− Rt L

I JK

LM 1 OP MMN FGH s + RL IJK PPQ

Further we have the property of the unit step function

bg

L [ f (t – a) u (t – a)] = e– as f s

bg

bg

where f s = L f t Taking

Then i.e., Also

bg

=

bg

= L−1

f s L−1 f s

f (t) =

bg

L−1 e − as f s

E s Ls + R

b

g

LM E OP MN s b Ls + Rg PQ E d1 − e i By Eqn. (3) R − Rt L

= f (t – a) u (t – a)

...(3)

416


i.e.,

LM MN

L−1 e − as

E s Ls + R

b

But

=

u (t – a) =

LM MN

L−1 e − as

∴

OP g PQ

when

E s Ls + R

b

OP g PQ

=

E 1 − e − Rbt − a g R

R|0 S|1 T

L

u( t − a )

in 0 < t < a if t ≥ a

E 1 − e − R b t −a g R

L

...(4)

t ≥ a = 0 in 0 < t < a

Using the results Eqns. (3) and (4) in Eqn. (2), we have i (t) =

E 1 − e − Rt R t ≥ a

Also,

i (t) =

i.e.,

i (t) =

when

when

E 1 − e − Rt R

E − Rbt − a g e R t ≥ a

L

L

L

in 0 < t < a

−

E 1 − e − Rb t −a g R

− e − Rt

...(5) L

L

...(6)

Thus Eqns. (5) and (6) represent the required i (t) in terms of t.

EXERCISE 8.5 d2x + 25 = 21 cos 2t. If the dt 2 particle starts from rest at t = 0, find the displacement at any time t > 0.

1. A particle undergoes forced vibrations according to the equation

Ans. x = cos 2t − cos 5t 2. A particle moves along a line so that its displacement x from a fixed point o at any time t is given by x″ + 2x′ + 5x = 52 sin 3t. If at t = 0 the particle is at rest at x = 0. Find the displacement at any time t > 0.

b

g b

Ans. x = e − t 6 cos 2t + 9 sin 2t − 2 3 cos 3t + 2 sin 3t

g

3. A particle of mass m moves along a line so that its displacement x at time t is given by mx″ + kx = f (t), where x (0) = a, x′ (0) = 0. Find x if (i) f (t) = f0 H (t – T ) where H (t – T ) is the Heaviside unit step function.

OP LMAns. x = af cos µt if t < T = af cos µt + b f k g m1 − cos µ bt − T grP MM PP MMif t > T PQ Nwhere µ = k m 0

0

2

0

417


(ii) f (t) = f0 δ (t – T) (Direct delta function).

LMAns. x = af = af N

0

0

OP cos µt + b f µ g sin µ bt − T g if t > T Q cos µt if t < T 0

d 2x dx +6 + 25x = 0. If 2 dt dt the initial position of the particle is at x = 20 and the initial speed is 10 find the displacement of

4. A particle is moving with damped motion according to the equation

b

Ans. x = 10e − 3t 2 cos 4t + sin 4t

the particle at any time t > 0.

g

ADDITIONAL PROBLEMS (From Previous Years VTU Exams.) s+1

b s − 1g b s + 2g

1. Find the Inverse Laplace transform of

Solution. Let

s+1

b s − 1g b s + 2g 2

=

2

A B + s−1 s−1

b g b 2

+

·

C s+2

g

s + 1 = A (s – 1) (s + 2) + B (s + 2) + C (s – 1)2

or Put

s = 1

∴

B=

2 3

Put

s = –2

∴

C=

–1 9

A=

1 9

Equating the coefficient of s2 on both sides We have

Now,

Thus

0 = A + C

L–1

L–1

LM s + 1 OP MN b s − 1g bs + 2g PQ 2

LM s + 1 OP MN b s − 1g bs + 2g PQ 2

∴

LM OP N Q

LM 1 MN b s − 1g

=

1 –1 1 2 L + L–1 9 s −1 3

=

1 t 2 t –1 1 1 − e –2 t e + e L 2 9 3 9 s

=

1 t 2 t 1 e + e ⋅ t − e –2 t 9 3 9

=

1 t 2 t 1 e + e ⋅ t − e –2 t . 9 3 9

LM OP N Q

OP − 1 L L 1 O PQ 9 MN s + 2 PQ –1

2

418


2. Evaluate L–1

LM 1 + MN s + 3 s

OP · b g PQ

s+3 1 − + 6s + 13 s − 2

2

3

Solution

LM 1 OP + L LM N s + 3Q N s L 1 OP = e L M N s + 3Q LM s + 3 OP MN bs + 3g + 4 PQ = e

We have L–1

–1

–1

L–1

2

OP Q

s+3 − L–1 + 6s + 13

3

–3t

–3t

2

L–1

s+ 3→ s

L–1

LM 1 OP MN bs − 2g PQ

LM 1 OP MN b s − 2g PQ 3

LM Ns

2

s − 22

OP Q

= e–3t cos 2t –1 2t = e L

LM 1 OP = e Ns Q

2t

3

t2 2!

e 2t t 2 2

=

–3t –3t Thus the required inverse Laplace transform is given by e + e cos 2t −

3. Find the inverse Laplace transform of (i)

2s − 1 2 s + 2s + 17

2s − 1 = s + 2 s + 17 2

L–1

LM Ns

2

2s − 1 + 2 s + 17

OP Q

=

=

=

Thus

L–1

LM Ns

2

2s − 1 + 2 s + 17

b s − 3g

(ii)

Solution (i)

e –2s

OP Q

=

2

·

b g b s + 1g + 4 bs + 1g + 4 L 2 bs + 1g − 3 OP L M MN bs + 1g + 4 PQ L 2s − 3 OP e L M Ns + 4Q R L s O − 3L FG e S2 L M T N s + 4 PQ H s 3 F I e G 2 cos 4t − sin 4t J · H K 4 2s − 1 2

2

=

2 s +1 − 3 2

2

–1

2

–t

2

–1

2

–t

–1

–1

2

–t

2

1 + 42

IJ UV KW

e 2t t 2 · 2

419


(ii)

L–1

Now,

L–1

Thus,

L–1

R| 1 U| e L F 1 I = e t = f t GH s JK b g b g S| bs − 3g V| = T W F e I GH bs − 3g JK = f (t – 2) u (t – 2) F e I e b g t −2 u t −2 GH bs − 3g JK = { b g} b g . 3t

3t

–1

2

2

−2 s

2

−2 s

3 t −2

2

4. Using Convolution theorem, find the inverse Laplace transform of —

Solution. Let

f (s) =

∴

bg g b sg

1 ; s+2

= f (t) = e–2t

L– 1

= g (t) = cos 3t

L

L–1

bg bg

f s ⋅g s

LM s MN b s + 2g ds

2

OP + 3 iP Q 2

2

z z

bg b g

t

=

f u g t − u du

u=0

b

t

=

g

e –2u cos 3t − 3u du

u=0

LM e m– 2 cos b3t − 3ug − 3 sin b3t − 3ugrOP N4 + 9 Q 1 oe b – 2g − b– 2 cos 3t − 3 sin 3t gt 13 –2 u

R.H.S. =

= –1 Thus L

u=0

LM s OP 1 MN b s + 2g ds + 9i PQ = 13 d2 cos 3t + 3 sin 3t − 2e i −2 u

2

d3y d2y dy 3 − +3 − y = t2 et given y (0) = 1, 3 2 dt dt dt

y′(0) = 0, y′′(0) = –2. Solution. The given equation is

bg

bg bg =t e

y ′′′ t − 3 y ′′ t + 3 y ′ t − y t

2

t

Taking Laplace transform on both sides, we have

bg

t

–2 u

5. Using Laplace transform method, solve

bg

2

—

We have Convolution theorem –1

b s + 2g d s + 9 i ·

s s + 32

—

g (s) =

L– 1 f s

s

bg

bg

bg

y ′′′ t − 3 L y ′′ t + 3 L y ′ t − L y t = L e t t 2

420


o bg

bg

bg

b gt o

bg

bg

b gt 2 3 ms L y b t g − y b 0gr – L y bt g = bs − 1g

s 3 L y t − s 2 y 0 − s y ′ 0 − y ′′ 0 − 3 s 2 L y t − s y 0 − y ′ 0 +

3

Using the given initial conditions, we have = (s3 – 3s2 + 3s – 1) L y (t) – s2 + 2 + 3s – 3 =

i.e.,

(s – 1)3 L y (t) = (s2 – 3s + 1) +

LM s − 3s + 1OP MN bs − 1g PQ LM b s − 1g − s OP MN b s − 1g PQ 2

Now,

L

3

2

L–1

3

2

i+

− 3s + 1 3

b s − 1g

3

2

b s − 1g

3

2

bs − 1g b s − 1g LM {b s − 1g + 2s − 1} − 3s + 1OP = L M PP bs − 1g MN Q L b s − 1g − b s − 1g − 1OP = L M MN b s − 1g PQ L s − s − 1OP = e L M N s Q F 1I F 1 I F 1 I = e L G J−L G J−L G J H sK H s K H s K F tI = e G1 − t − J H 2K F 2I t = 2e L G J = 2e H s K 5!

L y (t) =

–1

ds

2

6

...(1)

2

–1

3

2

–1

3

s −1→ s

2

+t

–1

3

t

–1

–1

–1

2

∴

Also

LM s − 3s + 1OP MN b s − 1g PQ L 2 OP L M MN b s − 1g PQ 2

L−1

3

–1

6

3

2

t

5

t

t

–1

6

et t 5 · 60 Thus by using these results in the R.H.S. of (1), we have =

RS T

y (t) = e t 1 − t −

UV W

t2 t5 · + 2 60

6. Solve using Laplace transform, the differential equation y (0) = 1 and

dy = 1 at x = 0. dx

d2y dy −3 + 2y = 1 – e2x given that 2 dx dx

421


Solution. We have to solve

bg

bg

bg

y ′′ x − 3 y ′ x + 2 y x

= 1 – e2x

bg

Subject to the conditions y (0) = 1 and y ′ 0 = 1 Taking Laplace transforms on both sides of the given equations, we have

bg

bg

bg

L y ′′ x − 3 L y ′ x + 2 L y x

= L [1 – e2x]

i.e., {s2 L y (x) – s y (0) – y′ (0)} – 3 {s L y (x) – y (0)} + 2L y (x) =

1 1 − s s−2

Using the given initial conditions, we have (s2 – 3s + 2) L y (x) – s – 1 + 3 = −

i.e.,

b g

(s2 – 3s + 2) L y (x) = (s – 2) –

2 s s−2

(s – 1) (s – 2) L y (x) = (s – 2) –

2 s s−2

L y (x) = ∴

Let

2 s s−2

b gb g

s s −1 s − 2

2

=

b g

1 2 − s −1 s s −1 s − 2

y (x) = L–1

2

b g

b gb g LM 1 OP − L LM 2 N s − 1Q MN s bs − 1gb s − 2g –1

A B C D + + + s s −1 s − 2 s−2

b g

2

OP PQ

...(1)

2

or

2 = A (s – 1) (s – 2)2 + B s (s – 2)2 + C s (s – 1) (s – 2) + D s (s – 1)

Put

s = 0,

–1 2 B =2

A=

s = 1, s = 2, D =1 3 Also by equating the coefficients of s on both sides, we get 0 = A+B +C

–1 Now L

LM 2 MN s bs − 1g bs − 2g

2

OP PQ

= –

=

FG IJ HK

∴ C=

FG IJ H K

–3 2

FG H

IJ K

1 –1 1 1 3 1 L – L–1 + 2 L–1 + L–1 2 2 s s −1 s−2

–1 3 ⋅ 1 + 2e x − e 2 x + e 2 x ⋅ x 2 2

F 1 I GH b s − 2g JK 2

422


Hence (1) becomes x y (x) = e +

Thus,

y (x) =

3 1 – 2e x + e 2 x – e 2 x ⋅ x 2 2

1 3 – e x + e 2 x – xe 2 x is the required solution. 2 2

OBJECTIVE QUESTIONS 1. L–1

d

1

i

s s +1 2

is

(a) 1 – cos t

(b) 1 + cos t

(c) 1 – sin t

(d ) 1 + sin t

de i

Ans. a

–3s

2. The inverse Laplace transform of

s3

is

(a) (t – 3) u3 (t)

(b) (t – 3)2 u3 (t)

(c) (t – 3)3 u3 (t)

(d ) (t + 3) u3 (t)

3. If Laplace transform of a function f (t) equals

de

–2 s

− e–s s

Ans. d

i , then

(a) f (t) = 1, t > 1 (b) f (t) = 1, when 1 < t < 2 and 0 otherwise (c) f (t) = –1, when 1 < t < 3 and 0 otherwise (d ) f (t) = –1, when 1 < t < 2 and 0 otherwise Ans. d 4. Inverse Laplace transform of 1 is (a) 1 (c) δ(t – 1) 5. The inverse Laplace transform of (a) sin kt (c) u (t – a) sin kt 6. For L–1

LM 1 OP Ns Q

(b) δ(t) (d ) u (t)

Ans. d

ke – as is s2 + k 2 (b) cos kt (d ) none of these

Ans. d

n

(a) n > –1

(b) n ≥ 1

(c) n = 1, 2, ....

(d ) n < 1

Ans. c

423


7. L

LM sin t OP Nt Q

(a)

=

1 s +1

(b) cot –1 s

2

(c) cot –1 (s – 1) 8. Given L–1

LM MN ds

OP sin at , then = a + a iP Q 1

2

2

(d ) tan–1 s L–1

(a) cos at

FG sin at IJ H a K LM s e OP = Ns + 9Q

LM Ns

2

s + a2

=

(b)

cosat a

(d )

sin at a

2

(c)

OP Q

Ans. b

Ans. a

–3π

–1 9. L

2

(a) cos 3t u (t – π)

(b) – cos 3t u (t – π)

(c) cos3t u

(d ) none of these

FG t − π IJ H 3K

10. The Laplace inverse of (a)

Ans. b

t is

π s

(b)

1 2

π s

π (c)

3 (d ) none of these 2s 2 11. A function f (t) is said to be exponential order if (b) f (t) . ekt = 1 (a) f (t) = et

bg

at (c) f t ≤ be

(d )

bg

f t > be at

Ans. c

Ans. c

12. The Laplace transform of a function f (t) exists of (a) it is uniformly continuous (b) it is piecewise continuous (c) it is uniformly continuous and of exponential order (d ) it is piecewise continuous of exponential order

bg

bg

Ans. d

13. If L [ f (t)] = f s , then L[e–at f t ] is

bg f b sg

b g f b s + ag

(a) – a f s

(b) f s − a

(c) e – as

(d )

Ans. d

424


14. Inverse Laplace transform of (s + 2)–2 is (a) t e–2t

(b) t e2t

(c) e2t

(d ) none of these

15. Inverse Laplace transform of

(a)

ds

1 2

i

+ 4 s + 13

1 –3t e sin 3t 2

1 –2 t e sin 3t 4 16. Laplace transform of f ′(t) is (c)

bg bg (c) s f b sg – f b sg L 1 OP is 17. L M MN bs + 3g PQ

Ans. a

is

(b)

1 –2 t e sin 3t 3

(d )

1 3t e sin 3t 2

Ans. b

bg bg

(a) s f s − f 0

(b) s f s + f 0

(d ) none of these

Ans. a

–1

5

(a)

e –3t t 4 24

(b)

(c)

e 3t 24

(d ) none of these

–1 18. L

FG 1 IJ Hs K n

e 2t t 2 3

is possible only when n is

(a) zero

(b) –ve integer

(c) +ve integer

(d ) negative rational

–1 19. L

LM s N

2

+ 3s + 7 s3

(a) 1 + 3t +

OP Q is

7t 2 2

(b) 13t +

(c) 1 – 3t + 7t2

bg

20. If L { f (t)} = f s ; then L–1

z z t

(a)

(c)

0

R| f b sg U| S| s V| T W

bg

Ans. c

t2 2

(d ) none of these

Ans. a

is

z

∞

f t dt

0

∞

Ans. a

(b)

bg

e – st f t dt

0

bg

e – t f t dt

(d ) none of these

Ans. a

425


21. If L–1 [φ (s)] = f (t), then L–1 [e–as φ (s)] is (a) f (t + a) u (t – a)

(b) f (t – a) u (t – a)

(c) f (t – a)

(d ) none of these

–1 22. L

LM Ns

OP is Q

1 + a2

2

(a)

sinat a

(b)

(c)

1 s

(d ) none of these

–1 23. L

LM Ns

π + π2

2

OP Q

cosat a

(b) sin πt

(c) cos πt

(d ) none of these

24. L

LM MM ds N

OP P +a i P Q s2

2

b

2 2

1 sin at + at cos at 2a

(c)

1 sin at – a cos at a

b

g

g

25. The inverse Laplace transform of

(b)

(c)

1 2t e 2

b

1 sin at – at cos at 2a

(d ) none of these

g Ans. a

1 is 2s − 7

7

1 2s e 2

Ans. b

is

(a)

(a)

Ans. a

is

(a) sin t

–1

Ans. b

3

(b)

1 2s e 2

7

26. L–1

LM Ns

2

s +9

(d ) none of these

OP Q

is

(a) cos 3t

(b) sin 3t

(c) e–3t

(d ) none of these

27. L–1

LM Ns

2

s – a2

(a) cos h at (c)

1 sin at a

Ans. c

Ans. a

OP is Q (b) sin h at (d ) eat f (t)

Ans. a

426


28. L–1

LM s OP is N s + 1Q 2

(a) cos t

(b) sin t

(c) sin ht

(d ) none of these

–1 29. L

Ans. a

LM 1 OP is N s + 1Q 2

(a) cos t

(b) sin t

(c) cos ht

(d ) none of these

Ans. b

30. L–1 [F (s)] is (a)

LM N

b gOPQ

–1 –1 d L F s t ds

(c) –t L–1 [F (s)]

(b)

L–1

LM d F bsgOP N ds Q

(d ) none of these

Ans. a GGG

MODEL QUESTION PAPER–I 06 MAT 21

Second Semester B.E. Degree Examination Engineering Mathematics–II Time : 3 hrs

Max. Marks : 100

Note: 1. Answer any five full questions selecting at least two questions from each part. 2. Answer all objective type questions only in first and second writing pages. 3. Answer for objective type questions shall not be repeated.

PART A 1. (a) (i) Lagranges mean value theorem is a special case of: (a) Rolle’s theorem (b) Cauchy’s mean value theorem (c) Taylor’s theorem (d ) Taylor’s series. [Ans. b] — (ii) The result, “If f ′ (x) = 0 V x in [a, b] then f (x) is a constant in [a, b]” can be obtained from: (a) Rolle’s theorem (b) Lagrange’s mean value theorem (c) Cauchy’s mean value theorem (d ) Taylor’s theorem [Ans. b] (iii) The rate at which the curve is called: (a) Radius of curvature (b) Curvature (c) Circle curvature (d ) Evolute [Ans. b] (iv) The radius of curvature of r = a cos θ at (r, θ) is: (a) a (b) 2a (c) 1/2a (d ) a2 [Ans. d] (04 marks) (b) Find the radius of curvature at the point


of the folium x3 + y3 = 3axy. [VTU, Jan. 2009] (04 marks)

Solution. Refer Unit I. (c) State and prove Cauchy’s mean value theorem. [VTU, Jan. 2009] (06 marks) Solution. Refer Unit I. (d ) If f (x) = log (1 + ex), using Maclaurin’s theorem, show that

x x2 x4 + − + ⋅⋅ ⋅⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅ 2 8 192 1 − cos x 2. (a) (i) The value of lim is x → 0 x log 1 + x log (1 + ex) = log 2 +

[VTU, Jan. 2008] (06 marks)

b g

(a) 1/2 (c) ∞

b g

(ii) The value of lim cos x x →0

(a)

−1 2

(c) e

1 x2

(b) 1/4 (d ) None of these.

[Ans. a]

is (b)

1 e

(d )

e.

427

[Ans. b]

428


(iii) The necessary conditions for f (x, y) = 0 to have extremum are (b) fxx = 0, fyy = 0, (a) fxy = 0, fyx = 0 (d ) fx = 0, fy = 0 and fxx > 0 fyy > 0. [Ans. c] (c) fx = 0, fy = 0 (iv) The point (a, b) is called a stationary point and the value f (a, b) is called (a) Stationary point (b) Stationary value (c) Maximum value (d ) Minimum value. [Ans. b] (04 marks)

b g

(b) Evaluate (i) lim cos x x →0

1 x2

,

(ii) lim

x →0

LM 1 − 1 OP · N sin x x Q

[VTU, Jan. 2009] (04 makrs)

Solution (i) Let

b g

k = lim cos x x →0

1 x2

[1∞] form

Taking logarithm on both sides, we get ⇒

loge k = lim

b g

LM 0 OP N0Q

log cos x

x →0

x

2

Applying L’ Hospital Rule, loge k = lim

x →0

=

− sin x cos x 2x

FG3 lim H

−1 −1 tan x = .1 lim 2 x →0 x 2

x→0

form

IJ K

tan x = 1 x

−1 2 k = e–1/2

loge k =

1 e

Thus

k =

(ii) Let

k = lim

LM 1 − 1 OP N sin x x Q L x − sin x OP lim M N x sin x Q

[∞ – ∞] form

x →0

=

x →0

By L’ Hospital Rule = lim x →0

LM 1 − cos x OP N x cos x + sin x Q

Again Applying L’ Hospital Rule = lim

x →0

Thus

k = 0.

sin x LM O N − x sin x + cos x + cos x PQ =

0 = 0 0+2

LM 0 OP N0Q

form

LM 0 OP N0Q

form

429

MODEL QUESTION PAPER—I

(c) Expand f (x, y) = tan–1 ( y/x) in powers of (x – 1) and ( y – 1) upto second degree terms. [VTU, Jan. 2009] (06 marks) Solution. Refer Unit II. (d ) Discuss the maxima and minima of f (x, y) = x3 y2 (1 – x – y) [VTU, Jan. 2009] (06 marks) Solution. Refer Unit II. 3. a. (i)

zz 1

2− x

0

x2

xy dx dy is equal to:

3 4 3 (c) 5

3 8 3 (d ) . 7 (b)

(a)

(ii)

zz 1

1− x

0

0

(a) (b) (c) (d )

dx dy represents .......

Area of the triangle vertices (0, 0), (0, 1), (1, 0) Area of the triangle vertices (0, 0), (0, 1) Both (a) and (b) None of these.

[Ans. a]

LM 1 , 1 OP = ............ N2 2Q

(iii) β

(a) 3.1416 (c) 1.1416 (iv)

[Ans. b]

zzz 2

0

3

1

2

1

(b) 2.1416 (d ) 4.236.

[Ans. a]

xy 2 z dz dy dx = ...............

(a) 28 (c) 30

(b) 26 (d ) 42.

(b) Evaluate the integral by changing the order of integration,

zz ∞

0

Solution

I =

z z

0

xe − x

2

y

dy dx.


∞

x

x=0

y =0

xe − x

2

y

dy dx

Y x=0

o

[Ans. b] (04 marks) x

Y=x

x Y=0

430


The region is as shown in the figure. On changing the order of integration we must have y = 0 to ∞, x = y to ∞

zz ∞

I =

∞

xe − x

2

y

dx dy

y =0 x = y 2

Put

Also when When

x y

x dx = ydt/2 x = y, t = y and x = ∞, t = ∞

zz z z ∞

∴

2x dx = dt y

= t ∴

I =

∞

y dt dy 2

e−t

y =0 t = y ∞

=

y =0

1 = 2 Applying Bernoulli’s rule,

y −t e 2

∞

∞ t=y

dy

ye − y dy

y =0

RS d i T b g

b g d i UVW

∞ ∞ 1 y −e − y − 1 e− y y =0 y =0 2 1 0− 0−1 = 2 1 I = · 2 (c) Find the volume of the sphere x2 + y2 + z2 = a2 using triple integration [VTU, Jan. 2009] (06 marks) Solution. Here, it is convenient to employ spherical polar coordinates (r, θ, φ). In terms of these coordinates, the equation of the given sphere is r2 = a2 or r = a.

=

z

r=a

o Y

x

431


In this sphere, r varies from 0 to a, θ varies from 0 to π and φ varies from 0 to 2π. Hence, the required volume is V =

zzz zzz z z

r 2 sin θ dr dθ dφ

v

π

a

=

2π

r 2 sin θ dr dθ dφ

r =0 θ=0 φ=0 π

a

=

0

z

2π

r 2 dr × sin θ dθ × 0

dφ

0

1 3 = a × (– cos π + cos 0) × 2π 3 4π 3 a· = 3 (d ) Express the following integrals in terms of Gamma functions.

(i)

z

1

dx

0

1− x

(ii)

4

0

Solution ⇒ (i)

z

[VTU, Jan. 2009] (06 marks) ∞

I =

z

x dx cx

dx

1

Put

1 − x4 x 2 = sin θ, i.e., x = sin1/2 θ

So that

dx =

0

When

1 –1/2 sin θ cos θdθ 2 x = 0, θ = 0

When

x = 1,

∴

I = =

=

z

θ= π2

0

1 2

1 2

z

1 sin −1 2 θ cos θ dθ ⋅ 2 1 − sin 2 θ

π2

0

π 2

sin −1 2 θ dθ

LM 1 + 1OP π 2 Q ⋅ N 2 LM − 1 + 2 OP ΓM 2 MN 2 PPQ Γ −

c

432


FG 1 IJ H 4K π ⋅ 4 F 3I ΓG J H 4K Γ

=

(ii)

I = = I =

Put

z z z

∞

0

xc dx cx xc

∞

0 ∞

e x log c

dx

e − x log c x c dx

0

x log c = t so that x =

∴

I = =

=

dx =

dt log c

t log c

z

∞

e−t

0

FG t IJ H log cK

1

b g Γ bc + 1g blog cg log c

c +1

c +1

z

∞

c

dt log c

t c e − t dt

0

·

4. (a) (i) If F (t) has a constant magnitude then: (a)

bg

d F t = 0 dt

bg

(c) F t ×

bg

dF t = 0 dt

bg

(b) F t ⋅

bg

bg

dF t dt

bg

(d ) F t − d F t dt

= 0

= 0.

[Ans. b]

(ii) Use the following integral work done by a force F can be calculated: (a) Line integral (b) Surface integral (c) Volume integral (d ) None of these. (iii) Green’s theorem in the plane is applicable to: (a) xy-plane (b) yz-plane (c) zx-plane (d ) All of these.

[Ans. a]

[Ans. d]

(iv) If all the surfaces are closed in a region containing volume V then the following theorem is applicable: (a) Stoke’s theorem (b) Green’s theorem (c) Gauss Divergence theorem (d ) Only (a) and (b). (04 marks) [Ans. c]

433


(b) Use the line integral, compute work done by a force F = (2y + 3) i + xz j + ( yz – x)

k when it moves a particle from the point (0, 0, 0) to the point (2, 1, 1) along the curve x = 2t2, y = t, z = t3. [VTU, Jan. 2009] (04 marks) Solution. Refer Unit IV. (c) Verify Green’s theorem for

zd c

i

xy + y 2 dx + x 2 dy where c is bounded by y = x and

y = x2 [VTU, Jan. 2009] (06 marks) Solution. Refer Unit IV. (d ) Prove that the cylindrical coordinate system is orthogonal. [VTU, Jan. 2009] (06 marks) Solution. Refer Unit IV.

PART B dy = y 2 is: dx (a) Linear (b) Non-linear (c) Quasilinear (d ) None of these

5. (a) (i) The differential equation

(ii) The particular integral of

d2y + y = cos x is: dx 2

1 1 1 1 x sin x sin x (b) cos x (c) x cos x (d ) 2 2 2 2 (iii) The general solution of the D.E. (D2 + 1)2 y = 0 is: (a) c1 cos x + c2 sin x (b) (c1 + c2 x) cos x + (c3 + c4 x) sin x (c) c1 cos x + c2 sin x + c3 cos x + c4 sin x (d ) (c1 cos x + c2 sin x) (c3 cos x + c4 sin x) (a)

(iv) The set of linearly independent solution of the D.E. (a) {1, x, ex, e–x} (c) {1, x, ex, xex} 3

(b) {1, x, e–x, xe–x} (d ) {1, x, ex, xe–x}

[Ans. d]

[Ans. b]

d4y d2y − = 0 is: dx 4 dx 2 [Ans. a] (04 marks)

2

d y d y dy = e–x + sin 2x +2 2 + 3 dx dx dx Solution. Here, the A.E. is m3 + 2m2 + m = 0 ⇒ m (m2 + 2m + 1) = 0 ⇒ m (m + 1)2 = 0 whose roots are 0, –1, –1 ∴ C.F. = C1 e0x + (C2 + C3 x) e–x ∴ C.F. = C1 + (C2 + C3 x) e–x (b) Solve

[Ans. b]

[VTU, July, 2008] (04 marks)

434


Also

P.I. =

1 (e–x + sin 2x) D + 2 D2 + D 3

sin 2 x e−x + 3 3 2 D + 2 D + D D + 2 D2 + D P.I. = P.I.1 + P.I.2 =

∴

P.I.1 = =

e− x D3 + 2 D2 + D

(D → –1)

e− x

b−1g + 2b−1g + b−1g 3

(Dr = 0)

2

Diefferentiate the denominator and multiply ‘x’, we get ∴

P.I.1 = =

xe − x 3D 2 + 4 D + 1

(D → –1)

xe − x

b g b g

(Dr = 0)

2

3 −1 + 4 −1 + 1

Again differentiate and multiply x. ∴

⇒

P.I.1 =

x 2e − x 6D + 4

P.I.1 =

x2e− x −2

P.I.2 =

sin 2 x D + 2 D2 + D

(D → –1)

(D2 → – 22 = – 4)

3

=

sin 2 x − 4D − 8 + D

=

sin 2 x −3 D − 8

= − = −

sin 2 x 3D − 8 × 3D + 8 3D − 8

b6 cos 2 x − 8 sin 2 xg

(D2 → – 22)

9 D 2 − 64

1 (6 cos 2x – 8 sin 2x) 100 ∴ The general solution is y = C.F. + P.I. ∴

P.I.2 =

y = C1 + (C2 + C3 x) e–x –

1 1 2 –x x e + (3 cos 2x – 4 sin 2x) 50 2

435


d2y − 4 y = cos h (2x – 1) + 3x (c) Solve dx 2 Solution. We have (D2 – 4) y = cos h (2x – 1) + 3x A.E. is m2 – 4 = 0 or (m – 2) (m + 2) = 0 ⇒ m = 2, – 2 ∴ C.F. = C1 e2x + C2 e–2x P.I. =

b


g

cos h 2 x − 1 + 3 x

LM MN

D −4 2

OP PQ

1 e 2 x −1 e b 2 1g 3x + + 2 = 2 2 2 D −4 D −4 D −4 −

x−

= P.I.1 + P.I.2 + P.I.3 P.I.1 = =

1 e 2 x −1 1 e 2 x −1 = 2 D2 − 4 2 22 − 4

(Dr = 0)

1 x e 2 x −1 1 x e 2 x −1 1 = = x e 2 x −1 2 2D 2 4 8 1 e b g 2 D2 − 4 − 2 x −1

P.I.2 =

(D → – 2)

1 e −b 2 −1g 2 −2 2 − 4 x

=

b g

(Dr = 0)

− 2 −1 1 xe b g 2 2D x

= P.I.2 =

− x – (2x e 8

(D → – 2)

– 1)

3x e b log 3g = P.I.3 = D2 − 4 D2 − 4 x

=

=

e b log 3g

blog 3g

2

(D → log 3)

x

−4

3x

blog 3g

2

−4

Complete solution: y = C.F. + P.I. = C1 e2x + C2 e–2x +

x 2x – 1 x – (2x – 1) e – e + 8 8

3x

blog 3g

2

−4

436


(d) Solve by the method of undetermined coefficients; (D2 + 1) y = sin x. [VTU, July 2008] (06 marks) Solution. Refer Unit V. 6. (a) (i) The homogeneous linear differential equation whose auxillary equation has roots 1, 1 and – 2 is: (a) (D3 + D2 + 2D + 2) y = 0

(b) (D3 + 3D – 2) y = 0

(c) (D3 – 3D + 2) y = 0

(d ) (D + 1)2 (D – 2) y = 0

(ii) The general solution of

(x2

D2

[Ans. c]

– xD) y = 0 is :

(a) y = C1 + C2

ex

(b) y = C1 + C2 x

(c) y = C1 + C2

x2

(d ) y = C1 x + C2 x2

[Ans. c]

(iii) Every solution of y′′ + ay′ + by = 0 where a and b are constants approaches to zero as x → ∞ provided. (a) a > 0, b > 0

(b) a > 0, b < 0

(c) a < 0, b < 0

(d ) a < 0, b > 0

[Ans. a]

(iv) By the method of variation of parameters the W is called. (a) Work done

(b) Wronskian

(c) Euler’s

(d ) None of these

[Ans. b] (04 marks)

(b) Solve (1 + x)2 y″ + (1 + x) y′ + y = 2 sin [log (1 + x)]. [VTU, Jan. 2009] (04 marks) Solution t = log (1 + x) or et = 1 + x

Put Then we have (1 + x)

dy = 1·Dy dx

d2y (1 + = 12·D (D – 1) y dx 2 Hence the given D.E. becomes x)2

[D (D – 1) y + D + 1] y = 2 sin t i.e, A.E. is ∴

(D2 + 1) y = 2 sin t m2 + 1 = 0 ⇒ m = ± i C.F. = C1 cos t + C2 sin t P.I. =

2 sin t D2 + 1

(D2 → –12 = –1)

=

2 sin t −1 + 1

(Dr = 0)

=

2 sin t 2 D × 2D 2D

437


=

4 cos t 4 D2

(D2 → –12 = –1)

4 cos t −4 P.I. = – cos t

=

∴ ∴ The complete solution is y = C.F. + P.I. y = C1 cos t + C2 sin t – cos t where t = log (1 + x) y = C1 cos [log (1 + x)] + C2 sin [log (1 + x)] – cos [log (1 + x)] (c) Solve x 3

FG H

IJ K

2 1 d 3y 2 d y 2 x + + 2 y = 10 x + · 3 2 x dx dx


Solution. Refer Unit VI.

d2y + y = tan x. dx 2

(d) Solve, by the method of variation of parameters Solution. Refer Unit VI. 7. (a) (i) The Laplace transform of t2 et is : (a)

(c)

2

b s − 2g

2

(b)

3

(d )

1

bs − 2g

2

b s − 2g

3

1

bs − 1g

[Ans. b]

3

(ii) L [e–t sin ht] is: (a)

(c)

1

bs + 1g

2

+1

1 s s+2

b g

(b)

(d )

1

bs − 1g

2

+1

s −1

bs − 1g

2

+1

[Ans. c]

(iii) L [e–3t cos 3t] = (a)

s−3 s − 6s − 18

(b)

s+3 s + 6s + 18

(c)

s+3 s − 6s + 18

(d )

s−3 s + 6s − 18

(iv) L

2

2

LM sin t OP = Nt Q

1 s +1 (c) cot–1 (s – 1) (a)

2

2

2

[Ans. b]

(b) cot–1 s (d ) tan–1 s

[Ans. b] (04 marks)

438


(b) Find the Laplace transform of cos 2t − cos 3t ⋅ t Solution. Refer Unit VII. 2t +

Put


a = 2,

Ans.

b= 3

s 2 + 32 1 1 + log 2 s − log 2 2 s + 22

(c) Find the Laplace transform of the periodic function with period

R|sin wt , 0 < t < π w f (t) = S . 2π π ||0,
2π w


Solution. Refer Unit VII.

R|cos t , 0 < t < π (d) Express f (t) = Scos 2t , π < t < 2π . |Tcos 3t , t > 2π Solution.

Let ⇒ and i.e.,

∴

But

f (t) = cos t + (cos 2t – cos t) u (t – π) + (cos 3t – cos 2t) u (t – 2π) L [f (t)] = L (cos t) + L [(cos 2t – cos t) u (t – π)] + L [(cos 3t – cos 2t) u (t – 2π)] ...(1) F (t – π) = cos 2t – cos t G (t – 2π) = cos 3t – cos 2t F (t) = cos 2 (t + π) – cos (t + π) G (t) = cos 3 (t + 2π) – cos 2 (t + 2π) F (t) = cos 2t + cos t G (t) = cos 3t – cos 2t

bg G b sg F s

=

s s + 2 s + 4 s +1

=

s s − 2 s +9 s +4

2

2

bg G b sg LM s + s OP N s + 4 s + 1Q LM s − s OP Ns + 9 s + 4Q

L [F (t – π) u (t – π)] = e–πs F s

and

L [G (t – 2π) u (t – 2π)] = e–2πs

i.e.,

− πs L [(cos 2t – cos t) u (t – π)] = e

and


L [(cos 3t – cos 2t) u (t – 2π)] = e −2 πs

2

2

2

2

439


Hence (1) becomes

Thus

LM N

OP Q

s s s s s + e −2 π s 2 − 2 + e − πs 2 + 2 s +1 s + 4 s +1 s +9 s +4

L [f (t)] =

5s e −2 πs 1 1 s − πs s e − + + s2 + 1 s2 + 4 s 2 + 1 s2 + 4 s2 + 9

8. (a) (i) Given L−1

LM Ns

2

2

1 + a2

OP = Q

FG sin at IJ H a K LM e OP is: Ns Q

LM N

sin at then L−1 a

(a) cos at

−1 (ii) L

bt − 1g

2

cos at a

(d )

sin at a

2

(b) u (t – 1)

2

t2 2

LM s e OP is: Ns + 9Q

[Ans. a]

bt − 1g

3

6

(d ) None of these

[Ans. a]

(b) – cos 3t u (t – π) (d ) e–as L [f (t – a)]

[Ans. b]

− sπ

2

(a) cos 3t u (t – π) (c) cos 3t u (t – π)/3 (iv) The Laplace inverse of

(c)

i

3

(c) u (t)

(a)

2

OP Q

−s

(a) u (t – 1)

(iii) L−1

OP Q d id LM s OP is: Ns + a Q

(b) 2

(c)

LM N

L [f (t)] =

t is:

π s π 2s

32

LM FG s + a IJ OP . N H s + bKQ F s + a IJ Solution. Let F b sg = log G H s + bK

(b) Find L−1 log

(b)

1 2

π s

(d ) Does not exist

[Ans. c] (04 marks)


= log (s + a) – log (s + b)

440


bg

−F ′ s

Now i.e., Thus,

LM N

b gOQP

L−1 − F ′ s

RS 1 − 1 UV Ts + a s +bW L 1 OP − L LM 1 OP L M Ns + bQ Ns + aQ

= −

=

−1

−1

t f (t) = e–bt – e–at f (t) =

e −bt − e −at · t

(c) Apply Convolution theorem to evaluate L−1

LM MM d s N

OP P⋅ +a i P Q s

2

2 2

[VTU, Jan. 2009] (06 marks) Solution. Refer Unit VIII. (d) Solve the differential equation by Laplace transform onethod, y″ + 4y′ + 3y = e–t and the initial conditions y (0) = y′ (0) = 1. [VTU, Jan. 2009] (06 marks) Solution. Refer Unit VIII. GGG

MODEL QUESTION PAPER–II 06 MAT 21

Second Semester B.E. Degree Examination Engineering Mathematics–II Time : 3 hrs

Max. Marks : 100

Note: 1. Answer any five full questions selecting at least two questions from each part. 2. Answer all objective type questions only in first and second writing pages. 3. Answer for objective type questions shall not be repeated.

PART A 2

1. (a) (i) The radius of curvature of y = e–x at (0, 1) is (a) 1 (b) 2

1 1 (d ) 2 3 (ii) The radius of curvature of the circle of curvature is (a) 1 (b) ρ (c)

(c)

1 ρ

[Ans. c]

(d ) ρ2

[Ans. b]

(iii) The first three non-zero terms in the expansion of ex tan x is (a) x + x2 +

1 3 x 3

(b) x +

x3 2 5 + x 3 5

5 3 x3 1 5 x + x (d ) x + 6 3 6 (iv) The radius of curvaluve r = a sin θ at (r, θ) is (a) 1 (b) 0 (c) 2 (d ) None of these 2 (c) x + x +

(b) Find the radius of curvature at x =

πa on y = a sec 4

[Ans. c]

FG x IJ . H aK

[Ans. d ] (04 marks)

[VTU, July 2008] (04 marks) Solution. Refer Unit I. (c) Verify Rolle’s theorem for the function f (x) = (x – a)m (x – b)n in [a, b] where m > 1 and n > 1. [VTU, Jan. 2008] (06 marks) Solution. Refer Unit I. (d ) Expand esin x up to the term containing x4 by Maclaurin’s theorem. [VTU, Jan. 2008] (06 marks) Solution. Refer Unit I. 441

442


2. (a) (i) The value of lim x log x is x→0 (a) 1

(b) 0

(c) 2

(d ) ∞

[Ans. b]

LM FG IJ OP cot (x – a) is N H KQ

x log 2 − (ii) The value of xlim →a a

−1 (b) 1 a (c) 2 (d) None of these [Ans. a] (iii) If f (x, y) has derivatives upto any order with in a neighbourhood of a point (a, b) then f (x, y) can be extended to the (a) Finite series (b) Infinite series (c) Some extension limits (d) None of these [Ans. b] 2 (iv) If AC – B < 0 then f has neither a maximum nor a minimum at (a, b) the point (a, b) is called : (a) Saddle point (b) Maximum at (a, b) (c) Minimum at (a, b) (d ) Both (a) and (b) [Ans. a] (04 marks) (a)

(b) Evaluate lim

x →0

FG 1 − cot xIJ . Hx K 2


2

Solution. Refer Unit II. (c) Expand ex log (1 + y) by Maclaurin’s theorem up to the third degree term. [VTU, Jan. 2008] (06 marks) Solution. Refer Unit II. (d ) Determine the maxima/minima of the function sin x + sin y + sin (x + y). [VTU, Jan. 2005] (06 makrs) Solution. Refer Unit II. 3. (a) (i) For (a) (c)

zz zz zz

b g f b x , y g dx dy f b x , y g dx dy

∞

∞

0

x

∞

∞

x

0

∞

y

0

0

f x , y dx dy , the change of order is

(ii) The value of the integral

(b)

z

(d ) 2

−2

dx is x2

zz zz ∞

∞

0

x

∞

x

0

0

b g f b x , y g dx dy

f x , y dx dy [Ans. c]

(a) 0 (b) 0.25 (c) 1 (d ) ∞ [Ans. d ] (iii) The volume of the tetrahedron bounded by the coordinate planes and the plane x y z + + = 1 is a b c

443

MODEL QUESTION PAPER—II

(iv)

(a)

abc 2

(b)

abc 3

(c)

abc 6

(d )

24 abc

bg Γ b5g

Γ7

[Ans. c]

is

(a) 30 (c) 48 (b) Find the value of x2.

zz

(b) 42 (d ) 17

b

[Ans. a] (04 marks)

g

xy x + y dx dy taken over the region enclosed by the curve y = x [VTU, Jan. 2008, July 2008] (04 marks)

and y = Solution. Refer Unit III.

x2 y2 z2 + + = 1. a 2 b2 c2 [VTU, Jan. 2008] (06 marks)

(c) Using the multiple integrals find the volume of the ellipsoid

Solution. Refer Unit III. (d ) With usual notation show that β (m, n) =

b g bg Γ b m + ng

Γ m . Γ. n

[VTU, Jan, 2008] (06 marks)

4. (a) (i) If all the surfaces are enclosed in a region containing volume V then the following theorem is applicable. (a) Stoke’s theorem (b) Green’s theorem (c) Gauss divergence theorem (d ) Only (a) and (b) [Ans. c] (ii) The component of ∇φ in the direction of a unit vector a is ∇φ · a and is called (a) The directional derivative of φ in the direction a . (b) The magnitude of φ in the direction a . (c) The normal of φ in the direction a . (d ) None of these

[Ans. a]

(iii) For a vector function F , there exists a scalar potential only when (a) div F = 0

(b) gred (div F ) = 0

(c) cur F = 0 (iv) Which of the following is true: (a) curl

FH A ⋅ BIK

(d ) F curl F = 0

= curl A + curl B

(b) div curl A = ∇ A

[Ans. c]

444


FH A ⋅ BIK

(c) div

= div A div B

(d ) div cur A = 0 (b) Evaluate

z

0


F ⋅ dr where F = x2 i + y2 j + z2 k and c is given by x = cos t, y = sin t,

z = t, 0 ≤ t ≤ π. Solution. Refer Unit IV. (c) Evaluate

zd


i

xy − x 2 dx + x 2 y dy where c is the closed formed by y = 0, x = 1 and

y = x (a) directly as a line integral (b) by employing Green’s theorem. [VTU, Jan. 2007] (06 marks) Solution. Refer Unit IV. (d ) If f and g are continuously differentiable show that ∇f × ∇g is a solenoidal. Solution. Refer Unit IV.

PART B 5. (a) (i) The general solution of the differential equation (D4 – 6D3 + 12D2 – 8D) y = 0 is (a) (b) (c) (d)

y y y y

= = = =

c1 + [c2 + c3 x + c4 x2) e2x (c1 + c2 x + c3 x2) e2x c1 + c2 x + c3 x2 + c4 x4 c1 + c2 x + c3 x2 + c4 e2x

[Ans. a]

2

(ii) The particular integral of (a)

x2 + 4x 3

d y dy + = x2 + 2x + 4 is dx 2 dx (b)

x3 +4 3

x3 x3 + 4x + 4x2 (d) 3 3 (iii) The particular integral of (D2 + a2) y = sin ax is (c)

(a)

−x cos ax 2a

(b)

[Ans. c]

x cos ax 2a

−ax ax cos ax (d) cos ax 2 2 (iv) The solution of the differential equation (D2 – 2D + 5)2 y = 0 is

(c)

(a) (b) (c) (d)

y y y y

= = = =

e2x {(c1 + c2 x) cos x + (c3 + c4 x) sin x} ex {(c1 + c2 x) cos 2x + (c3 + c4 x) sin 2x} (c1 ex + c2 e2x) cos x + (c3 ex + c4 e2x) sin x ex {4 cos x + c2 cos 2x + c3 sin x + c4 sin 2x}

[Ans. a]

[Ans. b] (04 marks)

445


(b) Solve:

d 3y − y = (ex + 1)2 dx 3 Solution. Refer Unit V. (c) Solve :

[VTU, July 2007] (04 marks)

d2y dy − 4 + 4 y = 3x2 e2x sin 2x 2 dx dx Solution. Refer Unit V. (d ) Solve by the method of undetermined coefficients y″ – 3y′ + 2y = x2 + ex Solution. Refer Unit V.



6. (a) (i) The general solution of (x2 D2 – xD) y = 0 is (a) y = c1 + c2 ex

(b) y = c1 + c2 x2

(c) y = c1 + c2 x2

(d ) y = c1 x + c2 x2

[Ans. c]

(ii) For the variation of parameters the value of W is (a) y1 y2′ − y 2 y ′

(b) y2 y2′ − y1 y2′

(c) y2 y1′ − y2 y2′

(d ) y2 y1′ − y 2 y1

[Ans. a]

(iii) The DE in which the conditions are specified at a single value of the independent variable say x = x0 is called (a) Initial value problem

(b) Boundary value problem

(c) Final value

(d ) Both (a) and (b)

[Ans. a]

(iv) The DE in which the conditions are specified for a given set of n values of the independent variable is called a (a) Intial value problem

(b) Boundary value problem

(c) Final value (d ) Both (a) and (b) [Ans. b] (04 marks) (b) Solve by the method of variation parameters y″ + y = tan x. [VTU, Jan. 2008] (04 marks) Solution. Refer Unit VI. (c) Solve:

d2y dy +x + 2 y = x cos (log x). 2 dx dx Solution. Refer Unit VI. (d) Solve: x2


d2y dy +4 + 3 y = e–x subject to the conditions is y (0) = y′ (0). 2 dx dx [VTU, Jan. 2008] (06 marks) Solution. Refer Unit VI.

446


7. (a) (i) The Laplace transform of sin2 3t is: (a)

(c)

3 s + 36

(b)

2

18

d

i

s s + 36 2

d

6

i

s s + 36 2

(d )

18 s + 36

[Ans. c]

2

(ii) L [e–3t cos 3t] is: (a)

s−3 s − 6 s − 18

(b)

s+3 s + 65 + 18

(c)

s+3 s + 6s + 18

(d )

s−3 s + 6s − 18

2

2

2

[Ans. b]

2

(iii) L [(t2 + 1) u (t – 1)]

d1 + x + s i

d1 + s + s i 2

2

(a) 2 e – s

(b) e – s

s3

d1 + s + s i

s3

2

(c) 2e

s

(d ) None of these s3 (iv) The Laplace transform of a function f (t) exists of

[Ans. a]

(a) It is uniformly continuous (b) It is piecewise continuous (c) It is uniformly continuous and of exponential order (d ) It is piecewise continuous of exponential order


(b) Prove that

z

∞

3 · 0 50 Solution. Refer Unit VII. (c) A periodic function f (t) of period 2a is defined by

e –3t t sin t dt =

f (t) =

RS a T–a

for for


0≤t
show that L {f (t)} =

FG IJ H K

a as tan b . s 2


Solution. Refer Unit VII. (d ) Express f (t) in terms of the Heavisides unit step function and find its Laplace transforms

R| t , S| 4t , T 8, 2

f (t) =

04

Solution. Refer Unit VII.


447


8. (a) (i) L–1

LM Ns

2

OP is: Q

1 + 32

(a)

1 sin 3t 3

(b) sin h 3t

(c)

1 sin h 3t 3

(d )

1 cos 3t 3

bg

(b)

t n−1 Γ n

t n+1 Γ n +1

(d )

t n−1 Γ n −1

(ii) L–1

LM 1 OP Ns Q

is:

n +1

tn (a) Γ n (c)

[Ans. a]

b g

(iii) Laplace transform of f ″ (t) is

bg b g

[Ans. b]

(a) s2 L {f (t)} – s f (0) – f ′ (0)

(b) s2 L {f (t)} – f ′ (0)

(c) s2 L {f (t)} + s f (0) + f (0)

(d ) None of these

(iv) For L−1

LM 1 OP Ns Q

[Ans. a]

is:

n

(a) n > –1

(b) n ≥ –1

(c) n = 1, 2, ....

(d ) n < 1

[Ans. c] (04 marks)

2s − 1 · s − 5s + 6


(b) Find the inverse Laplace transform of

2

Solution. Refer Unit VIII. (c) By employing convolution theorem; evaluate L–1

LM MN d s

OP + a id s + b i P · Q s2

2

2

2

2


Solution. Refer Unit VIII. (d ) Solve using Laplace transforms

d2y dy − 3 + 2 y = 4e2t 2 dt dt given that y (0) = –3 and y′ (0) = 5.


Solution. Refer Unit VIII. GGG

Ganesh - Engineering Mathematics II

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