Solution Manual for Advance Math

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS

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Final answers are shaded in green.


TECHNOLOGICAL INSTITUTE OF THE PHILIPPPINES #938 Aurora Blvd., Cubao, Quezon City

College of Engineering and Architecture Department of Electronics Engineering

FINAL PROJECT

In Partial Fulfillment of the Requirements Needed for the completion of the subject Advanced Engineering Mathematics for ECE (EC353)

Submitted By: Agustin, John Christopher V. Arrobang, Ma. Bernadette T. Moreno, Kendrick Kent L. Susbilla, Mark Anthony F.

Submitted To: Engr. Armil S. Monsura Instructor

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March 18, 2011



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UNIT 1

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COMPLEX NUMBERS



Example 1.1 Let z 8 j3 and z 9 j2 Find:

(a) The real part of z1 and z2.

(b) The imaginary part of z1 and z2.

(c) The sum z1 z2. Solutions:

(d) The product z1 z2

(a) The following are the real part of z and z

Re (z ) 8 Re (z ) 9

(b) The following are the imaginary part of z and z

Im (z ) 3 Im (z ) 2

(c) Using the general rule for addition of complex numbers,

z z (x x ) j ( y y ) z z (8 9) j ( 3 2 ) z z 17 j

(d) Using the general rule for multiplication of complex numbers,

z z (x x y y ) j (x y x y )

z z 0(8)(9)– (3)(2)2 j 0(8)(2) (9)(3)2)

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z z 78 j11



Example 1.2 Let z 8 j3 and z 9 j2. Find:

(a) The difference z – z Solutions:

(b) The quotient

56 57

(a) Using the general rule for subtraction of complex numbers, z z (x x ) j(y y ) z z (8 9) j(3 2) z z 1 j5

(b) Using the general rule for division of complex numbers,

z x x y y x y x y

j z x y x y

z 8(9) (2)(3) 9(3) 8(2)

j z 9 2 9 2 z 66 43

j z 85 85

Drill Problem 1.1 Direction: For items 1 to 9, let z 2 j3 and z 4 – j5. Showing the details of your work (in the rectangular form x jy): Page

7

1.) (5z 3z )


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Solution:

First, substitute the values for z 2 j3 and, z 4 – j5 then perform it algebraically (5z 3z ) 05(2 j3) 3(4 j5)2 (5z 3z ) 010 j15 12 j152 (5z 3z ) 22

(5z 3z ) 484 2.) z ? z ?

Solution:

Using the general rule for complex conjugate defined as z ? x jy , For z ? 2 – j3

and

z ? 4 j5

To find z ? z ? we can use FOIL method

z ? z ? (2 j3) (4 – j5)

z ? z ? 8 j10 j12 j 15 z ? z ? 8 15 j12 j10

3.) Re AZ7C

z ? z ? 23 j2

6

Solution :

Substituting the value of z 2 j3 to AD7 C,

Get the square of z 2 j3 by using FOIL method Final answers are shaded in green.

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1 1 E F ( 2 j3) z

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 1 E F (2 j3)(2 j3) z

1 1 E F z 4 j6 j6 j 9 1 1 E F z 4 9 j12 1 1 E F z 5 j12

After getting the simplified form of Z , we can now multiply it by the complex conjugate of the denominator By cross multiplication,

1 1 5 j12 E F . z 5 j12 5 j12

1 5 j12 E F z 25 j60 j60 j2 144

Since j2 -1, we can now simplify the denominator 1 5 j12 E F 25 144 z 1 5 j12 E F 169 z

Therefore,

1 5 Re E F 169 z

4.) Re (z ) , 0Re (z )2

z ( 4 j5) (4 j5)

z 16 j20 j20 j 25


9

Re (z ) Since z 4 – j5 we can get z by FOIL method Page

Solution:


z 16 25 j40 z 9 – j40

0Re (z )2

Re (z ) 9

Since z 4 – j5 , therefore Re (z ) 4 0Re (z )2 4

D7 D6

Solution:

z 4 j5 z 2 j3

Multiply both the numerator and the denominator by the complex conjugate of the denominator which is 2 – j3 4 j5 2 j3 z M 2 j3 2 j3 z

z 8 j10 j12 j 15 z 4 j6 j6 j 9 z 8 15 j10 j12 z 4 9

z 7 22 j z 13 33

10

Therefore,

z 7 j22 z 13

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5.)

0Re (z )2 16


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 6.)

D6 ? D7

?

, AD6 C D

7

?

Solution:

D6 ? D7 ?

Since

z ? 2 j3

and

z ? 4 j5

Multiplying the numerator and the denominator by the denominator’s complex conjugate, we have, z ? 2 j3 4 j5 M ? 4 j5 4 j5 z

z ? 8 j12 j10 j 15 z ? 16 j20 j20 j 25 z ? 8 15 j22 z ? 16 25

Therefore,

7.) (4z z )

z ? 7 j22 41 z ?

z ? 7 22 j ? z 41 41

Solution:

Substitute the values for Z1 and Z2 in the expression then solve it algebraically (4z z ) 04(2 j3) (4 j5)2

(4z z ) (4 j17)

(4z z ) 16 j68 j68 j 289 Final answers are shaded in green.

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Performing binomial expansion,

11

(4z z ) (8 j12 4 j5)


Therefore

D6

, D 6? D

6

Solution:

D6 ? D6

Since

z 2 j3

and

z ? 2 j3

Multiplying the numerator and the denominator by the denominator’s complex conjugate, we have z ? 2 j3 2 j3 M z 2 j3 2 j3

z ? 4 j6 j6 j 9 z 4 j6 j6 j 9 z ? 4 9 j12 4 9 z

Therefore

z ? 5 j12 z 13

z ? 5 12 j z 13 13

D6

D6 ?

Multiplying the numerator and the denominator by the denominator’s complex conjugate, we have


12

D6 ?

(4z z ) 273 j136

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8.)

(4z z ) 16 289 j136

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS z 2 j3 2 j3 M ? z 2 j3 2 j3

4 j6 j6 j 9 z z ? 4 j6 j6 j 9 z 4 9 j12 z ? 4 9

Therefore D6 PD7 D6 QD7

z 5 12

j ? z 13 13

Solution:

For the numerator: Perform addition of complex numbers z z 2 j3 4 j5 z z 6 j2

For the denominator: Perform subtraction of complex numbers z z 2 j3 4 j5 z z 2 j8

Substituting the values that we have in performing the addition and subtraction of complex number to the expression and multiplying the numerator and the denominator by the denominator’s complex conjugate, we have 6 j2 2 j8 z z M z z 2 j8 2 j8

z z 12 j4 j48 j 16 4 j16 j16 j 64 z z

13

z z 12 16 j48 j4 z z 4 64

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9.)

z 5 j12 ? z 13


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS z z 28 j44 z z 68

Therefore

z z 7 11 j z z 17 17

Direction: For items 10 to 13, let z x jy. Find the rectangular form. 10.) Im(z R ), 0Im(z)2R Solution:

Im(z R )

z x jy

z R (x jy) (x jy)

z R (x j2xy j y )(x jy) z R (x y j2xy)(x jy)

z R 0(x y ) j2xy2(x jy)

z R (x y )x j2x y (x y )jy j 2xy

Grouping like terms,

Therefore

z R x R xy 2xy j02x y y(x y )2

0Im(z)2R

Im(z R ) 3x y y R

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since Im(z) jy Final answers are shaded in green.

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 0Im(z)2R (jy)R 0Im(z)2R y R

0Im(z)2R y R 11.) Re A ? C

D

Solution:

Since z ? x jy

Multiply the numerator and the denominator by the denominator’s complex conjugate, we have S

Therefore

12.) Im0(1 j)U z 2

S

1 1 x jy T M ? z x jy x jy

1 x jy T ? x jxy jxy j y z S

S

1 x jy T ? z x j y

S

1 x jy T z? x y

1 x jy T

? z x y x y

Re S

1 x T ? z x y

Solution:

Perform first z by binomial expansion

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z x jxy jxy j y

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z (x jy)(x jy)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS z x j2xy j y z x j2xy y

Then, Im (z ) 2xy

Dealing with (1 j)U , we can say that it is just equal to 0(1 j) 2V in terms of laws of exponents, so we have: (1 j)U (1 j)MV

(1 j)U (1 j)(1 j) (1 j)U 1 j j j (1 j)U 1 1 j2 (1 j)U j2

(1 j)U (j2)V

(1 j)U (j)V (2)V (1 j)U (1)(16) (1 j)U 16

Multiplying the answers that we have

Im0(1 j)U z 2 16 (2xy)

7

Z?

C

Solution:

S

1 1 T 7 (x jy)2 z?

16

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13.) Re A

Im0(1 j)U z 2 32xy


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Performing the operations,

S

S

S

S

1 1 7T ? (x jy)2 z

1 1 7T ? (x jy)(x jy) z

1 1 7T 2 ? x jyx jyx j2 y2 z S

1 1 T 7 2 2 x y j2xy z?

Multiply the numerator and the denominator by the denominator’s complex conjugate, we have 1 1 x2 y2 j2xy S ?7 T 2 M x y2 j2xy x2 y2 j2xy z

x2 y2 j2xy 1 T 7 x4 x2 y2 j2x3 y y4 x2 y2 j2x3 y j2x3 y j2x3 y 4x2 y2 z?

S

We’ll have

Re S

S

1 x2 y2 j2xy T 7 x4 2x2 y2 4x2 y2 y4 z? S

x2 y2 j2xy 1 T 7 x4 2x2 y2 y4 z?

1 x2 y2 j2xy T

4 7 4 2 2 4 ? x 2x y y x 2x2 y2 y4 z

1 x y 1 x y T or Re S T 7 7 (x y ) x V 2x y y V z? z?

14.) Verify the following laws of conjugation

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(z z )? z ? z ?

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If Z 5 j3 and Z 2 j4 Final answers are shaded in green.


Solution:

(z z )? z ? z ?

0(5 j3) (2 j4)2? (5 j3) (2 j4)

7 j 7 j

Therefore

(z z )? z ? z ? is correct

(z z )? z ? z ?

If Z 5 j3 and Z 2 j4

(z z )? z ? z ?

(5 j3)(2 j4) (5 j3)(2 j4)

10 j20 j6 j 12 10 j20 j6 j 12 Therefore

10 12 j14 10 12 j14 22 j14 22 j14

(z z )? z ? z ? is correct.

15.) Show that j2 -1 , j3 -j , j4 1, and -j , Y7 -1 ,YZ j ,Y[ 1 . From the results of these

For j2 --1

Since j √1

\(1)⁄ ^ 1

For j3 -j

Since j √1

\(1)⁄ ^ 1 R

(1)(1)⁄ j

For j4 1

Since j √1

\(1)⁄ ^ 1 V

(1) 1

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Solution:

Y

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evaluate j2312 and its reciprocal j -2312


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For Y j

Since j √1

Getting the reciprocal of

√Q

√Q

.

b

√Q √Q

-j

√Q -j Q

For

YZ

1 j j

Getting the reciprocal 1 j . j j j Y

Y7

1

Since j2 -1 1 1 1

j

Since j2 -j

QY7

For

j

Since j2 -1

For

Yc

1

Since j4 1 1 1 1

1 j jR

To evaluate jR and jQR . We need to divide the exponent by 4 and if the results will be 0 ` then ja 1 1 ` then ja j

2 ` then ja 1

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3 ` then ja j


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For jR

2312 578 4

Since there is no remainder, we can say that jR 1

For jQR

By laws of exponents, we can say that

since jQR 1 Therefore,

jQR

1

jR

jQR 1

Example 1.3 For z 1 j, z 1 j and zR 3 j3√3 find

(a) The modulus r, the principal argument θ and express each in polar form.

(b) All possible arguments

(c) The plot of each one in complex plane Solutions

For z

r √2

20

r f(1) (1)

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(a) The modulus r



The principal argument

Expressing in polar form,

(b) All possible arguments

or

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(c) Graph in the complex plane


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For z

(a) The modulus r

r f(1) (1)

The principal argument θ

r √2

y Arg z tanQ A C x

Arg z tanQ S Arg z

π 4

1 T 1

Since θ is directed angle from the positive axis to the terminal point z. Here, all angles are measured in the counter clockwise sense. Thus, Arg z

Expressing in polar form,

π π 4

3 Arg z π 4

3 3 3 z √2 Scos π j sin πT or z √2 j S πT 4 4 4

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3 arg z π k 2πn ( n k1, k2, … ) 4

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(b) All the possible arguments


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS (c) Graph in the complex plane

For

(a) The modulus r

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The principal argument



Expressing in polar form, (b) All the possible arguments

(c) Graph in the complex plane

and

(a) Find the product of form.

and the quotient

in rectangular form, without converting to polar

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Given

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Example 1.4


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS (b) Find the product z z and the quotient rectangular form Solutions:

D6

Z7

by converting first to polar form, then back to

(a) Product

Perform distribution

z z (2 j2)(j3) z z j6 j 6

Quotient

z z 6 j6 z 2 j2 z j3

Multiplying –j3 to both numerator and denominator, we have z 2 j2 j3 M z j3 j3

Since, j 1

Therefore

z j6 j26 z j 9 z 6 j6 z 9

z 2 2 j z 3 3

(b) Convert z and z into their polar form

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r fx y

r f(2) (2)

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Convert z first


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS r 2√2

y x 2 Arg z θ tanQ 2 Arg z θ tanQ

since all angles must be in counter clockwise sense, Arg z θ

Now, convert z

π

π 4

Arg z θ

3π 4

r fx y r f0 3 r 3

y x 3 Arg z θ tanQ S T 0 Arg z θ tanQ Arg z θ

p 2

Now that we have our values converted in polar form, let us get first the product of z and z z z r r 0cos(θ θ ) j sin(θ θ )2

3π π 3π π

T j sin S Tr 4 2 4 2 5π 5π

j sin T 4 4

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z z 6√2 Scos

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z z 2√2 (3) qcos S



z z 6√2 s

√2 √2

Ej Ft 2 2

z z 6 j6

Then, we can also get the quotient of z and z

z r 0cos(θ θ ) j sin(θ θ )2 z r

z 2√2 3π π 3π π qcos S T j sin S Tr z 3 4 2 4 2 z 2√2 √2 √2 E j F 3 2 2 z z 2 2 j 3 z 3

Example 1.5 Direction: Evaluate the following, expressing the answer in rectangular form (a) (3 j4)R Solution:

z R (3 j4)R z (3 j4)

First, get the magnitude and the principal argument r f3 4 r 5

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4 θ tanQ S T 3

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θ 0.93


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Using De Moivre’s Formula,

z a r a (cosnθ jsin nθ) r a jθ

We can now substitute the values that we computed z R 0 5 (cos 0.93 j sin 0.93 )2

z R 5R 0cos(3) 0.93 j sin (3)0.932 (b) (1 j) Solution:

z R 117 j43

z ( 1 j) z (1 j)

First, get the magnitude and the principal argument r f1 1 r √2

1 θ tanQ S T 1 Using De Moivre’s Formula,

θ

π 4

z a r a (cosnθ jsin nθ) r a jθ

z 0 u√2 v ( cos

π π

j sin ) 2 4 4

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z 64



Example 1.6 Find all the roots of the following in rectangular form and plot them (a) fj

(b) f5 j12 Solutions: (a) fj

Using De Moivre’s Formula

w √z √r Scos

w

r 1

θ

π 2

θ 2kπ T n n2

k 0, 1

If k 0 we can obtain the 1st root

π π

2π (0)

2π (0) 2 y 1st root √1 xcos

j sin 2 2 2 √2 √2 1st root 1 E j F 2 2 1st root

√2 √2 j 2 2

If k 1 we can obtain the 2nd root

π π

2π (1)

2π (1) 2 y 2nd root √1 xcos

j sin 2 2 2

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√2 √2

j ) 2 2

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2nd root 1 (



Graph of the Roots:

(b)

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If k 0 we can obtain 1st root


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS If k 1 we can obtain the 2nd root

Graphs of the roots

Drill Problems 1.2 Direction: For items 1 to 8, represent the following numbers in polar form. Show the details of your work. We must first obtain the magnitude and the principal argument

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Solution:

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1.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS r 3√2

y θ tanQ A C x

θ tanQ S θ

3 T 3

π 4

Since the formula to convert to polar form is

z r(cos θ j sin θ) rjθ

We can substitute the values that we have in the equation

2.) j2, j2

z 3√2 zcos A

π π π C j sin A C{ or z 3√2 j 4 4 4

Solution: We must first obtain the magnitude and the principal argument r f(0) (2) r √4 r2

y θ tanQ A C x θ

π 2

Since the formula to convert to polar form is

z r (cos θ j sin θ) rjθ Final answers are shaded in green.

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2 θ tanQ S T 0

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(a) j2

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS We can substitute the values that we have in the equation π π π z 2 zcos A C j sin A C{ or z 2 j 2 2 2

3.) 5

Solution:

We must first obtain the magnitude and the principal argument r f(5) 0 r √25 r5

y θ tanQ A C x 0 θ tanQ S T 5 θ π

Since the formula to convert to polar form is z r (cos θ j sin θ) rjθ

We can substitute the values that we have in the equation z 5(cos 0 j sin 0) or 5 jπ

jVπ

Solution:

We must first obtain the magnitude and the principal argument 1 1 | r S T S πT 2 4

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r 0.93

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4.)



θ tanQ

θ tan

Q

y x

1 4π 1 2

θ 1.004

Since the formula to convert to polar form is z r(cos θ j sin θ) rjθ

We can substitute the values that we have in the equation z 0.93(cos 1.004 j sin 1.004) or z 0.93 j1.004

R√PY

7 Z

Q√QYA C

We must first obtain the magnitude and the principal argument For the numerator,

3√2 j2

r }u3√2v (2)

r √22

θ tanQ

θ tan1

y x

2

3√2

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Solution:

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5.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS θ 0.44 For the denominator,

2 √2 j S T 3

r |u√2v S

r

√22 3

θ tanQ

θ tanQ

2 T 3

y x

2 3

√2

θ 2.70

We can now perform the division in polar form, z r jθ θ z r

z √22 j0.44 2.70 z √22 3 z 3 j 2.26 z

Converting to its polar form, we can come up to

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35

z z 30cos(2.26) j sin(2.26)2or 3 j 2.26 z z


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS R√PY

7 Z

Q√QYA C


3√2 j2

r }u3√2v (2)

r √22

θ tanQ

θ tanQ

y x

2

3√2

θ 0.44

For the denominator,

2 √2 j S T 3

r |u√2v S

r

√22 3

θ tanQ

θ tan

Q

2 T 3

y x

2 3 √2

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Solution:

θ 2.70

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6.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Using the general equation for finding the equation for the division of complex numbers in polar form z r jθ θ z r

Therefore

YR

Solution:


6 j5

r f(6) (5) r √61

θ tanQ

θ tan1

y x

5 6

θ 0.69 π θ 2.45

For the denominator,

r f(0) (3) r √9 r3

θ tanQ

y x

37

Q~PY

z z 30cos(π) j sin(π)2 or 3 jπ z z

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7.)

z √22 j0.44 2.70 z √22 3


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS θ tan1 θ

π 2

3 0

Using the general equation for finding the equation for the division of complex numbers in polar form z r jθ θ z r

z √61 π j2.45 3 z 2

z √61 z √61 0cos (0.87) j sin(0.87)2 or j0.87 z 3 z 3 PYR PYV


2 j3

r f(2) (3) r √4 9 r √13

θ tanQ

θ tanQ

θ 0.98

y x

3 2

38

Solution:

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8.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For the denominator,

5 j4

r f(5) (4) r √25 16 r √41

θ tanQ

θ tan1

y x

θ 0.67

4 5

Using the general equation for finding the equation for the division of complex numbers in polar form z r jθ θ z r

z √13 j0.98 0.67 z √41

z z √533 0cos(0.31) j sin(0.31)2 or j0.31 z z 41

For items 9 through 11, find all the roots of the given expression in rectangular form. Plot the roots in the complex plane. Let z1,

We must first find the values for r and θ, r fx y

r f(1) 0 r1

Arg z θ tanQ

y x

Arg z θ tanQ Q Arg z θ π


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c

39

9.) √1

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS r1

θπ

n4,

Using De Moivre’s formula,

w √z √r Scos

w

We can now find the first root, if k0,

k 0, 1, 2, & 3

θ 2πk θ 2πk

j sin T n n

1st root √1 Ecos c

π 2π(0) π 2π(0)

j sin F 4 4

1st root Acos 1st root

For the second root, if k 1,

2nd root √1 Ecos c

π π

j sin C 4 4

√2 √2

j 2 2

π 2π(1) π 2π(1)

j sin F 4 4

2nd root Scos

2nd root

3π 3π

j sin T 4 4 √2 √2

j 2 2

For the third root, c

π 2π(2) π 2π(2)

j sin F 4 4

3rd root Scos

5π 5π

j sin T 4 4

3rd root

√2 √2 j 2 2

40

3rd root √1 Ecos

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if k 2,


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For the fourth root, if k 3,

Graph in the complex plane

Let

41

We must first find the values for r and ,

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10.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS r5

r5

n4,


w √z √r Scos

w

We can now find the first root, if k 0,

1st root √5 Ecos Z

Arg z θ 0.927295218

θ 0.927295218

k 0, 1, 2, & 3

θ 2πk θ 2πk

j sin T n n

0.927295218 2π(0) 0.927295218 2π(0)

j sin F 3 3

1st root 1.628937 j0.5201745


2nd root √5 Ecos Z

3rd root √5 Ecos Z

0.927295218 2π(2) 0.927295218 2π(2)

j sin F 3 3

3rd root 0.36398 j1.670788

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if k 2,

2nd root 1.26495 j1.1506

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For the third root,

0.927295218 2π(1) 0.927295218 2π(1)

j sin F 3 3


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Graph in the complex plane

We must first find the values for r and ,


43

Let

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11.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS We can now solve for the first root, if k 0

1st root √1 Ecos

0 2π(0) 0 2π(0)

j sin F 8 8

1st root (cos 0 j sin 0) 1st root 1


2nd root √1 Ecos

0 2π(1) 0 2π(1)

j sin F 8 8

π π 2nd root Acos j sin C 4 4

2nd root

√2 √2

j 2 2

For the third root, if k 2,

3rd root √1 Ecos

0 2π(2) 0 2π(2)

j sin F 8 8

3rd root Acos

0 2π(3) 0 2π(3)

j sin F 8 8


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4th root √1 Ecos

44

3rd root j

For the fourth root, if k 3,

π π

j sin C 2 2

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 4th root Scos

For the fifth root, if k 4,

4th root

5th root √1 Ecos

3π 3π

j sin T 4 4

√2 √2

j 2 2

0 2π(4) 0 2π(4)

j sin F 8 8

5th root (cos π j sin π) 5th root 1

For the sixth root, if k 5,

6th root √1 Ecos

0 2π(5) 0 2π(5)

j sin F 8 8

6th root Scos 6th root

5π 5π

j sin T 4 4

√2 √2 j 2 2

For the seventh root,

0 2π(6) 0 2π(6)

j sin F 8 8

7th root Scos

3π 3π

j sin T 2 2

7th root j

45

7th root √1 Ecos

Page

if k 6,


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For the eighth root, if k 7,

Graph in complex plane

the

For items 12 & 13, evaluate the following, expressing the final answers in rectangular form. We must first find the values for r and ,

46

Let

Page

12.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS r f9 9

Arg z θ tanQ

r 9√2

Arg z θ


π 4

9 9

z a r a (cos nθ j sin nθ) r a jnθ

R π π (9 j9)R u9√2v zcos 3 A C j sin 3 A C{ 4 4 R

(9 j9)R u9√2v E

(9 j9)R 729 (2)

13.) (2 j6)

√2 √2

j F 2 2

√2 R E 2

b

(9 j9)R 1458 j1458

√2 F 2

Let z 2 j6

We must first find the values for r and θ, r fx y

r f(2) 6 r 2√10


Arg z θ tanQ

y x

6 2 Arg z θ 1.249045772 Arg z θ tanQ

z a r a (cos nθ j sin nθ) r a jnθ

(2 j6) u2√10v 0cos 2(1.249045772) j sin 2(1.249045772)2 Page

47

(2 j6) 32 j24


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For items 14 & 15, prove the following trigonometric identities using De Moivre’s formula. 14.) cos 2θ cos θ sin θ 15.) sin 2θ 2 cos θ sin θ

Solution for numbers 14 and 15

Using the De Moivre’s formula, (cos θ j sin θ)a cos nθ j sin nθ

where in n 2

(cos θ j sin θ) cos 2θ j sin 2θ

cos θ j2 cos θ sin θ j sin θ cos 2θ j sin 2θ

(cos θ sin θ) j2 cos θ sin θ cos 2θ j sin 2θ Since in trigonometry, cosine is located at the x-axis where your x-axis (real axis) is considering only the real part of the equation, we can already say that: cos 2θ cos θ sin θ

Also in trigonometry, sine is located at the y-axis where it is called imaginary axis because it only considering the imaginary part of the equation, therefore we can say that: sin 2θ 2 cos θ sin θ

Example 1.7 Evaluate the following expressions, expressing answers in rectangular form. a.) cos(1 j)

Solution:

cos(1 j) cos 1 cos j sin 1 sin j


Page

cos(A B) cos A cos B sin A sin B

48

Using the trigonometric identity of sum of two angles of cosine,

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Since,

cos j cosh 1 sin j sinh j

cos (1 j) cos 1 cosh 1 j sin 1 sinh 1 cos (1 j) 0.8337 j 0.9889

b.) sinh(4 j3)

Solution:

Using the trigonometric identity of difference of two angles of hyperbolic sine, sinh (A B) sinh A cosh B cosh A sinh B

sinh(4 j3) sinh 4 cosh(j3) cosh 4 sinh(j3) cos j cosh 1

Since,

sin j sinh j

sinh(4 j3) sinh(4) cos(3) j cosh(4) sin(3) sinh(4 j3) 27.0168 j 3.8537

Example 1.8 Evaluate the following logarithms, expressing the answers in rectangular form.

ln 1

Since it doesn’t consider the principal argument, we will use the equation Final answers are shaded in green.

Page

Solution:

49

a.) ln 1, Ln 1


Let z 1

ln z ln r jθ k j2πn

r fx y

r

f1 0

r1

Ln 1

Arg z θ 0

ln 1 ln 1 j(0) k j2πn

ln 1 0 kj2πn ; n 0,1,2, …

Since it consider the principal argument, we will use the equation Ln z ln|z| j Arg z

Let z 1

r fx y

r f1 0

Ln 1 ln 1 j(0)

50

Ln 1 ln 1 j(0)

Page

r1

Arg z θ 0


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS b.) ln(3 j4) , Ln (3 j4) Solution:

ln(3 j4)

Since it doesn’t consider the principal argument, we will use the equation Let z 3 j4


y x V Q Arg z θ tan R

r fx y

Arg z θ tanQ

r f3 (4) r5

Arg z θ 0.927295

ln(3 j4) ln 5 j(0.927295) k j2πn

ln(3 j4) 1.609 j 0.927 k j2πn ; n 0,1,2, …

Ln (3 j4)

Since it consider the principal argument, we will use the equation Ln z ln|z| j Arg z

Let z 3 j4

r fx y

r f3 (4) r5

Arg z θ tanQ 4 Arg z θ tanQ 3

Arg z θ 0.927295

Ln(3 j4) ln 5 j(0.927295) Ln(3 j4) 1.609 j 0.927

a.) jY


Page

Evaluate the following, expressing the answers in rectangular form.

51

Example 1.9


The general power of a complex number z x jy are defined by the formula, z e a D Let

zj

and

cj

We must first solve for the value of r and θ

y x 1 θ Arg z tanQ 0 π θ 2

r fx y

θ Arg z tanQ

r f0 1 Then solve for ln z

Therefore,

ln z ln r jθ k j2πn π ln j ln 1 j A C k j2πn 2 π ln j j A C k j2πn 2

z e a D

YAYA CkYaC

z e

7 ACkY7 a

z eY

jY eQ a

Using the general power for a complex number, a 5 r fx y

r f1 1 r √2

and

c 2j

θ Arg z tanQ

θ Arg z tanQ θ

π 4

1 1


52

Let z 1 j

Page

b.) (1 j)(QY)

r1

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Solve for the value of ln z,


Then,

π ln(1 j) ln √2 j A C k j2πn 4 z e a D

(1 j)(QY) e(QY) a √PYA V CkYa

7 Y7 aC

(1 j)(QY) eA a √PY PYVaQY a √QY Since

(1 j)(QY)

V eA a √P V kaC eYA C eQY a √

eY cos θ j sin θ

eQY cos θ j sin θ

π π (1 j)(QY) 2e V ka Acos j sin C ucos ln √2 j sin ln √2v 2 2

(1 j)(QY) 2e V ka jucos ln √2 j sin ln √2v

(1 j)(QY) 2e V ka j cos ln √2 j sin ln √2

(1 j)(QY) 2e V ka sinuln √2v j cosuln √2v

Page

53

1 1 (1 j)(QY) 2e V ka sin S ln 2T j cos S ln 2T 2 2



Drill Problem 1.3 For number 1 through 4, find the principal value of ln z, rectangular form, when z equals, 1.) -10

Solution:

|z| r fx y

|z| r f(10 ) (0) |z| r 10

Using the formula:

θ Arg z tanQ

θ Arg z tanQ θπ

0 10

ln z ln|z| j Arg z

ln(10) ln 10 jπ 2.) 2 j2

|z| r f2 2 |z| r 2√2

Using the formula:

y x 2 θ Arg z tanQ 2 θ Arg z tanQ θ

ln z ln|z| j Arg z

ln(2 j2) ln 2√2 j

1 π ln(2 j2) ln 8 j 2 4

π 4

π 4

54

|z| r fx y

Page

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 3.) 2 j2

Solution:

|z| r fx y

|z| r f2 (2)

|z| r 2√2

θ Arg z tanQ

θ Arg z tanQ θ

π 4

y x

2 2

Using the formula: ln z ln|z| j Argz

ln(2 j2) ln 2√2 j

π 4

1 π ln(2 j2) ln 8 j 2 4

|z| r fx y

|z| r f0 (e) | z| r e

Using the formula:

θ Arg z tanQ

y x

θ Arg z tanQ θ

π 2

e 0

ln z ln|z| j Arg z ln(je) ln e j

π 2

55

Solution:

Page

4.) – je



ln(je) 1 j

π 2

For number 5 through 8, evaluate the following answers in rectangular form. 5.) ln e

Solution:

Let z e

Then, find the values for r and θ r fx y

θ Arg z tanQ

r fe 0

θ Arg z tanQ

re

θ0

y x

0 e

Since it doesn’t consider the principal value, we will use the formula ln z ln r jθ k j2πn

ln e ln e j0 k j2πn

6.) ln eQY

ln(e) 1 k j2πn ; n (0,1,2, … )

Solution:

Considering the value of z e and c j, we use the formula

r fx y

r fe 0

θ Arg z tanQ

y x

θ Arg z tanQ

0 1


56

Solve for the values of r and θ

Page

Let z e

z e a D

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS re

θπ

Since it doesn’t consider the principal value, we will use the formula, ln z ln r jθ k j2πn

ln(e) ln e 1 jπ k j2πn z e a D

eQY eQY(PYkYa)

ln eQY ln eQY(PYkYa)

ln eQY j(1 jπ k j2πn)

ln eQY (j j π) kj2πn ln eQY (j π k j2πn)

ln eQY (π 1 k 2πn)j ; n (0,1,2, … )

7.) ln(4 j3)

Solution:

Since it doesn’t consider the principal value, we will use the formula ln z ln r jθ k j2πn

Let z 4 j3, find the values for r and θ

y x 3 θ Arg z tanQ 4

r fx y

θ Arg z tanQ

r5

θ 0.6435011

r f4 3


Page

ln(4 j3) 1.6094 j0.6435 k j2πn ; n (0,1,2, … )

57

ln(4 j3) ln 5 j0.6435011 k j2πn

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 8.) ln eYR

Solution:

Considering the value of z e and c j3, we use the formula z e a D

Let us solve for the value of r and θ r fx y

r fe 0 re

y x 0 θ Arg z tanQ e θ Arg z tanQ


θ0

ln e ln e j(0) k j2πn ln e 1 k j2πn

eQYR eYR(kYa)

ln eQYR ln eYR(kYa)

ln eQYR j3 k j2πn ; n (0,1,2, … )

For numbers 9 through 12, find the principal value in rectangular form. 9.) jY , j2Y

jY Considering the value of z j and c j2, we use the formula

58

z e a D

Page

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Let us solve for the value of r and θ r fx y

r f0 1

θ Arg z tanQ

θ Arg z tanQ

r1

θ

Using the general equation,

π 2

y x

1 0

ln z ln r j Argz ln j ln 1 j ln j j

π 2

π 2

jY eYAY C

jY eY jY eQ

j2Y

Considering the value of z j2 & c j, we use the formula z e a D

r f0 2 r2

θ Arg z tanQ

θ Arg z tanQ θ

π 2

2 0

y x


59

r fx y

Page

Let us solve for the value of r and θ

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Using the general equation,

ln z ln r jθ

ln j2 ln 2 j j2Y eY a PY

7 7

π 2

j2Y eYAa PY C

using logarithmic exponent rule eP e e

j2Y eY a eQ

since, eY cos θ j sin θ, therefore

j2Y eQ (cos ln 2 j sin ln 2)

10.) 4(RPY)

Solution:

Considering the value of z 4 & c (3 j), we use the formula z e a D

Let us solve for the value of r and θ r fx y

r f4 0

θ Arg z tanQ

y x

θ Arg z tanQ

r4

θ0

0 4

Using the general equation for natural logarithm of a complex number, ln z ln|z| jθ

ln(4) ln 4 j(0)


Page

4(RPY) e(RPY) a V using the logarithmic exponent rule eP e e

60

ln(4) ln 4

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 4(RPY) eR a V eY a V Since eY cos θ j sin θ

eY 4R (cos ln 4 j sin ln 4)

z 64 0cos(ln 4) j sin(ln 4)2 11.) (1 j)(PY)

Solution:

Considering the value of z 1 j and c 1 j, we use the formula r fx y

z e a D

r f1 (1)

θ Arg z tanQ

y x

θ Arg z tanQ

r √2

θ

Using the formula for natural logarithm of z,

π 4

1 1

ln z ln r jθ

ln z ln √2 j

π 4

(1 j)(PY) e(PY)Aa √QY V C

7 C

Using the logarithmic exponent rule, eP e e


61

V

Page

(1 j)(PY) eAa √QY V PY a √QY


(1 j)(PY) eAa √Q V C eYAa √Q V C

Since eY cos θ j sin θ

π π eY ea √ eQ V zcos Aln √2 C j sin Aln √2 C{ 4 4

12.) (1)(QY)

π π z √2eQ V zcos Aln √2 C j sin Aln √2 C{ 4 4

Solution:

Considering the value of z 1 and c (1 j2), we use the formula z e a D

r fx y

θ Arg z tanQ

r f(1) 0

y x

θ Arg z tanQ

r1

θπ

Using the general formula of natural logarithm of z,

0 1

ln z ln r jθ

ln z ln 1 jπ ln z jπ

(1)(QY) e(QY)(Y)

(1)(QY) eY eQY

7

(1)(QY) eY e

Page

62

since eY cos θ j sin θ


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS eY e (cos 2π j sin 2π) z e

For numbers 13 through 15, solve for z in rectangular form 13.)

ln z A2 j C π

Solution:

To eliminate the natural logarithm, we need to equate on both sides, AQY C

ea D e

AQY C

ze

since, eY cos θ j sin θ

z e eQY

π π z e Acos j sin C 2 2

ln z 0.3 j0.7 Solution:

To eliminate the natural logarithm, we need to equate on both sides, ea D e(.RPY.) z e(.RPY.)

63

z e.R eY.

Page

14.)

z e


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS since, eY cos θ j sin θ

ln z e jπ Solution:

z 1.032 j0.870

To eliminate the natural logarithm, we need to equate on both sides, ea D e(QY) z e eQY

since, eY cos θ j sin θ

z e (cos π j sin π) z e

64

z 15.154

Page

15.)

z e.R (cos 0.7 j sin 0.7)



UNIT 2

Page

65

LAPLACE AND INVERSE LAPLACE TRANSFORM



Example 2.1 Direction: Using the Laplace integral, find the Laplace transform of the following: a.) f(t) 1

Solution:

Using the general equation of the function defined for s ,

F(s) eQ f(t)dt

T

F(s) lim eQ f(t)dt T

T

F(s) lim eQ (1)dt T

T

F(s) lim eQ dt T

T 1 F(s) lim 0eQ 12 T 0 s 1 F(s) lim 0eQT 12 T s 1 F(s) 0eQ() 12 s

1 s

Solution:

using the general equation of the function defined for s ,


66

b.) f(t) e

F(s)

Page

Therefore,

1 F(s) 00 12 s


F(s) eQ f(t)dt

T

F(s) lim eQ f(t)dt T

T

F(s) lim eQ (e )dt T

T

F(s) lim eQP dt T

T

F(s) lim eQ(Q) dt T

1 Q(Q) T e 1r T sa 0 T 1 lim \eQ(Q) 1^ F(s) 0 s a T F(s) lim q F(s)

1 lim \eQT(Q) 1^ s a T

F(s)

1 \eQ(Q) 1^ sa

F(s)

1 00 12 sa

F(s)

1 sa

c.) f(t) t

Solution:


67

F(s) eQ f(t)dt T

F(s) lim eQ f(t)dt

Page

T


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS T

F(s) lim eQ (t)dt T

Integrating by parts we let:

Q¡¢ ()

1 1 T F(s) lim qt S eQ T eQ dtr T s s 0 1 T t F(s) lim q eQ (eQ )dtr T s 0 s t 1 1 T F(s) lim q eQ S eQ Tr T s s s 0 T t 1 F(s) lim q eQ (eQ )r T s 0 s t T 1 F(s) lim q eQ (eQ )r T s s 0

1 T 1 F(s) lim q eQT ueQT v r T s s s

T 1 1 F(s) lim q eQT r lim q ueQT vr lim q r T T s T s s F(s)

1 1 lim 0eQT 2 0 s T s

1 1 F(s) 0eQ() 2 s s F(s)

1 s

1 s

68

F(s) 0

Page

Then,

1 Q¡¢ £


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS d.) f(t) cos ωt , ω is a constant Solution:


F(s) eQ f(t)dt

T

F(s) lim eQ f(t)dt T T

F(s) lim eQ (cos ωt)dt

Integrating by parts we let

T

¥¦£ §

Q¡¢ 1 Q¡¢ £

£¨© §

Then,

F(s) S

cos ωt Q 1 e T S eQ T ω sin ωt dt s s

F(s)

1 Q T ω T e cos ωt 2 eQ sin ωt dt s 0 s

Integrating by parts for ª eQ sin ωt dt, we let: T

£¨© §

§ ¥¦£ §

Then,

Q¡¢

1 Q¡¢ £

e

Q

1 Q T ω T Q cos ωt dt e sin ωt 2 e cos ωt dt s 0 s


Page

T

69

T T 1 T 1 eQ cos ωt dt eQ sin ωt2 ( eQ ) ω cos ωt dt s 0 s

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS T

eQ cos ωt dt T

eQ cos ωt dt

T

eQ cos ωt dt T

e

Q

«sE1

1 Q T ω T ω T e cos ωt 2 0eQ sin ωt 2 eQ cos ωt dt s 0 s 0 s

ω T Q T ω 1 T cos ωt dt e cos ωt dt eQ cos ωt 2 0eQ sin ωt 2 s s 0 s 0

T ω 1 Q T ω Q T Q 0e t q r¬ F e cos ωt dt e cos ωt

sin ωt2 s 0 s 0 s T

e

Q

ω 1 T s eQ cos ωt 2 T s 0eQ sin ωt 2

cos ωt dt ω ω S1 T S1 T s s

1 ω S1 T s

Then, applying the limits,

ω 1 s eQT cos ωT s 0eQT sin ωT 2 F(s)

ω ω S1 T S1 T s s

Since

ω QT s e sin ωT s eQT cos ωT s F(s)

s ω s ω

cos ∞ 1

sin ∞ 0

eQ 0

Applying the limits: lim F(s)

T

s ω eQ sin ω(∞) Q e cos ω(∞)

s ω s ω F(s)

s

s

ω

70

1 Q T ω T ω T e cos ωt 2 0eQ sin ωt 2 eQ cos ωt dt 0 s s 0 s

Page

1 Q T ω 1 T ω T e cos ωt 2 0 eQ sin ωt 2 eQ cos ωt dt s 0 s s 0 s


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS e.) f(t) sin ωt , ω is a constant Solution:

Using the general equation of the function defined for s

F(s) eQ f(t)dt

T

F(s) lim eQ f(t)dt T T

F(s) lim eQ (sin ωt)dt

Integrating by parts we let:

T

£¨© §

¥¦£ § –

eQ sin ωt dt

1 Q T ω T e sin ωt 2 eQ cos ωt dt s 0 s

Integrating by parts for ª eQ cos ωt dt, we let: T

¥¦£ §

T

T

eQ cos ωt dt

eQ sin ωt dt

T

eQ sin ωt dt

E1

du § £¨© §

Q¡¢ ¡ Q¡¢

1 Q T ω T Q e cos ωt 2 e sin ωt dt s 0 s

1 Q T ω 1 T ω T e sin ωt 2 q eQ cos ωt r eQ sin ωt dt 0 s 0 s s s

1 Q T ω 1 T ω T e sin ωt 2 q eQ cos ωt r eQ sin ωt dt 0 s s 0 s s

T ω 1 Q T ω Q T ω T Q Q 0e 2 F e sin ωt dt e sin ωt 2 cos ωt e sin ωt dt s s 0 s 0 s


71

T

1 Q¡¢ £

Page

Then,

Q¡¢


T

eQ sin ωt dt S

1 Q T ω T 1 e sin ωt2 0eQ cos ωt 2 T ( ) s 0 s ω 0 s s

ωs Q sin ωt 2 T 0e 1 s Q T s Q e sin ωt dt ( ) E e sin ωt F 2 s s ω s ω 0 T

Applying the limits: F(s)

Since

seQT sin ωT ω eQT cos ωT s ω s ω

ω QT s e sin ωT s eQT cos ωT s F(s)

s ω s ω

cos ∞ 1

F(s)

sin ∞ 0

s

ω

ω

eQ 0

Example 2.2 Direction: Using linearity theorem and the previously obtained Laplace transform pairs, find the Laplace transform pairs, find the Laplace transform of a.) cosh at

Solution:

Since

cosh at

e eQ 2

Page

72

We can now use the linearity theorem, e eQ ®0cosh at2 ® 0 2 2 Final answers are shaded in green.


Since

1 ®0cosh at2 ®0e eQ 2 2 ®0e 2

1 sa

® 0eQ 2

and

1 1 1 r ® 0cosh at2 q

2 sa s a

1 s a

1 s a sa r ® 0cosh at2 q s a 2 1 2s r ® 0cosh at2 q 2 s a

b.) sinh at

F(s)

s

s a

Solution:

Since

sinh at

e eQ 2

We can now use the Linearity theorem

e eQ ® 0sinh at2 ® 0 2 2

®0e 2

1 sa

1 ®0e eQ 2 2

and

® 0eQ 2

1 1 1 r ® 0sinh at2 q 2 sa s a

1 s a

73

1 s as a r ® 0sinh at2 q 2 s a

Page

Since

® 0sinh at2


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2a 1 r ® 0sinh at2 q 2 s a F(s)

c.) cos ωt

Solution:

Since

cos ωt

a s a

eY eQY 2

We can now apply the Linearity theorem

eY¯ eQY¯ ® 0cos ωt2 ® 0 2 2

®\eY¯ ^

1 s jω

and

® \eQY¯ ^

1 s jω

1 1 1 r

® 0cos ωt2 q 2 s jω s jω 1 s jω s jω r ® 0cos ωt2 q s j ω 2 1 2s r ® 0cos ωt2 q 2 s ω s

s

ω

74

F(s)

Page

Since

1 ® 0cos ωt2 ®0eY¯ eQY¯ 2 2


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS d.) sin ωt

Solution:

Since

eY eQY j2

sin ωt

Using the Linearity theorem

eY eQY ® 0sin ωt2 ® 0 2 j2

Since

® 0sin ωt2 ®\eY¯ ^

1 s jω

® 0sin ωt2 ® 0sin ωt2

1 ®0eY eQY 2 j2 ® \eQY¯ ^

and

1 1 1 q r

j2 s jω s jω

1 s jω

1 s jω s jω q r j2 s j ω

® 0sin ωt2 F(s)

2jω 1 q r j2 s ω

s

ω

ω

Example 2.3 Direction: Apply the shifting theorem and the previously obtained Laplace transform of the following:

® 0cos ωt2 °

s2

s

ω2

±

. s sa

75

Solution:

Page

a.) e cos ωt



b.) e sin ωt Solution:

®0cos ωt2

s2

(s a)2 ω2

ω

® 0sin ωt2 °

®0sin ωt2

sa

ω2

±

. s sa

ω

(s a)2 ω2

Example 2.4 Direction: Find the Laplace transform of the following functions using the table (variables other than t are considered constant). a.) t 2t

Solution:

By linearity,

Since Then,

b.) cos πt

® 0t 2t2 ® 0t 2 2 ® 0t2 ® 0t 2

2! sR

and ® 0t2

t 2t

1 s

2! 1 R s s

® 0t 2t2 2 S

1 1 T R s s

76

Solution:

Page

Since Final answers are shaded in green.

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS ® 0cos πt2

Then,

s

s

ω

® 0cos πt2

s s π

® 0cosh t2

s s a

c.) e cosh t Solution:

Using shifting theorem,

s . ± s 1 s s 2

® 0e cosh t2 °

® 0e cosh t2

® 0e cosh t2

s

® 0e cosh t2

Solution:

Using logarithmic properties Then,

s2 2s 4 1

s

s2 2s 3

eP e e

eRQ eR ® 0eQ 2

Since eR is a constant, and ® 0eQ 2

P

77

d.) eRQ

s2 (s 2) 1

Page

Then,



®0eRQ 2

eR s 2b

e.) cos(ωt θ) Solution:

Using a trigonometric identity (sum of two angles of cosine) cos(α β) cos α cos β sin α sin β

So,

cos(ωt θ) cos ωt cos θ sin ωt sin θ

Since cos θ and sin θ are constant, and ® 0cos ωt2

Then,

s

s

ω

and

® 0cos(ωt θ)2

Simplifying further, we get

® 0sin ωt2

s

ω

ω

s cos θ ω sin θ s ω s ω

® 0cos(ωt θ)2

s cos θ ω sin θ s ω

Example 2.5 Direction: Find the inverse Laplace transform of the following functions using he table (variables other than s are constants). a.) ® Q z 7P7 { VQR

78

Solution:

Page

By Linearity theorem, Final answers are shaded in green.

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS ® Q q

Since

s

Then,

b.) ® Q z

ZQR7 P [

Solution:

{

4s 3π 4s 3π r ® Q q r ® Q q r s π s π s π

s ® 0cos ωt2

ω

and

s

ω ® 0sin ωt2

ω

4s 3π r 4 cos πt 3 sin πt ® Q q s π

Expanding the term,

7PVPµ

Solution:

n!

saP

Then

c.) ® Q z

sR 3s 12 £R 3£ 12 Q s t Q s t t ® Q q r ® ® s £ £ s

® Q s {

ta

sR 3s 12 3 1 t 1 t tV s 2 2

Completing the square of the denominator we’ll have 15 15 s 4s 29 (s 2) 25 e sin ωt

ω (s a) ω

Page

Since

79

Since

® Q s

sR 3s 12 £ R 3£ 12 £ £ £ s



Then Therefore,

15 15 Q 5 r r ® Q q ® q (s 2) 25 5 (s 2) 5 15 r 3eQ sin 5t ® Q q (s 2) 25

d.) ® Q z7PV { U

Solution:

Performing partial fraction expansion

8 A B q r s(s 4)

s s 4 s(s 4) 8 A(s 4) B(s) s: 0 A B s : 8 4A

We’ll have the value of A and B as,

A 2; B -2

8 2 2 r ® Q q r ® Q q r ® Q q s(s 4) s s 4

√vuP√v

Solution:

r

Performing partial fraction expansion s

1

us √2vus √5v

A

us √2v

B

us √5v

t us √2vus √5v Final answers are shaded in green.

80

e.) ® Q quQ

8 r 2 2eQV ® Q q s(s 4)

Page

Therefore,


1 A us √5v Bus √2v

By elimination method, Then, ®

Q s

A

s :

s: 0 A B

1 A√5 B√2 B

√Q √R

1

us √2vus √5v

Therefore,

® Q s

t ®

√5 √3 2 · s

Q ¶

1

us √2vus √5v

t

√Q √R

®

Q

¶

√5 √3 2 · s 4

e√R eQ√ √5 √3

Drill Problem 2.1 Direction: Find the Laplace transform of the following functions. Variables other than t are constants. Performing binomial expansion,

(t 3) t V 6t 9

Then, applying the distributive property of Laplace,

®(t 3) ®0t V 2 ®06t 2 ®092 Final answers are shaded in green.

81

®(t 3) ®(t V 6t 9)

Page

1.) (t 3)

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS ®(t 3)

4! (2!) 9 6 R

s s s

®(t 3)

2.) sin 4t

24 12 9 R

s s s

Using the trigonometric identity for double angle formula of sine,

1 sin A (1 cos 2A) 2

® (sin 4t)

® (sin 4t)

1 ( ® 012 ® 0cos 8t2 ) 2

® (sin 4t)

®(sin 4t)

1 1 s S T 2 s s 64

1 s 64 s E F 2 s (s 64)

®(sin 4t)

5 . ¸ s 25 s s 1

®(sinh5t) °

®(eQ sinh 5t) 4.) sin A3t C

32

64)

(s

5 (s 1) 25

Using trigonometric identity for the sum of two angles of sine,

sin(A k B) sin A cos B k cos A sin B


82

Using the shifting theorem,

s

3

Page

3.) eQ sinh 5t

1 ® (1 cos 8t) 2


1 1 1 sin S3t T sin 3t cos cos 3t sin 2 2 2

1 1 1 ® qsin S3t Tr cos ® 0sin3t2 sin ®0cos 3t2 2 2 2 1 3 cos 0.5 s sin 0.5 ® Ssin S3t TT 2 s 9

5.) 8 sin 0.2t

® (8 sin 0.2t ) 8 ® 0sin 0.2t2 since, sin ωt

s

ω

ω

® (8 sin 0.2t) 8 S ® (8 sin 0.2t)

s

s

0.2 T

0.04

1.6

0.04

4

6.) sin t cos t

Using the trigonometric identity for product formula of the angles sine and cosine,

1 1 sin(t t) sin(t t) 2 2

sin t cos t

1 1 sin 2t sin 0 2 2

®(sin t cos t)

®(sin t cos t)

1 ® 0sin 2t2 2

1 2 S T 2 s 4 Final answers are shaded in green.

83

sin t cos t

1 1 sin(A B) sin(A B) 2 2

Page

sin A cos B


7.) (t 1)R

®(sin t cos t)

1 s 4

(t 1)R t R 3t 3t 1

By applying the linearity theorem,

®(t 1)R ® 0t R 2 3®0t 2 ® 012 ®(t 1)R

9.) 3t V eQ.

By shifting theorem,

10.) 5eQ sin ωt


3.8 . ® (3.8t) ° ¸ s (s 2.4) s ®(3.8te.V )

3.8 (s 2.4)

3(4!) . ®(3t V ) ° ¸ s s (s 0.5) ®(3t V eQ. )

®(5 sin ωt) °

s

72 (s 0.5)

5ω . ¸

ω s (s a)


84


6 6 2 1

sV sR s s

®(t 1)R

Page

8.) 3.8te.V

3! (2!) 2 1

3 R

V s s s s


®(5eQ sin ωt)

5ω (s a) ω

II. Direction: Find the Inverse Laplace transform of the following functions. Variables other than s are constant.

P

1 1 1 1 r ® Q q r ® Q q r ® Q q s 5 s 5 s 5 s 5

Multiplying

12.)

P~ 7Q~

® Q q

Since,

® Q z

13.)

¡P√

√

√

in 7 P in order to satisfy ®(sin ωt)

¯

7 P¯7

1 1 1 r ® Q q sin √5 t eQ s 5 s 5 √5

2s 16 2s 16

s 16 s 16 s 16

25 16 2s 16 r ® Q q r ® Q q r s 16 s 16 s 16

s

s { cosh ωt and ω

z

s

ω { sinh ωt ω

2s 16 r 2 cosh 4t sinh 4t ® Q q s 16

10

2£ √2

10 ¹ 2

1

√2 £ 2

º

85

Page

11.) 7 P



By Linearity theorem, ® Q q

10

2£ √2 ® Q q

14.)

aL

L77 Pa77

10

1 10 Q · ® ¶ 2 √2 £

2

2s √2

nπL nπ ¹ L s n π L

®

r

nπ 1 L º π º ¹ n π n s s L L

nπL r ® Q ¹ L s n π

Q q

√2

r 5e 2 t

nπ L º n π s L

nπL nπ r sin t ® Q q L s n π L

By Partial fraction, q

20 A B r (s 1)(s 3)

(s 1)(s 3) (s 1) (s 3) 20 A(s 3) B(s 1) if s ` 0 A B

s : 20 3A B

By elimination method we will have the value for A 10 and B-10 20 10 10 r ® Q q r ® Q q r ® Q q (s 1)(s 3) s 1 s 3


86

(P)(PR)

Page

15.)

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 20 r 10eQ 10eQR ® Q q (s 1)(s 3)

20 r 10 (eQ eQR ) ® Q q (s 1)(s 3)

16.)

UQ µ7 Q

18s 12 18s 12 9s 1 9s 1 9s 1

18s 12 18 Q s 12 Q 1 r ¬ ¬ ® Q q ® « ® « 9s 1 9 9 s 19 s 19 ® Q q

18s 12 1 1 r 2 cosh t 4 sinh t ® Q q 9s 1 3 3

(P)(P)

By partial fraction expansion,

1 A B r (s a)(s b)

(s a)(s b) (s a) (s b)

Coefficient of

1 A(s b) B(s a) s: 0 A B

s : 1 bA aB

Performing elimination method, we get 1 1 B and A ab ab Final answers are shaded in green.

87

q

Page

17.)

1 s 18s 12 3 ¬ Q « Q « r ¬ 2 ® 4 ® 1 9s 1 s 9 s 19


®

1 1 a b 1 a by Q Q q y ® x r ® x (s a)(s b) s a s b 1 1 r ® Q q (eQ eQ ) (s a)(s b) ba

VQ

7 Q~PU

Performing completing the square in the denominator, we will arrive at: s

4s 2 4s 2 s 6s 9 18 9 6s 18 4s 2 4s 2 (s 3) 9 s 6s 18

In order to satisfy the form of the numerator to obtain its inverse Laplace transform, we must get the value of x. We will able to form the equation, 4s 2 4(s 3) x 4s 2 4s 12 x x 2 12 x 10

4s 2 4 (s 3) 10 r ® Q q r ® Q q r ® Q q s 6s 18 (s 3) 9 (s 3) 9

sa r e cosh ωt (s a) ω Therefore, ® Q q

and ® Q q

ω r e sinh ωt (s a) ω

4s 2 10 R r 4eR cosh 3t

® Q q e sinh 3t s 6s 18 3


88

Since,

Page

18.)

Q


7 PPV7

Performing completing the square in the denominator of the equation, s

π π

10πs 24π s 10πs 25π 24π 25π s

π π (

10πs 24π s 5π) π

π π r ® Q q (s 5π) π

10πs 24π s

Q~

7 QVQ

π e sinh πt

10πs 24π

Performing completing the square in the denominator of the equation, s

2s 56 2s 56 4s 12 s 4s 4 12 4 s

2s 56 2s 56 4s 12 (s 2) 16

in order to satisfy the form of the numerator to obtain its inverse Laplace transform we must get the value of x. We will be able to form the equation: 2s 56 2(s 2) x 2s 56 2s 4 x x 4 56 x 52

2s 56 2 (s 2) 4 Q q r ® Q q r r ® Q q 13 ® (s 2) 16 s 4s 12 (s 2) 16 2s 56 r 2e cosh 4t 13e sinh 4t ® Q q s 4s 12

2s 56 r e (2 cosh 4t 13e sinh 4t) ® Q q s 4s 12


89

20.)

s

Page

19.)


Example 2.6 Direction: Find the Laplace transform of the following using the differentiation property.

a.) te¼

Solution: When

f(t) te¼ F(s) f(0) 0

f ½ (t) kte¼ e¼

f ½ (t) k f(t) e¼

®0f(t)2 k f(s)

1 sk

DP: ®0f(t)2 s f(s) f(0)

k f(s)

1 s f(s) f(0) sk

k f(s)

1 s f(s) sk

(s k)f(s)

f(t) t sin ωt F(s) f(0) 0

f(t) ωt cos ωt sin ωt

90

Solution:

1 (s k)

f ½ (0) 0

Page

b.) t sin ωt

f(t)

1 sk



f ½½ (t) ω t sin ωt ω cos ωt ω cos ωt f ½½ (t) ω t sin ωt 2ω cos ωt ®0f ½½ (t)2 ω F(s)

2ωs

ω

s

DP: ®0f(t)2 s F(s) s f(0) f ½ (0) ω F(s)

2ωs s F(s)

ω

s

(s ω )F(s)

f(t)

f(t) sin ωt F(s) f(0) 0

f ½ (t) 2ω0ω sin ωt ω cos ωt2 f ½½ (t) 2ω sin ωt 2ω cos ωt f¾¾(0) 0

®0f ½½ (t)2 2ω F(s) 2ω cos ωt f¾¾(t) 0

s 1 1 ®0f ½½ (t)2 2ω F(s) 2ω S T 2 s s ω¿ DP: ®0f ½½ (t)2 s F(s) s f(0) f ½ (0) Final answers are shaded in green.

91

Solution:

2ωs

ω )

Page

c.) sin ωt

(s

2ωs

ω

s

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS ®0f ½½ (t)2 s F(s) 0 0

1 s 2ω F(s) ω S T s f(s) s s 4ω¿ 1 s (s 2ω )F(s) ω S T s s 4ω¿ F(s)

1 ω A s

s C s 4ω¿ (s 2ω )

ω us 4ω¿ v ω s s(s 4ω¿ ) F(s) (s 2ω ) F(s)

F(s)

ω s 4ω ω s sR 6ω sR 8ωV s

2ω (s 2ω ) (sR 4ω s)(s 2ω )

F(s)

2ω (sR 4ω s)

F(s)

2ω2 s(s2 4ω2 )

Example 2.7 Direction: Find the inverse Laplace transform of the following using the integration property.

(7 P ¯7)

F(s)

1 G(s) ( s s ω )

ω 1 ·

ω ω G(s) F(s) s

G(s)

s

92

Solution:

Page

a.)



f(t) g(τ)dτ g(τ)

1 sin ωτ ω

f(t) g(τ)dτ

f(t)

1 sin ωτ dτ ω

1 ° f(t) cos ωτr ω

f(t)

1 (cos ωt 1) ω

f(t)

7 (7 P ¯7 )

F(s)

F(s)

s (s

1

ω )

1 s · s(s ω )

G(s)

s(s

F(s)

1

ω )

G(s) s

f(t) g(τ)dτ

g(τ)

1 (1 cos ωt) ω

f(t) g(τ)dτ

f(t)

1 (1 cos ωt)dτ ω

93

Solution:

Page

b.)

1 (1 cos ωt) ω



f(t)

f(t)

1 1 qτ sin ωτr ω ω

1 1 q(t 0) (sin ωτ 0)r ω ω f(t) ¯7 ¯Z sin ωt

Example 2.8

Direction: Find the general solution of the differential equation. y ½½ 2y ½ 2y 0 for y (0) 1 and y ½ (0) 3

s Y(s) sY(0) Y ½ (0) 2sY(s) y(0) 2Y(s) 0 s Y(s) s 3 2sY(s) 2 2Y(s) 0 (s 2s 2)Y(s) s 1 Y(s)

Y(s)

Y(s)

s

s

Y(s)

s1

2s 2

s1

2s 1 1

s 111 (s 1) 1

2 s 1 (s 1) 1 (s 1) 1

y(t) eQ cos t 2eQ sin t

94

Solution:

Page

a.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS b.) y ½½ y t for y(0) y ½ (0) 1 s Y(s) sY(0) Y ½ (0) Y(s) s Y(s) s(1) 1 Y(s) (s 1)Y(s)

Y(s)

1

s 1 s

1 s 1

1) s 1

s (s

By Partial Fraction Expansion,

1 s

1 s

A B C D 1

s s s1 s 1 1)

s (s

1 As(s 1) B(s 1) C(s )(s 1) D(s )(s 1) 1 A(sR s) B(s 1) C(sR s ) D(sR s ) sR : 0 A C D s : 0 B C D s: 0 A

A0

s : 1 B

B 1

C 12

D 12

1 1 1 1 1 r ® Q q r ® Q « 2 ¬ ® Q « 2 ¬ ® Q q r ® Q q s (s 1) s s1 s 1 s1

95

y(t) e sinh t t

Page

Solution:



Drill Problem 2.2

Direction: Use the differentiation property to find the Laplace transform of the following: 1.) t cos 5t

Solution:

f(0) 0

f ½ (t) 0t(5 sin 5t) cos 5t2 f ½ (t) 5t sin 5t cos 5t f ½ (0) 1

f ½½ (t) 50t(5 cos 5t) sin 5t2 5 sin 5t f ½½ (t) 25t cos 5t 5 sin 5t 5 sin 5t f ½½ (t) 25t cos 5t 10 sin 5t

®0f ½½ (t)2 25f(t) 10 sin 5t ®f ½½ (t) 25F(s)

s

50

25

DP: s F(s) sF(s) f ½ (0)

s F(s) sF(s) f ½ (0) 25F(s) s F(s) 25F(s) 1

s F(s) 25 F(s)

50

25

50

25

s 25 50 s 25

(s 25) F(s) s 25

F(s)

s

s

s 25 s 25

s 25 (s 25)

Page

96

2.) cos πt Final answers are shaded in green.


f(0) 1

f ½ (t) 2 cos πt(π sin πt) f ½ (t) 2π cos πt sin πt f ½ (0) 0

f ½½ (t) 2π0cos πt π cos πt π sin πt sin πt2 f ½½ (t) 2π 0cos πt sin πt2 f ½½ (t) 2π 0f(t) sin πt2

1 1 s ®0f ½½ (t)2 2π qF(s) S Tr 2 s s 4π ®0f ½½ (t)2 2π sF(s) ®0f ½½ (t)2 2π F(s)

2π t s(s 4π )

4πV s(s 4π )

DP: s F(s) sF(s) f ½ (0)

s F(s) sF(s) f ½ (0) 2π F(s)

(s s)F(s) 2π F(s)

(s 2π )F(s) s

(s 2π )F(s)

4πV s(s 4π )

4πV s(s 4π )

4πV s(s 4π )

s (s 4π ) 4πV s(s 4π )

sV 4π s 4πV s(s 4π ) (s 2π )

(s 2π )F(s)

(s 2π )(s 2π ) s(s 4π )(s 2π ) (s 2π ) s(s 4π )

97

F(s)

Page

F(s)



3.) sinh at

Solution:

f(0) 0

f ½ (t) 2 sinh at a cosh at f ½ (t) 2a sinh at cosh at f ½ (0) 0

f ½½ (t) 2a0sinh at a sinh at cosh at a cosh at2 f ½½ (t) 2a 0sinh at cosh at2 f ½½ (t) 2a 0f(t) cosh at2

1 1 s ®0f ½½ (t)2 2a qF(s) S Tr 2 s s 4a 1 2s 4a t¬ ®0f ½½ (t)2 2a «F(s) s 2 s(s 4a ) ®0f ½½ (t)2 2a F(s) a s

2s 4a t s(s 4a )

DP: s F(s) sF(s) f ½ (0)

s F(s) sF(s) f ½ (0) 2a F(s) a s (s 2a )F(s) 2a q s 2a

Solution:

2a s(s 4a )

f(0) 1

98

s 2a r s(s 4a )

Page

4.) cosh t

F(s)

2s 4a t s(s 4a )


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 1 1 f ½ (t) 2 cosh t S sinh tT 2 2 2 1 1 f ½ (t) cosh t sinh t 2 2 f ½ (0) 0 1 1 1 1 f ½½ (t) q cosh t sinh tr 2 2 2 2 1 1 1 ®0f ½½ (t)2 f(t) sinh t 2 2 2 1 1 1 s ®0f ½½ (t)2 qF(s) S Tr 2 2 s s 1 1 2 1 ½½ ( )2 y ®0f t xF(s)

s(s 1) 2 ®0f

1 4 1 t)2 F(s)

2 s(s 1)

½½ (

DP: s F(s) sF(s) f ½ (0)

1 4 1 s F(s) sF(s) f 0) F(s)

2 s(s 1) ½(

1 1 4 s F(s) F(s) s 2 s(s 1)

1 s (s 1) 4 1 Ss T F(s) s(s 1) 2

1 sV s 4 1 Ss T F(s) 2 s(s 1) 1 As C 1 2 As C F(s) 2 s(s 1) 1 As C 2

99

F(s)

1 s 2

s(s 1)

Page



5.) sinV t

Solution:

f(0) 0

f ½ (t) 4 sinR t cos t f ½ (0) 0

f ½½ (t) 40sinR t cos t2

f ½½ (t) 40sinR t ( sin t) cos t (3 sin t)(cos t)2 f ½½ (t) 40 sinV t 3 sin t cos t2 f ½½ (t) 4 sinV t 12 sin t cos t

®0f ½½ (t)2 4f(t) 12 sin t cos t

®0f ½½ (t)2 4F(s) 12 sin t cos t ®0f ½½ (t)2 4F(s) 3 sin 2t

s 3 1 r ®0f ½½ (t)2 4F(s) q 2 s s 16

3 s 16 s t ®0f ½½ (t)2 4F(s) s 2 s(s 16) ®0f ½½ (t)2 4F(s)

24 ( s s 16)

DP: s F(s) sF(s) f ½ (0)

s F(s) sF(s) f ½ (0) 4F(s)

24 ( s s 16)

24 (s 4)F(s) s(s 16) s 4

24

4)(s 16)

100

s(s

Z Q7

Page

6.)

F(s)



F(s)

G(s) s

f(t) g(τ)dτ

10 πs 10 F(s) s (s π) F(s)

F(s)

sR

10 s(s)(s π)

G(s)

10 ( s s π)

g(t) 10e

f(t) g(τ)dτ

f(t) 10eÃ dτ

f(t) q

g(t)

10eÃ t r π 0

10e 10 π π

f(t) g(τ)dτ

f(t)

10 Ã (e 1)dτ π

f(t)

10 Ã (e 1)dt π

f(t)

10e 10t 10 π π π

Page

101

10 eÃ q f(t) tr π π


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS f(t)

10e 10πt 10 π

f(t)

c P7

F(s)

G(s) s

f(t) g(τ)dτ

F(s)

1

1)

s (s

G(s)

1 s(s 1)

g(t) sin t

f(t) g(τ)dτ

f(t) sin τ dτ

f(t) 0 cos τ2

t 0

f(t) cos t cos 0 g(t) cos t 1

f(t) g(τ)dτ

f(t) ( cos τ 1)dτ

t f(t) \° sin τ τ2^ 0

102

Solution:

f(t) sin t t

Page

7.)

10(e πt 1) π



Z Q

Solution:

F(s)

G(s) s

f(t) g(τ)dτ

5 5) 5 G(s) s 5

F(s) g(t)

5

√5 « ¬ √5 s u√5v

s(s

f(t) cosh √5t 1 g(t)

5

√5

sinh √5t

f(t) g(τ)dτ

f(t) S f(t)

f(t)

5

√5 5

5

√5

sinh √5τT dτ

usinh √5τvdτ

1

q cosh √5τr √5 √5 5 t f(t) \cosh √5τ^ 5 0 t f(t) \cosh √5τ^ 0 ( ) f t cosh √5t cosh 0

c QV7

F(s)

G(s) s

103

Solution:

f(t) g(τ)dτ

Page

9.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 4) 1 F(s) s(s)(s 4) 1 G(s) s(s 4) F(s)

s (s

1 G(s) (sinh 2t) 2

f(t) g(τ)dτ

1 f(t) (sinh 2τ) dτ 2 1 1 f(t) q cosh 2τr 2 2 cosh 2τ t r f(t) q 4 0 cosh 2t 1 f(t) 4 4

f(t) g(τ)dτ

1 f(t) (cosh 2τ 1)dτ 4 1 sinh 2τ f(t) q τr 4 2 1 sinh 2t f(t) S tT 4 2

Solution:

F(s)

G(s) s

f(t) g(τ)dτ

F(s)

s(s

2

3 )

104

Z Pµ

Page

10.)

1 1 f(t) sinh 2t t 8 4


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2(3)

3 )

G(s)

(s

2 G(s) sin 3t 3

f(t) g(τ)dτ

f(t)

2 sin 3τ dτ 3

2 ( ) f t sin 3τ dτ 3 2 cos 3τ r f(t) q 3 3 2 cos 3t 2 cos 0 f(t)

9 9 2 f(t) ( cos 3t 1) 9

11.) y ½ 4y 0; y(0) 2.8

2 f(t) (1 cos 3t) 9

Solution:

12.) y ½ y 17 sin 2t ; y(0) 1

sY(s) y(0) 4 Y(s) 0 sY(s) 2.8 4Y(s) 0 (s 4)Y(s) 2.8 2.8 Y(s) s 4 1 Y(s) 2.8 S T s 4 y(t) 2.8eQV

Page

13.) y ½½ y ½ 6y 0; y(0) 6, y ½ (0) 13

105

Solution:



Solution:

s Y(s) sy(0) y ½ (0) 0sY(s) y(0)2 6Y(s) 0 s Y(s) 6s 13 sY(s) 6 6Y(s) 0 (s s 6)Y(s) 6s 7 6s 7 s6 6s 7 Y(s) (s 3)(s 2) Y(s)

Y(s) q

s

6s 7 A B r (s 3)(s 2)

(s 3)(s 2) s 3 s 2 Y(s) 6s 7 A(s 2) B(s 3) if s 3 25 A(s 2) 25 5A A5 if s 2 5 B(s 3) 5 B(2 3) B1 1 1 Y(s) 5 S T

s3 s 2 y(t) 5eR eQ

14.) y ½½ 4y ½ 4y 0; y(0) 2.1, y ½ (0) 3.9 s y(s) sy(0) y ½ (0) 40sy(s) y(0)2 4y(s) 0 s y(s) 2.1s 3.9 4sy(s) 4(2.1) 4y(s) 0 (s 4s 4)y(s) 2.1s 4.5 2.1s 4.5 s 4s 4 2.1s 4.5 Y(s) (s 2)(s 2)

106

Y(s)

Page

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2.1s 4.5 A B r (s 2) Y(s) q

(s 2)(s 2) s 2 (s 2) 2.1s 4.5 A(s 2) B s : 0 0 s: 2.1 A A 2.1 s : 4.5 A(2) B 4.5 2.1(2) B 4.5 4.2 B B 0.3

2.1 0.3 r

q (s 2) s2 1 1 r T 0.3 q Y(s) 2.1 S ( s 2) s2 Y(s) 2.1e 0.3te Y(s)

y(t) e (2.1 0.3t)

15.) y ½½ ky ½ 2k y 0; y(0) 2, y ½ (0) 2k

s y(s) sy(0) y ½ (0) k0sy(s) y(0)2 2k y(s) 0 s y(s) 2s 2k ksy(s) 2k 2k y(s) 0 (s ks 2k )y(s) 2s 4k 2s 4k

ks 2k 2s 4k Y(s) (s 2k)(s k) 2(s 2k) Y(s) (s 2k)(s k) 2 Y(s) sk 1 Y(s) 2 S T sk Y(s)

s

107

y(t) 2e¼

Example 2.8

Page

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Directions: Write the following function using unit step functions and find its transform. t

Æ Å Ä cos t

0 È È 1 . 1 È È π ° . t É π

Solution:

2 1 f(t) Ê2 t cos t

0ÈÈ1 1 È È π2° t É π2

since ®0f(t)u(t a)2 eQ ®0f(t a)2 for 2

2 2eQ ®0t2 s s

for 2

for;

for;

2 2eQ s s

1 1 π t zt u(t 1) t u At C{ 2 2 2

1 1 Q π t Ëe ®0(t 1) 2 eQ ® qAt C rÌ 2 2 2


108

2

Page

a.) f(t)

Ç Å


for ;

1 1 Q π t Íe ®0t 2t 12 eQ ® st πt tÎ 4 2 2

1 1 1 1 Q 1 π π Q for; t S R T e E R F e 2 s s 2s s 2s 8s for ; cos t cos t u At

for; cos t eQ

for ; cos t eQ

®0f(t)2

π C 2

π ® zcos At C{

π ® zcos t cos

for; cos t eQ

Therefore,

2

π

sin t sin { 2 2

®0sin t2

eQ for; cos t s 1

2 2eQ 1 1 1 1 π π 1

S R T eQ E R F eQ/ eQ/ s s s s 2s s 2s 8s s 1

Example 2.9

Direction: Find the inverse Laplace Transform of

Solution:

F(s)

eQ eQ eQR

s π (s 2)

We will first solve on the first term, by expanding the term, we can get two terms,

109

eQ s π

Page

For


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS F (s)

s

1

π

f (t) z sin πt{ t (t -1) f (t) Then,

f (t)

1 sin(πt π) π

1 (sin πt cos π cos πt sin π) π f (t) For

1 (sin πt) π eQ s π

F (s)

s

1

π

f (t) z sin πt{ t (t -2) f (t)

f (t)

1 sin(πt 2π) π

1 (sin πt cos 2π cos πt sin 2π) π

Now, for the second term,

f (t)

1 (sin πt) π

eQR (s 2)

f (t) 0teQ 2 t (t -3)

0ÈÈ1

1ÈÈ2

1ÈÈ3 tÉ3

°


110

Æ Å Ä

0 sin πt π 0 (t 3)eQ(QR)

Page

f(t)

Ç Å

f (t) \(t 3)eQ(QR) ^


Drill Problems 2.3

Direction: Sketch or graph the given function (assumed to be zero outside the given interval). Represent it using unit step function. Find its transform. 1.) t (0 È È 1) Solution:

® 0 f(t) u(t a)2 eQ ®0 f(t a)2

® 0 t u(t 0)2 ® 0 t u(t 1)2 e ®0 t2 eQ ®0 t 12 ® 0 t u(t 0)2 ® 0 t u(t 1)2

2.) sin 3t (0 È È p) Solution:

1 1 1 Q e S

T s s s

® 0 t u(t 0)2 ® 0 t u(t 1)2

1 eQ eQ s s

® 0 f(t) u(t a)2 eQ ®0 f(t a)2

® 0 sin 3t u(t 0)2 ® 0 sin 3t u(t π)2 e ®0 sin 3t2 eQ ®0 sin 3(t π)2

® 0 sin 3t u(t 0)2 ® 0 sin 3t u(t π)2 ®0 sin 3t2 eQ ®0 sin 3t cos 3π cos 3t sin 3π2 ® 0 sin 3t u(t 0)2 ® 0 sin 3t u(t π)2 ®0 sin 3t2 eQ ®0 sin 3t2 ® 0 sin 3t u(t 0)2 ® 0 sin 3t u(t π)2

® 0 sin 3t u(t 0)2 ® 0 sin 3t u(t π)2

s

3 (1 eQ

9


111

3 3eQ

s 9 s 9

Page

® 0 sin 3t u(t 0)2 ® 0 sin 3t u(t π)2

3 3 QÏ¡

£ 9 £ 9

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 3.) t (t É 3) Solution:

® 0 t u(t 3)2 eQR ®0 (t 3) 2

® 0 t u(t 3)2 eQR ®0 t 6t 92 ® 0 t u(t 3)2 eQR q

2! 6 9

r sR s s

® 0 t u(t 3)2 eQR q 4.) 1 eQ (0 È È p)

2 6 9

r R s s s

Solution:

® 0 (1 eQ ) u(t 0)2 ® 0 (1 eQ ) u(t u)2 e ®0 (1 eQ )2 eQ ®\ u1 eQ(P) v^ ® 0 (1 eQ ) u(t 0)2 ® 0 (1 eQ ) u(t u)2 e ®0 (1 eQ )2 eQ ®0 (1 eQ eQ )2 ® 0 (1 eQ ) u(t 0)2 ® 0 (1 eQ ) u(t u)2 ®0 (1 eQ )2 eQ ®0 (1 eQ eQ )2 ® 0 (1 eQ ) u(t 0)2 ® 0 (1 eQ ) u(t u)2

® 0 (1 eQ ) u(t 0)2 ® 0 (1 eQ ) u(t u)2

¯

)

® q sin ωt u St

® q sin ωt u St

~ 6π 6π Tr eQ ¯ ® q sin ω(t )r ω ω

~ 6π Tr eQ ¯ ®0sin ωt cos 6π cos ωt sin 6π ω


112

Solution:

~

1 eQ (eQ eQ ) 1

s s 1

Page

5.) sin ωt (t É

1 1 eQ eQ eQ

s s 1 s s 1

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS ~ 6π Tr eQ ¯ ®0sin ωt2 ω ~ 6π ® q sin ωt u St Tr eQ ¯ ®0sin ωt2 ω

® q sin ωt u St

~

6π ωeQ ¯ ® q sin ωt u St Tr ω s ω

Direction : Find and sketch or graph f(t) is F(s) equals 6.)

Ð Ñ

7 P¯7

Solution:

seQ s s ω s ω

s 0cos ωt2 t (t 1) s ω

7.)

7

eQ A7 C

Solution:

f(t) Ë

0 cos ω(t 1)

F(s)

tÈ1 ° tÉ1

1 1

s s

F(s) 0t 12 (. Q)

F(s) t 0(t 1) 12 u(t 1) F(s) t (t 1 1) u(t 1) F(s) t (t) u(t 1)

ÐÑ

7 PP

Page

8.)

0ÈÈ1 ° tÉ1

113

t f(t) Ò 0



F(s)

By completing the square,

F(s)

s

1

2s 2

1 (s 1) 1

F(s) 0eQ sin t2

. ( Q )

F(s) eQ(Q) sin(t π) u(t π)

F(s) eQ(Q) 0sin t cos π cos t sin π2 u(t π) F(s) eQ(Q) 0sin t2 u(t π)

f(t) Ò

9.)

0 eQ(Q) sin t

tÈp ° tÉp

(QÐÑÓÔ) Q¼

Solution:

Expanding the term, we get

F(s)

F(s)

1 eQP¼ sk sk

1 eQ e¼ sk sk

F(s) e¼ 0e¼ e¼ 2 (. Q)

F(s) e¼ 0e¼ e¼(Q) 2u(t 1)

ÐZ.Ñ QÐ7.ÕÑ

Page

10.) 2.5

0ÈÈ1 ° tÉ1

114

¼ f(t) Ëe 0



Solution:

F(s) 2.5 s

eQR.U eQ.~ t s s

F(s) 2.5 q0t2 (.QR.U) – 012

. r ( Q.~)

F(s) 2.5 0u(t 3.8) u(t 2.6)2 2.5 f(t) Ò 0

2.6 È È 3.8 ° elsewhere

Example 2.10

Direction: Determine the response of a system described by the differential equation For y (0) y’(0) 0 and inputs r(t).

y ½½ 3y ½ 2y r(t)

a.) r(t) u(t 1) u(t 2) may GRAPH DITO

r(t) u(t 1) u(t 2)

1 s Y(s) sy(0) y ½ (0) 3sY(s) 3y(0) 2Y(s) (eQ eQ ) s Final answers are shaded in green.

115

y(0) y’(0) 0

Page

Solution:

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 s Y(s) 3sY(s) 2Y(s) (eQ eQ ) s 1 Y(s)0s 3s 22 (eQ eQ ) s Y(s)

By partial fraction expansion

(eQ eQ ) s0s 3s 22

eQ For s0s 3s 22

1 A B C

s(s 1)(s 2) s 1 s 2 s

1 A(s 4s 5) B(2s 4)(s 1) C(s 1) 1 As(s 2) Bs(s 1) C(s 3s 2)

1 A(s 2s) B(s 1s) C(s 3s 2) s : 0 A B C s:

0 2A B 3C

s : 1 2C

A 1 B C

1 1 1 2

2 F(s) s 1 s 2 s

1 2 1 2

eQ s0s 3s 22

1 A B C

s(s 1)(s 2) s 1 s 2 s Final answers are shaded in green.

Page

For

116

1 1 f(t) eQ(Q) eQ(Q)

2 2

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 A(s 4s 5) B(2s 4)(s 1) C(s 1) 1 As(s 2) Bs(s 1) C(s 3s 2)

1 A(s 2s) B(s 1s) C(s 3s 2) s : 0 A B C s:

0 2A B 3C

A 1

s : 1 2C

B C

1 1 1 2 F(s)

2 s 1 s 2 s

y(t)

b.) r(t) δ(t 1)

Ç Å Æ Å Ä

1 2 1 2

1 1 f(t) eQ(Q) eQ(Q)

2 2

0 0ÈÈ1 1 1 eQ(Q) eQ(Q) 1 È È 2° 2 2 1 1 eQ(Q) eQ(Q) eQ(Q) eQ(Q) tÉ2 2 2

Drill Problem 2.4

Direction: Showing the details, find and graph the solution.

s Y(s) sy(0) y ½ (0) Y(s) eQ s Y(s) 10s Y(s) eQ Y(s)0s 12 eQ 10s Y(s)

eQ 10s

s 1 s 1

117

Solution:

y(0) 1, y’(0) 0

Page

1.) y ½½ y δ(t 2π)



For

Ð7Ñ 7 P

F(s)

s

:

1

1

f(t) 0sin t2 t (t – 2p)

f(t) sin(t 2π) u(t 2π)

Then,

Y(s) 10 cos t sin(t 2π) u(t 2π)

Y(s) 10 cos t 0sin t cos 2π cos t sin 2π2 u(t 2π) y(t) 10 cos t sin t u(t 2π)

2.) y ½½ 2y ½ 2y eQ 5 δ(t 2) Solution:

y(0) 0, y’(0) 1

s Y(s) sy(0) y ½ (0) 2sY(s) 2y(0) 2Y(s) s Y(s) 1 2sY(s) 2Y(s)

1

5eQ 1 s 1

1 5eQ 1

(s 1)0s 2s 22 0s 2s 22 0s 2s 22

(P)07PP2

:

A(2s 2) B C 1

(s 1)0s 2s 22 s 2s 2 s 1

1 A(2s 2)(s 1) B(s 1) C(s 2s 2) Final answers are shaded in green.

118

For

1

5eQ s 1

Page

Y(s)

Y(s)0s 2s 22

1

5eQ s 1

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 A(2s 4s 2) B(s 1) C(s 2s 2) s: A

Then,

Y(s)

s : 0 2A C

0 4A B 2C

s : 1 2A B 2C 1 , 2

1 ( 2 )(2s 2) s 2s 2

B0 ,

C1

1 5eQ 1

s 1 0s 2s 22 0s 2s 22

Completing the square of the denominators: Y(s)

(s 1) 1 5eQ 1

(s 1) 1 s 1 (s 1) 1 (s 1) 1

y(t) eQ sin t eQ cos t eQ 5eQ(Q) sin(t 2) u (t 2)

3.) y ½½ y 10 δ At C 100 δ(t 1) y(0) 10, y’(0) 1 Solution:

s Y(s) sy(0) y ½ (0) Y(s) 10eQ 100eQ

Y(s)0s 12 10eQ 100eQ 10s 1

10eQ 100eQ 10s 1 Y(s)

0s 12 0s 12 0s 12 0s 12 6 Ð Ñ

010 sinh t2 ( – 6) .

119

6 Ð Ñ

7 7 , 07 Q2 07Q2

7

Page

For



10eQ 1 1 10 sinh( t ) u St T 0s 12 2 2 For

ÐÑ 07 Q2

,

ÐÑ 07Q2

0100 sinh t2 (. – )

100eQ 1 100 sinh( t 1) u St T 0s 12 2

1 1 1 y(t) 10 sinh( t ) u St T 100 sinh( t 1) u St T 10 cosh t sinh t 2 2 2

4.) y ½½ 3y ½ 2y 10 sin t 10δ(t 1) Solution:

y(0) 1, y’(0) 1

s Y(s) sy(0) y ½ (0) 3sY(s) 3y(0) 2Y(s) s Y(s) s 1 3sY(s) 3 2Y(s) Y(s) 0s 3s 22 Y(s) 0s 3s 22

10

10eQ

1

s

10

10eQ s 1

10

10eQ s 2

1

s

10

10eQ s 2

1

s

5.) y ½½ 4y ½ 5y 01 u(t 10)2eQ e δ(t 10) Solution:

y(0) 0, y’(0) 1

s Y(s) sy(0) y ½ (0) 4sY(s) 4y(0) 5Y(s) eQ eQ u(t 10) 1 1

1 eQ eQ S

1T s1 s1

Page

Y(s)0s 4s 52

1 1 eQ eQ S 1T s1 s1

120

s Y(s) 1 4sY(s) 5Y(s)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Y(s)0s 4s 52

1 s1 1 s1 eQ eQ S T s1 s1

Y(s)0s 4s 52

Y(s)

s s C eQ eQ A s1 s1

s (eQ eQ )(s) (s 10)0s 4s 52 (s 1)0s 4s 52

By partial fraction expansion:

s

A B(2s 4) C

s1 s 4s 5

s A(s 4s 5) B(2s 4)(s 1) C(s 1) s A(s 4s 5) B(2s 2s 4) C(s 1) s: A

s : 0 A 2B

1 4A 2B C

s : 0 5A 4B C

1 , 10

B

1 , 20

C

7 10

1 7 1 20 (2s 4) 10 10

s s1 s 4s 5 s 4s 5

y(t) e eQ cos t eQ sin t z e(Q) e(Q) cos (t 10) eQ(Q) { e u(t 10)

6.) y ½½ 2y ½ 3y 100 δ(t 2) 100 δ(t 3) y(0) 1, y’(0) 0 Solution:

s Y(s) sy(0) y ½ (0) 2sY(s) 2y(0) 3Y(s) 100eQ 100eQR

Y(s)

100eQ 100eQR s 2

0s 2s 32 0s 2s 32

Completing the square of the denominator:


Page

Y(s)0s 2s 32 100eQ 100eQR s 2

121

s Y(s) s 2sY(s) 2 3Y(s) 100eQ 100eQR


Y(s)

100eQ 100eQR s 2

(s 3)(s 1) (s 3)(s 1)

By partial fraction expansion: s 2

A B

(s 3) (s 1)

s 2 A(s 1) B(s 3) 1 A B , 2 a 3B A

1 , 4

B

3 4

100 A(s 1) B(s 3)

0 A B , 100 a 3B A 25 ,

B 25

3 1 25 25 25 25 4 4 r eQ 100 q r eQR Y(s)

100 q

(s 3) (s 1) (s 3) (s 1) (s 3) (s 1)

1 3 y(t) eQR e 025eQR(Q) 25eQ(Q) 2u(t 2) 025eQR(QR) 25eQ(QR) 2u(t 3) 4 4 1 y(t) 0eQR 3e 2 25\eQRP~ e(Q) ^u(t 2) 25 0eQRPµ eQR 2(t 3) 4

7.) y ½½ 2y ½ 10y 1001 u(t 4)2 10δ(t 5) y(0) 1, y’(0) 1 Solution:

Y(s)0s 2s 102 10eQ

10 1 4 QV 10e S

T s s s

10 10eQV 40eQV s s s


Page

s Y(s) s 1 2sY(s) 2 10Y(s)

122

s Y(s) sy(0) y ½ (0) 2sY(s) 2y(0) 10Y(s) 010 10µ(t 4)2 10δ(t 5)


Y(s)

For 7 (7PP)

10 10eQV 40eQV 10eQ

s (s 2s 10) s (s 2s 10) s(s 2s 10) s 2s 10 s (s

10 A B Cs D

2s 10) s s s 2s 10

10 As(s 2s 10) B(s 2s 10) Cs(s ) Ds

10 A(sR 2s 10s) B(s 2s 10) C(sR ) Ds

sR : 0 A C

s : 0 2A B D

A 2 ; s (s

For 7 (7PP) ÐcÑ

For (7 PP)

B1 ;

s: 0 10A 2B C2

;

s : 10 10B

D3

10 2 1 2s 3

2s 10) s s s 2s 10 s 2s 10

10eQV 2 1 2s 3 s (s 2s 10) s s s 2s 10 s 2s 10

VÐcÑ

40eQV A Bs C s(s 2s 10) s s 2s 10

40 A(s 2s 10) Bs Cs

s : 0 A B

A 4 ;

B4 ;

s : 40 10A C8

2s 3 2 1 2s 3 2 1

TS T s s s 2s 10 s 2s 10 s s s 2s 10 s 2s 10 4 1 4s 8 10

S

T S TÌ s s s 2s 10 s 2s 10 s 2s 10 Final answers are shaded in green.

123

40eQV 4 1 4s 8

s(s 2s 10) s s s 2s 10 s 2s 10

Page

Y(s) ® Q ËS

s: 0 2A C

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Competing the square of the denominator Y(s) ® Q Ís

2 1 2(s 1) 3 2 1 2(s 1) 3 ts t

(s 1) 9 (s 1) 9 (s 1) 9 (s 1) 9 s s s s 4 1 10 4(s 1) 2 t q rÎ

s

(s 1) 9 (s 1) 9 s 2s 10 s s

y(t) 2 t 2eQ cos 3t eQ sin 3t \2 (t 4) 2eQ(QV) cos 3(t 4)^u(t 4)

z4 4eQ cos 3(t 4) sin 3(t 4){ u(t 4) sin 3(t 5) u(t 5) R

8.) y ½½ 5y ½ 6y δ At πC u(t π) cos t Solution:

R

y(0) y’(0) 0

1 y ½½ 5y ½ 6y δ St πT u(t π) cos t y(0) y’(0) 0 2

s Y(s) sy(0) y ½ (0) 5sY(s) 5y(0) 6Y(s) eQ Y(s)0s 5s 62 eQ

s

eQ ( ) s 1

s

eQ ( ) s 1

eQ seQ Y(s)

0s 5s 62 (s 1)0s 5s 62 eQ Ø0cos(t π)2

eQ 0cos t cos π sin t sin π2 eQ 0cos t2

Completing the square of the denominator

eQ seQ 1 1 Y(s) S T eQ (s 2)(s 3) (s 1)(s 2)(s 3) (s 2) (s 3)

A(2s) B C D

(s 1) (s 2) (s 3)

s A(2s)(s 5s 6) B(s 5s 6) C(s 1)(s 3) D(s 1)(s 2)

s A(2sR 10s 12s) B(s 5s 6) C(sR 3s s 3) D(sR 2s s 2) Final answers are shaded in green.

Page

s

124


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS A

1 , 20

B

1 , 10

C

2 , 5

D

3 10

2 1 1 3 (2s) 1 1 Q 5 20 10 10 re x y eQ Y(s) q

(s 2) (s 2) (s 2) (s 3) (s 2) (s 3) y(t) (eQ eQ R ) (. – ) ( cos t sin t eQ eQ R ) (. – ) 7

R

y(t) 0eQ AQ7 eQ RAQ7 2u(t A cos(t π) sin(t π) eQ (Q) eQ R(Q) C u(t π)

C

C

R

y(t) 0eQ AQ 7 C eQ RAQ 7 C 2u(t A cos (t π) sin(t π) eQ (Q) eQ R(Q) C u(t π)

R

9.) y ½½ 2y ½ 5y 25t 100δ(t π) y(0) 2, y’(0) 5 Solution:

s Y(s) sy(0) y ½ (0) 2sY(s) 2y(0) 5Y(s) s Y(s) 2s 5 2sY(s) 4 5Y(s) Y(s)0s 2s 52

Y(s)

25 100eQ s

25 100eQ s

25 100eQ 2s 1 s

25 100eQ (2s 1)

s 0s 2s 52 0s 2s 52 0s 2s 52


2s 1

A(2s 2) B 0s 2s 52

2s 1 A(2s 2) B A B C(2s 2) D

s s 0s 2s 52

25 A(s)(s 2s 5) B(s 2s 5) C(2s 2)(s ) D(s )


125

25

B3

Page

A 1 ,

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 25 A(sR 2s 5s) B(s 2s 5) C(2sR 2s ) D(s ) sR : 0 A 2C

s : 0 2A B 2C D s: 0 5A 2B

A 2 ,

s : 25 5B

B 5,

C 1,

D3

(2s 2) 2(2s 2) 3 2 5 3

0s 2s 52 0s 2s 52 s s 0s 2s 52 0s 2s 52 2 5 100eQ ( ) Y s 0s 2s 52 s s

y(t) 2 5t 50eQ(Q) sin 2(t π) u(t π) 10.) y ½½ 5y 25t 100δ(t π)

s Y(s) sy(0) y ½ (0) 5Y(s) s Y(s) 2s 5 5Y(s) Y(s)0s 52

25 100eQ s

25 100eQ 2s 5 s

25 100eQ (2s 5)

0s 52 s 0s 52 0s 52


2s 5

A(2s) B 0s 52

126

Y(s)

25 100eQ s

2s 5 A(2s) B

Page

Solution:

y(0) 2, y’(0) 5


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS A 1 , 25

B5

A B C(2s) D

0s 52 s s

25 A(s)(s 5) B(s 5) C(2s)(s ) D(s ) 25 A(sR 5s) B(s 5) C(2sR ) D(s ) sR : 0 A 2C s : 0 B D s: 0 A

A 0,

s : 25 5B

B 5,

C0,

D 5

(2s) 100eQ 5 1 1 q r

5

0s 52 0s 52 0s 52 s (s 5)

y(t) cos √5t 5t 20√5 sin √5(t π) u(t π)

Example 2.11

Direction: Find the response of a mass-spring system without damping to the following inputs a.) Hammerblow input βδ(t)at t 0,zero initial conditions Solution:

We are given the conditions,

x(0) x ½ (0) 0 2γ 0

Page

127

β γ ω


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Then,

x ½½ (t) 2γx ½ (t) β x(t) F (t) x ½½ (t) β x(t) F (t)

s X(s) sx(0) x ½ (0) β X(s) F (s) s X(s) β X(s) F (s) X(s)0s β 2 F (s) X(s)

F (s) 0s β 2

x(t) sin βt

b.) no input, but with non-zero initial conditions (x(0) x ½ (0) Û 0) Solution:

We are given the condition,

x(0) x ½ (0) Û 0

Then,

x ½½ (t) 2γx ½ (t) β x(t) F (t)

s X(s) sx(0) x ½ (0) 2γsX(s) 2γx ½ (0) β X(s) F (s) X(s)0s 2γs β 2 sx(0) x ½ (0) 2γx ½ (0) F (s) sx(0) x ½ (0) 2γx ½ (0) F (s)

0s 2γs β 2 0s 2γs β 2 X(s)

sx(0) x ½ (0) 2γx ½ (0)

0(s γ) ω 2 0(s γ) ω 2

x(t) x(0)eQÜ cos ωt

0x ½ (0) 2γx ½ (0)2 eQÜ sin ωt ω


128

X(s)

sx(0) x ½ (0) 2γx ½ (0) 0(s γ) (β γ )2

Page

X(s)

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS let C x(0) and C

Therefore

x ½ (0) 2γx ½ (0) ω

x(t) eQÜ (C cos ωt C sin ωt)

c.) sinusoidal driving force A sin ωt with ω Û β, x(0) x ½ (0) Û 0. Solution:


x(0) x ½ (0) Û 0 β Û γ

F (s) A sin ωt

Then,

x ½½ (t) 2γx ½ (t) β x(t) F (t)

s X(s) sx(0) x ½ (0) 2γsX(s) 2γx ½ (0) β X(s) F (s) X(s)0s 2γs β 2 sx(0) x ½ (0) 2γx ½ (0) F (s) X(s) X(s)

sx(0) x ½ (0) 2γx ½ (0) F (s)

0s 2γs β 2 0s 2γs β 2

sx(0) x ½ (0) 2γx ½ (0) Aω

(s γ) (β γ ) 0s 2γs β 2(s ω )


Aω Ms(s 2γs β ) N(s 2γs β ) Os(s ω ) P(s ω ) sR : 0 M O


129

Aω Ms N Os P

2γs β 2(s ω ) s ω s 2γs β

Page

0s

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS s : 0 2γM N P

s:

0 β M 2γN ω O M0

x(t) x(0)eQÜ cos βt

O0

s : Aω β N ω P N

Aω

γ

β

P

Aω

γ

β

x¾(0)sin βt Aω Aω 1 1 r r

® Q q ® Q q β γ ω β γ s ω s 2γs β

Therefore

x(t) x(0)eQÜ cos βt x¾(0)sin βt

Aω Aω sin ωt sin βt

γ β γ

β

d.) sinusoidal driving force A sin ωt with ω β, x(0) x ½ (0) Û 0. Solution:


x(0) x ½ (0) Û 0 β γ

F (s) A sin ωt

Then,

x ½½ (t) 2γx ½ (t) β x(t) F (t)

s X(s) sx(0) x ½ (0) 2γsX(s) 2γx ½ (0) β X(s) F (s) X(s)0s 2γs β 2 sx(0) x ½ (0) 2γx ½ (0) F (s)

sx(0) x ½ (0) 2γx ½ (0) Aω

(s γ) (ω ) (s γ) (s ω )


130

X(s)

F (s) sx(0) x ½ (0) 2γx ½ (0)

0s 2γs β 2 0s 2γs β 2

Page

X(s)

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS By partial fraction expansion: (s

γ)

Aω Ms N Os Os

(s ω ) s ω s γ (s γ)

Aω Ms(s 2γs β ) N(s 2γs β ) O(sR γs ω s ω γ) P(s ω ) sR : 0 M O

s : 0 2γM N γO s:

0 γ M ω O

s : Aω γ N ω γ O ω P M

At 2β

O0

x(t) x(0) cos βt

Therefore,

N

A 2β

P0

A x¾(0)sin βt At Q s β r ® Q q r

® q 2β βt s β s β β

x(t) x(0) cos βt

x¾(0)sin βt A

(sin βt βt cos βt) β 2β

Example 2.12

Solving for the needed values to be substituted to the general equation, Final answers are shaded in green.

Page

Solution:

131

1.) A spring is such that a 5-lb weight is attached, the spring reaches equilibrium, then the weight is pulled down 3 in. below the equilibrium and started off with an upward velocity of 6 ft/sec. Find an equation giving the position of the weight at all subsequent times.

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS x 6 in

1 ft 2

1 x(0) 3 in ft 4

x ½ (0) 6 ft/sec

Solving for k:

Fs kx

Solving for β and 2γ:

1 5 lb k S ftT 2 k 10 lbs/ft

β

kg (10lbs/ft)(32) 64 w 5 lbs

2γ

bg (0)(32) 0 w 5 lbs

Now substituting the conditions to the equation

x ½½ (t) 2γx ½ (t) β x(t) F (t) x ½½ (t) β x(t) 0

Getting its Laplace Transform

s X(s) sx(0) x ½ (0) β X(s) 0

Substitute the given:

s X(s)

1 s 6 64X(s) 0 4

X(s)0s 642

X(s)

1 s6 4

1 s 6 4 0s 642 0s 642

Taking the inverse Laplace transform

132

3 1 cos 8t sin 8t 4 4

Page

x(t)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 0cos 8t sin 8t2 4

x(t)

2.) (a.) A spring is stretched 1.5 in. by a 2 lb weight. Let the weight be pushed up 3 in. above the equilibrium point and then released. Describe the motion. (b.) For the same mass-spring system, let the weight be pulled down 4 in. below the equilibrium point and given a downward initial velocity of 8 ft/sec. Describe the motion Solution:

a.) Let us first solve for the needed values, Solving for k:

Fs kx



lbs kg A16 ft C (32) β 256 2 lbs w

2γ

bg (0)(32) 0 w 2 lbs

Now, substituting it to the equation,

x ½½ (t) 2γx ½ (t) β x(t) F (t) x ½½ (t) 256x(t) 0

s X(s) sx(0) x ½ (0) 256X(s) 0 s X(s)

1 s 256X(s) 0 4

1 X(s)0s 2562 s 4 X(s)

1 s 4 0s 2562

Page

133

1 x(t) cos 16t 4


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS For the amplitude,

1 | amplitude S T 3

1 amplitude of varibration t 4

For the frequency of vibration,

β 2πf ; f

√256 2π

frequency of vibration

8 Hz π

The equation of the motion of the spring is x(t) V cos 16t. That means that the amplitude of

vibration is ft above and below the equilibrium point and the frequency of the vibration is V

f

b.) Let us solve for the needed values

V

Hz

1 x(0) 4 in ft 3 x ½ (0) 8

Solving for k:

ft sec

Fs kx

1 2 lb k S ftT 8

kg (16 lbs/ft)(32) 256 w 2 lbs

134

β

lbs ft

Page


k 16


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2γ

bg (0)(32) 0 w 2 lbs

Substituting the computed values to the equation,

x ½½ (t) 2γx ½ (t) β x(t) F (t) x ½½ (t) 256x(t) 0

s X(s) sx(0) x ½ (0) 256X(s) 0 s X(s)

1 s 8 256X(s) 0 3

1 X(s)0s 2562 s 8 3

s 8 1 r

X(s) q 3 0s 2562 0s 2562 x(t) R cos 16t sin 16t

amplitude of vibration 2√13 in frequency of varibration

8 Hz π

The equation of the motion of the spring is x(t) R cos 16t sin 16t. The amplitude of vibration

is 2√13 in in above and below the equilibrium point and the frequency of vibration is still f Hz U

Example 2.13 (a) Using x ½½ (t) 2γx ½ (t) β x(t) F (t) we can see that it is the same in the mass spring system equation in the underdamped motion. The formula would be

x(t) eQÜ (c cos ωt c sin ωt). Since c 0 and c ÞL because LC VL7 is positive, then R7

135

E QÜ e sin βt βL

Page

ί(t)

E


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS (b) Using x ½½ (t) 2γx ½ (t) β x(t) F (t) we can also see that it is the same in the mass spring system equation in the critically damped motion. The formula would be

x(t) eQÜ (c c t). Since c 0 and c L because LC VL7 is positive, then ί(t)

E

E QÜ te L

R7

(c) Using x ½½ (t) 2γx ½ (t) β x(t) F (t) we can find out that it is the same in the mass spring system equation in the overdamped motion. The formula would be

x(t) eQ(ÜPâ) eQ(ÜQâ) . Since LC VL7 is negative, then c c ÞL.

R7

So the formula would be

ί(t)

E

E 0e(ÞQÜ) eQ(ÞQÜ) 2 2βL

Example 2.14 An iron ball whose weight is 98 N stretches a spring 1.09 m. determine the equation of motion of the object when it is pulled down 16 cm. from its equilibrium and for the following damping parameters.

a.) b 0

b.) b 10 kg/sec c.) b 60 kg/sec d.) b 100 kg/sec Solution:

a.) Solve for the needed values,

bg w

136

2γ

Page

For 2γ and β

b0 x(0) 0.16 m x ½ (0) 0


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2γ 0

β

β

kg w

90(9.81) 98

β 9

For k

F kx

98 k(1.09) k 90

Then, substituting to the equation,

x ½½ (t) 9x(t) 0

s X(s) sx(0) x ½ (0) 9X(s) 0 s X(s) 0.16s 9X(s) 0 s X(s) 9X(s) 0.16s (s 9)X(s) 0.16s X(s)

0.16s (s 9)

x(t) 0.16 cos3t 0undamped motion2 kg m bg S10 s T A9.8 s C 2γ w 98 N 2γ 1 β 9

x(0) 0.16 m

Page

For 2γ and β

137

b.) Solve for the needed values


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS x ½ (0) 0

Then, substitute the values to the equation

x ½½ (t) 2yx ½ (t) β x(t) 0 x ½½ (t) x ½ (t) 9x(t) 0

s x(s) sx(0) x ½ (0) sx(s) x(0) 9x(s) 0 x(s)0s s 92 0.16s 0.16 x(s)

® Q sx(s)

x(s)

0.16s 0.16 s s 9

0.16s 0.16 (s 0.5) (2.96)

0.16(s 0.5) 0.16 (0.16)(0.5) t

(s 0.5) (2.96) (s 0.5) (2.96)

x(t) 0.16e. cos 2.96 ® Q s

(0.16)(0.5) t (s 0.5) (2.96)

x(t) 0.16eQ. cos 2.96 0.027eQ. sin 2.9t

x(t) eQ. (0.16 cos 2.96 0.027 sin 2.9t )0underdamped motion2

c.) Let us solve for the needed values

bg 60(9.8) w 98 2y 6 kg 90(9.8) β w 98 β 9 x(0) 16 cm or 0.16 m x ½ (0) 0

Then, substitute the values in the equation

x ½½ (t) 6x ½ (t) 9x(t) FE t

s x(s) sx(0) x ½ (0) 6sx(s) 6x(0) 9x(s) FE (s) Final answers are shaded in green.

138

2y

Page

For 2γ and β

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS (s 6s 9)x(s) 0.16s 0.96 0.16s 0.96 A B r (s 3) x(s) q

(s 3) s 3 (s 3) 0.16s 0.96 A(s 3) B s½ : 0.16 A s : 0.96 3A B 0.96 3(0.16) B B 0.96 0.48 B 0.48

x(t)

0.16 0.48

s 3 (s 3)

x(t) 0.16eQR 0.48teQR 0critically undamped2

d.) Let us solve for the needed values

F kx

98 N k(1.09 m) N k 90 m bg kg m 2y S100 T A9.8 C w s s 10 2y s m N A90 C A9.81 C kg m s β 98 N w 9 β s x(0) 0.16 m

Then, substitute the values in the equation

x ½½ (t) 10x ½ (t) 9x(t) FE (t)

s x(s) 0.16s 10sx(s) 1.6 9x(s) 0 (s 10s 9)x(s) 1.6 0.16s


Page

s x(s) 0.16s 100sx(s) 0.162 9x(s) 0

139

s x(s) 5x(t) x ½ (0) 10sx(s) x(0) 9x(s) 0

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS x(s)

x(s)

1.6 0.16s (s 10s 9) 1.6 0.16s (s 9)(s 1)

A B 1.6 0.16s q r (s 9)(s 1)

(s 9)(s 1) s 9 s 1 1.6 0.16s A(s 1) B(s 9) if s 1

1.6 0.16(1) B(8) B

1.44 8

B 0.18

if s 9

1.6 1.44 A(9) A 0.02

x(s)

0.02 0.18

s 9 s 1

x(t) 0.02eQµ 0.18eQ

x(t) 0.18eQ 0.02eQµ 0overdamped2

Drill Problem 2.5 Direction: Solve each of the problems completely using Laplace transform.

1.) A spring mass is such that a 4-lb weight stretches it 6 in. an impressed force cos 8t is acting on the spring. If the 4-lb weight started from the equilibrium point with an imparted upward velocity 4 ft/sec, determine the position of the weight as a function of time. Let us solve for the needed values

Page

Solution:

140


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS x ½ (0) 4

Solving for k:

Fs kx

ft sec

1 4 lb k S ftT 2 Solving for β and 2γ:

k8

lbs ft

lbs kg A8 ft C (32) β 64 w 4 lbs

2γ

bg (0)(32) 0 w 2 lbs

Substituting the values in the equation, we get

x ½½ (t) 2γx ½ (t) β x(t) F (t) 1 x ½½ (t) 64x(t) cos 8t 2

1 s C s X(s) sx(0) x ½ (0) 64X(s) A 2 s 64 1 s C s X(s) 4 64X(s) A 2 s 64 1 s C4 X(s)0s 642 A 2 s 64

Performing partial fraction for

(7P~V)7

:

1 As B Cs D 2s

(s 64) s 64 (s 64) Final answers are shaded in green.

141

1 s 4 0s 642 2 (s 64)

Page

X(s)

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 s As(s 64) B(s 64) Cs D 2 1 s A(sR 64s) B(s 64) Cs D 2 sR : 0 A

s:

Therefore,

s : 0 B

1 64A C; 2

s : 0 64B D; X(s)

C

1 2

D 0

1 s 4 0s 642 2 (s 64)

1 1 x(t) t cos 8t sin 8t 2 2

1 x(t) (t cos 8t sin 8t) 2

2.) A spring is such that is stretched 6 in by a 12-lb weight. The 12-lb weight is pulled down 3 in below the equilibrium point and then released. If there is an impressed force of magnitude 9 sin 4t, describe the motion. Solution:

Let us solve for the needed values

1 x(0) 3in ft 4 x ½ (0) 0 Fs kx

142

1 12 lb k S ftT 2

Page

Solving for k:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS k 24


lbs ft

lbs A24 C (32) kg ft β 64 w 12 lbs 2γ

bg (0)(32) 0 w 2 lbs

Then substituting the values in the equation, we would get, x ½½ (t) 2γx ½ (t) β x(t) F (t) x ½½ (t) 64x(t) 9 sin 4t

s X(s) sx(0) x ½ (0) 64X(s) 9 S

s

1 36 s X(s) s 64X(s) S T 4 s 16 X(s)0s 642 S

X(s)

s

4 T

16

36 1 T s

16 4

s 36 1

(s 16)0s 642 4 0s 642

Performing partial fraction expansion: (s

(s

36

16)0s 642

36 As B Cs D

16)0s 642 s 16 s 64

36 As(s 64) B(s 64) Cs(s 16) D(s 16) 36 A(sR 64s) B(s 64) C(sR 16s) D(s 16)

s: 0 64A 16C;

B

3 4

C 0

143

s : 0 B D;

A0

Page

sR : 0 A C


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS s : 36 64B 16D; Therefore,

D

3 4

1 s 4 r X(s) q 0s 642 2 (s 64) 1 1 x(t) t cos 8t sin 8t 2 2

1 x(t) (t cos 8t sin 8t) 2

3.) A spring is such that a 2-lb weight stretches ft. An impressed force sin 8 is acting upon V

the spring. If the 2-lb weight is released from a point 3 in below the equilibrium point, determine the equation of the motion. Solution:

Let us first find the needed values

1 x(0) 3in ft 4 x ½ (0) 0

Solving for k:

Fs kx

1 2 lb k S ftT 2 kg (4 lbs/ft)(32) 64 w 2 lbs

2γ

bg (0)(32) 0 w 2 lbs

Now that we have the values, we can substitute it to the equation x ½½ (t) 2γx ½ (t) β x(t) F (t)


144

β

Page


k 4 lbs/ft

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 x ½½ (t) 64x(t) sin 8t 4

8 1 T s X(s) sx(0) x ½ (0) 64X(s) S 4 s 64 1 2 T s X(s) s 64X(s) S 4 s 64 X(s)0s 642 S X(s)

(s

x(t)

x(t)

s

2 1 T s

64 4

2 s 1 0s

64)

642 4

1 1 tsin 8t cos 8t 4 4

1 (t sin 8t cos 8t) 4

4.) A spring is such that a 16-lb weight stretches it 1.5 in. The weight is pulled down 4 in below the equilibrium point and given an initial down ward velocity of 4 ft/sec. an impressed force of 360 cos 4t lb is applied. Find the position and velocity of the weight at time t U sec. Solution:

Let us first find the values needed

x 1.5 in

1 ft 8

1 x(0) 4in ft 3 x ½ (0) 4

Fs kx

145

1 16lb k S ftT 8 k 128 lbs/ft

Page

Solving for k:

ft sec


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Solving for β and 2γ:

β

kg (128lbs/ft)(32) 256 16lbs w

2γ

bg (0)(32) 0 w 2 lbs

Substituting the values we get in the equation

x ½½ (t) 2γx ½ (t) β x(t) F (t) x ½½ (t) 256x(t) 360 cos 4t

s X(s) sx(0) x ½ (0) 256X(s)

360s s 16

1 360s s X(s) s 4 256X (s) 3 s 16

X(s)

X(s)0s 2562

s

360 1

s4

16 3

s

360s 1 s 4

160s 2562 3 0s 2562 0s 2562

Performing partial fraction expansion: s

360s As B Cs D

160s 2562 s 16 s 256

360s As(s 256) B(s 256) Cs(s 16) D(s 16)

360s A(sR 256s) B(s 256) C(sR 16s) D(s 16)

s: 360 256A 16C; Therefore,

s : 0 256B 16D;

3 2

B0

C

3 2

D 0

3 s 3 s 1 s 4 X(s) S T S T

2 s 16 2 s 256 3 0s 2562 0s 2562 Final answers are shaded in green.

146

s : 0 B D;

A

Page

sR : 0 A C


if t

U

3 7 1 x(t) S T cos 4t S T cos 16t sin 16t 2 6 4

, x(t) ~ ft

x¾(t) sin 4t S

56 T sin 16t 4 cos 16t 3

x ½ (t) v 2 ft/sec

5.) A 20-lb weight stretches a certain spring 10 in. Let the spring first be compressed 4 in, and then the 20-lb weight attached and given an initial downward velocity of 8 ft/sec. Find how far the weight would drop. Solution:

Let us first find the needed values

1 x(0) 43 in ft 3 x ½ (0) 8ft/sec

Solving for k:

Fs kx



lbs A24 C (32) 192 kg ft β w 20 lbs 5 2γ

bg (0)(32) 0 w 2 lbs

Substituting the values we get in the equation, 192 x(t) 0 5

Page

x ½½ (t)

147

x ½½ (t) 2γx ½ (t) β x(t) F (t)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS s X(s) sx(0) x ½ (0)

192 X(s) 0 5

1 192 X(s) 0 s X(s) 8s

3 5 X(s) qs

X(s)

192 1 r 8s 5 3

8s 1 1 ã ä 192 192 zs

{ 3 s

5 5

x(t) 8 cos

8√15 8√15 √15 t sin t 15 72 15

6.) Consider an underdamped motion of a body mass m 2 kg. If the time between two consecutive maxima is 2 second the maximum amplitude decreases to V of its initial value after 15 cycles, what is the damping constant of the system? Solution:

β 2πf 2π S

15 T 15π 2

β 225π β

225π

kg w

k(9.81) 2(9.81)

k 450π

450π γ

γ 450π

γ 47.044

2γ 94.09

15 2

15 2

148

Page

β γ ω where ω }


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2γ

94.09

bg w

b(9.81) 2(9.81)

b 188.18

kg r

7.) A certain straight line motion is determined by the differential equation d x dx

2γ 169x 0 dt dt

and the conditions that when t 0, x 0 and x ½ 8 . (a) Find the value of ç that leads to æ

critical damping and determine x in terms of t. (b) Use γ 12. Find x in terms of t. (c) γ 14. Find x in terms of t. Solution:

a.) β y

β 169 β 13 y 13

x ½½ 2(13)x ½ 169x 0 x ½½ 26x ½ 169x 0

s x(s) sx(0) x ½ (0) 26sx(s) 26x(0) 169x (s) 0 x(s)(s 26s 169) 8 x(s)

s

x(s)

8

26s 169 8 (s 13)

x ½½ 2(12)x ½ 169x 0

Page

b.) y 12

149

x(t) 8teQR


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS s x(s) sx(0) x ½ (0) 24sx(s) 24x(0) 169x (s) 0 x(s)(s 24s 169) 8 x(s)

x(s)

s

8

24s 169

8 s 24s 144 25

8 5 r x(s) q 5 (s 12) 25 8 x(t) eQ sin 5t 5

c.) y 14

x ½½ 2(14)x ½ 169x 0

s x(s) sx(0) x ½ (0) 28sx(s) 28x(0) 169x (s) 0 x(s)(s 28s 169) 8 x(s)

8 s 28s 196 27

x(s)

x(t)

x(t)

8

8 (s 14) 27

3√3 t 3√3 (s 14) 27 8

s

3√3

eQV sinh 3√3 t

8√3 QV e sinh 3√3 t 9

150

x(s)

8

28s 169

Page

x(s)

s


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 8.) A spring is such that a 2-lb weight stretches it

ft. An impressed forced

V

sin 8 and a

damping force of magnitude |FD | |v| (v is the velocity of the object) are both acting on the spring. The weight starts

V

ft below the equilibrium point and with an imparted upward

velocity of 3 ft/sec. Find a formula for the position of the weight at time t. Solution:

x(0) FD bx ½

|FD | |x|

FD bFD b1

FS kx

1 2 kS T 2

k4

1 ½ x (0) 3 4

2y

2y

bg w

1(32) 2

2y 16

β

β

kg w

4(32) 2

β 64

1 x ½½ 16x ½ 64x sin 8t 4

1 8 s x(s) sx(0) x ½ (0) 16sx(s) 16x(0) 64x(s) S T 4 s 64 2 1 s 3 4 r

q

(s 64)(s 8) 4 (s 8) (s 8) (s 8)

(s

2 As B C D

64)(s 8) s 64 s 8 (s 8)

2 (As B)(s 8) C(s 64)(s 8) D(s 64)


151

x(s)

2 1 1 ( )

s

3

16 S T s 64 4 4

Page

x(s)(s 16s 64)

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2 (As B)(s 16s 64) C(sR 8s 64s 512) D(s 64)

AsR 16As 64As Bs 16Bs 64B CsR 8Cs 64Cs 512C Ds 64D sR : 0 A C

s : 0 16A B 8C D s: 0 64A 16B 64C

A Y(s) S

s : 2 64B 512C 64D

1 512

B0

C

1 512

D

1 64

1 Q 1 1 s 1 1 { S r S T ® Q q r T® z T ® Q q (s 8) 512 s 64 512 s 8 64 y(t)

1 1 1 cos 8t eQU S tT 8 512 64

9.) A spring is such that a 4-lb weight stretches 0.4 ft. The 4-lb weight is attached to the spring (suspended form a fixed support) and system is allowed to reach equilibrium. Then the weight is started from the equilibrium position with an imparted upward velocity of 2 ft/sec. assume that the position takes place in a medium that furnishes a retarding force of magnitude numerically equal to the speed in feet per second of the moving weight. Determine the position of the weight as a function of time.

FD bFD b1

FS kx

4 k(0.4) k 10

bg w 1(32) 2y 4 2y

2y 8 β

β

kg w

10(32) 4

β 80


152

FD bx ½

Page

Solution:


x ½½ 8x ½ 80x 0

s x(s) sx(0) x ½ (0) 8sx(s) 8x(0) 80x(s) 0 x(s)(s 8s 80) 2

x(s)

s

x(s)

x(s)

2

8s 16 64

2 (s 4) 8

2 8 q r 8 (s 4) 64

1 x(t) eQV sin 8t 4

10.) A particle started along x-axis according to the law

d x dx

6

25x 0 dt dt

If the particle stated x 0 with an initial velocity of 12 ft/sec to the left, determine (a) x in terms of t, (b) times at which stops occur, and (c) the ratio between the numerical values of x at successive stops. Solution:

x ½½ 6x ½ 25x 0

s x(s) sx(0) x ½ (0) 6sx(s) 6x(0) 25x(s) 0 x(s)(s 6s 25) 12 x(s)

12

6s 9) 16

12 4 q r 4 (s 3) 4

153

x(t) 3eQR sin 4t

Page

x(t)

(s



Example 2.16 Find the current i(t) in a series RLC circuit with ê 11 Ω, L 0.1H, C 10Q F which is connected to a source voltage E(t) 100 sin 400t. Assume that the current and charge are zero when t 0. Solution:

We are given the values R 11Ω

C 10Q F

Substituting the values we have to the equation,

L 0.1H

1 R i(t) L i½½ (t) i(τ)dτ V(t ì ) E(t) C

11 i(s) 0.1 i½½ (s)

1 400 i(s) 100 S T Q 10 s s 400

i(s) q11 0.1s

i(s)

i(s)

4000 100 r s 400 s

4000 0.1s 11s 100 r 0s 400 2 q s 0s

4000s

110s 10002

400 20s

4000s 0sR 100s (160x10R )s 16x10~ 2 B0sR 10s (160x10R )s 1.6x10~ 2

C(sR 110s 1000s) D(s 110s 1000) sR : 0 A B C

s : 0 100A 10B 110C D

s: 40x10R (160x10R )A (16x10R )B (1x10R )C 110D s : 0 (16x10~ )A (1.6x10~ )B 1x10R

i(s)

0s

D 258.68

4000s

100)(s 10)

400 2(s


154

C 2.3368

B 2.6144

Page

A 0.277

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS By partial fraction expansion: 0s

4000s A B Cs D

100)(s 10) s 10 s 100 s 400

400 2(s

4000s A(s 100)(s 400 ) B(s 10)(s 400 ) Cs(s 10)(s 100) D(s 10)(s

100) i(t) 0.277eQ 2.6144eQ 2.3368 cos 400t 0.6467 sin 400t A

Drill Problem 2.6 1. Find the current of a series LC circuit when L 0.9 H, C 0.05 F and E sin t V, assuming zero initial current and charge. Solution:

We are given the values C 0.05 F

L 0.9H

Let us substitute the values to the equation Li" (t)

1 i (t) E(t) c

Ls i(s) i(0)

0.2s i(s)

S0.2s

1 1 i(s) s 1 Cs

1 1 i(s) 0.05s s 1

20 1 T i(s) s s 1

0.2s 20 1 E F s s 1

5s (s 1)(s 100)

155

i(s)

s (s 1)(0.2s 20)

Page

i(s)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 5s As B Cs D

(s 1)(s 100) (s 100) (s 1)

5s As (s 1) B (s 1) Cs (s 100) D(s 100) 5s A(sR s) B(s 1) C(s 100s) D( s 100) sR ;

s ; s;

s ;

A µµ

0 A C

0 B D

5 A 100C

0 B 100D

B 0 C µµ

D 0

s 5 s 5 A C

( ) 99 s 1 99 s 100

5 5 cos 10t

cost 99 99

i(t)

5 ( cost cos10t)A 99

2.) What are the conditions for an RLC circuit to be overdamped, critically damped and over damped? 3.) Find the steady-state current in the RLC circuit for the given data a.) R 8 Ω

Solution:

C 0.5 F

E 100 sin 2t

Li½½ (t)

100(2) 1 (i)t Ri(t) C s 4

S0.5s

1 200

8 T I(s) 0.1s s 4

1 200 I(s) 8 I(s) 0.1s s 4

Page

0.5s I(s)

156

Since we are given the data, let us substitute them into the equation, and solve it simultaneously


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 0.5s 10 8s 200 E F I(s) s s 4

I(s)

(s

(s

I(s)

200s

4 )(0.5s 8s 10)

(s

400s

4 )(s 16s 20)

400s As B Cs D

4 )(0.5s 16s 20) s 4 s 16s 20

400s As (s 16s 20) B (s 16s 20) Cs(s 4) D(s 4)

400s A (sR 16s 20s) B(s 16s 20) C(sR 4s) D(s 4) sR :

s : s:

s ;

(s

Therefore,

400 20A 16B D 0 20B 4D

A 5 , B 20, C -5 , D -100

20 5s 100 5s

4 ) (s 4 ) s 16s 20 s 16s 20

5s 20 5s 100

4 ) (s 4 ) (s 8) 44 (s 8) 44

i(t) 5 cos2t 10sin2t 5eQU cos√44t

b.) R 1Ω

Solution:

70√11 sinh√πt 11

Steady state 5 cost 10 sin2t A

L O. 25 H

C 5x10Q F

E 110 V

Since we are given the data, let us substitute them into the equation, and solve it simultaneously


157

I(s)

(s

0 16A B D

Page

I(s)

0 A C

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Ls I(s)

0.25 I(s)

1 110 i(s) R I(s) cs s

110 1 I(s) I(s) Q 5x10 s s

0.25s 20000 s 110 s s

I(s)

I(s)

s(0.25s

110s

s 20000)

110 0.25(s 4s 80000)

I(s)

s

I(s) 440

440

s 80000 4

440 (s 2) 79996

√79996

eQ sin√79996t

i(t) 1.556 eQ sin 1.556 t

c.) R 2Ω

Solution:

steady state 0A

L 1H

C 0.05F

E

µ

sin 3t V

Since we are given the data, let us substitute them into the equation, and solve it simultaneously Ls I(s)

s 20 2s 157 1 F I(s) S T s 3 s 9

I(s)

157 s S T 3 (s 9)(s 2s 20) Final answers are shaded in green.

158

E

1 157 1 I(s) 2I(s) S T 0.05s 3 s 9

Page

s I(s)

1 157 3 i(s) R I(s) S T cs 9 s 9

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 157 As B Cs D 3 s

(s 9)(s 2s 20) s 9 s 2s 20

157 s As(s 2s 20) B (s 2s 20) Cs (s 9) D(s 9) 3 sR :

s:

s :

0 A C

0 2A B D

157 20A 2B 9C 3 s : 0 20B 9D A C

0 2(C) B D 0 2C B D

157 20(C) 2B 9C 3 157 11C 2B 3 0 20B 9D

I(s)

B 6,C

11 11 40 ,A ,D 3 3 3

11 S 6 11 s 40 1 C S S T A T 3 s 9 s 9 3 s 2s 20 3 s 2s 20

29 11 11 Q i(t) cos3t 2sin3t e cos√19 t 3 eQ sin√19 t 3 3 √19 R

cos3t 2 sin3t A

4.) Find the transient current in the RLC circuit for the given data

Solution:

Page

a.) R 6Ω , L 0.2H, C 0.02F, E 110 sin10t V

159

steady state


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Since we are given the data, let us substitute them into the equation, and solve it simultaneously Ls I(s)

1 10 I(s) R I(s) 110 S T cs s 100

0.2 I(s)

1 1100 I(s) 6 I(s) 0.025s s 100

0.2s 40 6s 1100 E F I(s) s s 100

I(s)

I(s)

(s

i(s)

(s

1100s (s 100)(0.2s 40 6s)

i(s)

1100s

100) (0.2)(s 30s 200)

(s

(s

5500s

100)(s 30s 200) 5500s

100)(s 10)(s 20)

5500s A B Cs D

(s 10) (s 20) (s 100)

100)(s 10)(s 20)

5500s A(s 20)(s 100) B(s 10)(s 100) Cs (s)(s 30s 200) D((s 30s

200)

5500s A(sR 1000s 20s 2000) B(s 100s 10s 1000) C (sR 30s 200s)

D(s 30s 200) 0 A B C

0 20A 10B 30C D

5500 100A 100B 200C 30D 0 2000A 1000B 200D A B C

0 20(B C) 30C D

0 20B 20C 30C D

0 20B 10C D

eq. 1

5500 100(B C) 100B 200C 30D


160

s ;

Page

s;

s ;

sR ;

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 5500 100B 100C 100B 200C 30D 5500 100C 30 D eq. 2

0 2000(B C) 100B 200D

0 2000B 2000C 1000B 2000D 0 1000B 2000C 2000D eq. 3

B

110 9

C

55 6

D

1375 385 A 9 18

385 110 55 1379 s 18 9 6 9 I(s)

s 10 s 20 s 100 s 100

i(t)

Therefore,

b.) R 0.2Ω Solution:

275 385 Q 110 Q 55 e

e

cos 10t

sin 10t 9 6 18 18

transient current

L 0.1H

C 2F

385 Q 110 Q e

e A 18 9 E 757 sin 0.5t


1 0.5 I(s) R I(s) 754 ã ä 1 cs s 4

0.1 I(s)

1 377 I(s) 0.2 I(s) ã ä 1 2s s

4

1 2 5 s 1 5 s I(s) ã 377 ä 1 2s s 4

161

754s 1 1 2 As C ( s s 1) 4 5 5

Page

I(s)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS I(s)

3770s 1 As C (s 2s 5) 4

3770s As B Cs D

1 1 As C (s 2s 5) s 4 s 2s 5 4 1 1 3770s As( s 2s 5) B((s 2s 5) C SsR sT D Ss T 4 4 s :

sR :

0 A C

0 2A B D

1 s: 3770 5A 2B C 4 s :

1 0 5B D 4

A C

0 2C B D eq. 1

1 3770 5C 2B C 4

3770 2B B 80

0 5B

C 760

19 C eq. 2 4

1 D eq. 3 4

D 1600

A 760

80 760s 1600 760s

1 1 s 4 s 4 s 2s 5 s 2s 5

760s 80 760s 1600

1 1 (s 1) 4 (s 1) 4 s 4 s

4 760(s 1) D 760s 1600

760s 760 D 760s 1600 D 1600 760

Page

162

D 840



760s 80 760 (s 1) 840

1 1 (s 1) 4 (s 1) 4 s 4 s 4

1 1 i(t) 760 cos 160 sin t 760eQ cos 2t 420eQ sin 2t 2 2

c.) R Ω

i(t) transient current 760eQ cos 2t 420eQ sin 2t A

Solution:

L H

C

R

F

E eQV (1.932 cos t 0.256 sin t)

Since we are given the data, let us substitute them into the equation, and solve it simultaneously E

483 QV 1 123 QV 1 e cos t

e sin t 250 2 500 2

1 s 4 123 483 1 2 x y

x y Ls I(s) I(s) R I(s) 250 (s 4) 1 500 (s 4) 1 Cs 4 4 966 123 483 s 125 1000 1 1 1 250 s I(s)

I(s) I(s) 100 1 2 10 (s 4)

13 s 4 7851 483 s 1000 1 13 1 250 S s

s T I(s) 1 2 100 10 (s 4)

4

7851 483 s 1000 50s 13 10s 250 E F I(s) 1 100s (s 4)

4

483 7851 s 500 125 I(s) 13 1 65 As s C As 8s C 5 50 4


Page

966 7851 s 10 5 I(s) 65 1 13 50 As 5 s 50C As 8s 4 C

163

483 7851 C 100s A250 s

1000 I(s) 1 As 8s 16 C (50s 10s 13) 4

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 483 7851 As B Cs D 125 s 500

1 13 13 1 65 65 As s C As 8s C s 5 s 50 s 8s

50 5 4 4

483 7851 65 65 1 13 1 13 s

As Ss 8s T B Ss 8s T Cs Ss s T D Ss s T 125 500 4 4 5 50 5 50

483 7851 65 65 1 13 1 13 s

A SsR 8s sT B Ss 8s T C SsR s sT D Ss s T 125 500 4 4 5 50 5 50 sR ; 0 A C

s;

1 s ; 0 8A B C D 5

483 65 13 1 A 8B C D 125 4 50 5 s ;

7851 65 13 B D 500 4 50 A C

1 0 8C B D 5

0 13

39 C D eq. 1 5

483 65 13 1 C 8B C D 125 4 50 5

483 1599 1 8B C D eq. 2 125 100 5 309 16 63 16 ,C ,D ,A 325 65 65 65

16 309 65 s 325

16 63 s 65 65

1 1 13 1 1 s 5 s 100 50 100 (s 4) 4


164

309 16 63 s 325 65 65

1 13 1 13 65 65 s 5 s 50 s 5 s 50 s 8s

s 8s 4 4

Page

16 65 s

B

7851 65 13 B D 500 4 50

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 16 309 65 s 325

63 16 s 65 65

1 1 (s 2) 1 As C

4 10 4

16 309 16 1 s

Ss T B 65 325 65 10

16 309 16 468 s

s

B 65 325 65 325 B

317 325

16 63 16 (s 2) D s

65 65 65

16 63 16 32 s

s S T D 65 65 65 65 D

transient current

31 65

16 1 317 16 31 (s 2)

65 As C

10 325 65 65 I(s)

1 1 1 (s 2)

As C

4 10 4

16 Q 1 634 Q 1 16 1 31 16 Q 1 e cos t

e sin t eQ cos t

e sin t A 65 2 325 2 65 2 130 65 2

5.) Find an expression for the current at any time t in the RLC circuit given the data

a.) R 4Ω

Solution:

L 0.1H

C 0.025F

E 10 sin 10t

Since we are given the data, let us substitute them into the equation, and solve it simultaneously 0.1s I(s)

E

1 100 I(s) 4 I(s) s 100 0.025

0.1s 40 4s 100 F I(s) s s 100


165

1 10 I(s) R I(s) 10 S T Cs s 100

Page

Ls I(s)


I(s)

100s (s 100)(0.1s 4s 40) (s

I(s)

1000s

100)(s 40s 400) (s

1000s

100)(s 20)

1000s A B Cs D

(s 100)(s 20) s 20 (s 20) s 100

1000s A(s 20)(s 100) B(s 100) Cs(s 20) D(s 20)

1000s A(sR 20s 100s 2000) B(s 100) C(sR 40s 400s) D(s 40s 400) sR : 0 A C

s : 0 20A B 40C D

s: 1000 100A 400C 40D s : 0 2000A 100B 400D A C

0 20C B 40C D 0 B 20C D eq. 1

1000 100C 400C 40D 1000 300C 40D eq. 2

0 2000C 100B 400D eq. 3 6 6 B 40, C , D 16, A 5 5

6 5

6 s 40 16 5

s 20 (s 20) s 100 s 100

L 1H

C 0.04F

E 600(cos t 4 sin t)


Page

b.) R 6Ω

2 6 i(t) (3 cos 10t 4 sin 10t) eQ S 40tT A 5 5

166

6 6 8 i(t) eQ 40teQ cos 10t sin 10t 5 5 5


Solution:


1 s 4 I(s) R I(s) 600 S

T Cs s 1 s 1

s I(s)

1 600s 2400 I(s) 6I(s) 0.04s s 1

s 25 6s 600s 2400 E F I(s) s s 1 I(s)

600s 2400s (s 26s 25)(s 1)

600s 2400s As B Cs D

(s 26s 25)(s 1) s 1 s 26s 25

600s 2400s As(s 26s 25) B(s 26s 25) Cs(s 1) D(s 1) 600s 2400s A(sR 26s 25s) B(s 26s 25) C(sR s) D(s 1) sR : 0 A C

s : 600 6A B D

s: 2400 25A 6B C s : 0 25B D A C

600 6C B D eq. 1 2400 25C 6B C

2400 24C 6B eq. 2 0 25B D eq. 3

A 100, B 0, C 100, D 0 100s 100(s 3) D

167

100s 100s 300 D

Page

D 300



I(s)

Solution:

100s 100s

s 1 (s 3) 16

100s 100(s 3) 300

s 1 (s 3) 16 (s 3) 16

I(s)

c.) R 3.6Ω

100s 100s

s 1 s 6s 25

i(t) 100 cos t 100eQR cos 4t 75eQR sin 3t A

L 0.2H

C 0.0625F

E 164 cos 10t


0.2s I(s)

1 164s I(s) R I(s) Cs s 100

1 164s I(s) 3.6 I(s) s 100 0.0625s

(0.2s 16 3.6s)I(s) 164s s s 100

I(s)

164s (s 100)(0.2s 3.6s 16)

I(s)

I(s)

820s (s 100)(s 18s 80) 820s (s 100)(s 8)(s 10)

820s A B Cs D

(s 100)(s 8)(s 10) s 8 s 10 s 100

820s A(sR 10s 100s 1000) B(sR 8s 100s 800) C(sR 18s 80s)

D(s 18s 80) sR : 0 A B C

Page

s : 820 10A 8B 18C D

168

820s A(s 10)(s 100) B(s 8)(s 100) Cs(s 18s 80) D(s 18s 80)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS s: 0 100A 100B 80C 18D s : 0 1000A 800B 80D A B C

820 10(B C) 8B 18C D

820 10B 10C 8B 18C D 820 2B 8C D eq. 1

0 100(B C) 100B 80C 18D

0 100C 100B 100B 80C 18D 0 20C 18D eq. 2

0 1000 (B C) 800B 80D

0 1000B 1000C 800B 80D 0 200B 1000C 80D eq. 3 B 205, C 45, D 50, A 160

I(s)

160 205 45s 50

s 8 s 10 s 100 s 100

Page

169

i(t) 160eQU 205eQ 45 cos 10t 5 sin 10t A



UNIT 3

Page

170

FOURIER SERIES, INTEGRALS AND TRANSFORMS



Example 3.1 Find the Fourier coefficients of the periodic function k f(t) Ò ° k

and determine its Fourier series expansion. Solution:

π È t È 0 0ÈtÈ π

Since the graph of the function is symmetrical to the origin, the function is odd. ï a 0 and aa 0

ba

1 . ba f(t) sin nt dt π ð

1 s k sin nt dt k sin nt dtt π Q

ba

cos nt 1 ° cos nt sk ¸ °k ¸ t π n Q n

ba

ba

n 1; b

k 02 2 cos nπ2 nπ

ba 4k π

2k 01 cos nπ2 nπ

n 2; b 0


171

since, cos(nπ) cos nπ

k 0cos 0 cos(nπ) cos nπ cos 02 nπ

Page

ba

k ° 0cos nt|Q °cos nt| 2 nπ


Drill Problem 3.1 Direction: Find the Fourier series expansion of the given f(t) which is assumed to have the period 2π. In the expansion, terms involving cos5t and sin5t must be included. 1.)

Page

172

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS aa

1 . f(t)cos nt dt π P

1 aa « cos nt dt cos nt dt¬ Q π

1 sin nt 0 sin nt π2 s° ¸ π ° ¸ t π n n 0 2 1 πn πn aa zsin 0 sin

sin sin 0{ 2 π 2 1 πn aa z2sin A C{ 2 π 2 πn aa zsin { 2 π aa

a 0

aR

ba

2 3π

1 . f(t) sin nt dt π P

aV 0

a

2 5π

1 ba «Qsin nt dt sin nt dt¬ π

nt 0 cos nt π2 1 cos s ° ¸ π ° ¸ t n n 0 π 2 1 πn πn ba zcos 0 cos cos

cos 0{ π 2 2 ba

f(t)

ba 0

1 2 1 1

Scos t cos 3t cos 5t ò T 2 π 3 5

173

2 π

Page

a



Page

174

Solution:



3.)

Page

175

Solution:



4.)

Page

176

Solution:



1 « tdt tdt¬ aì Q 2π

π 1 t 0 t 2 ós t π s t ô aì 2π 2 2 2 0

π π A C A C 1 ¶0 2 aì

2 0· 2π 2 2 aa

aì 0


1 aa «Qtcos nt dt tcos nt dt¬ π

ba

aa 0


1 ba «Qtsin nt dt cos nt dt¬ π

2 1 1 1 1 Ssin t sin 3t sin 5t ò T S sin 2t sin 4t ò T π 9 25 2 4


Page

f(t)

177

1 tcos nt 0 sin nt 0 tcos nt π2 sin nt π s ° ¸ π ° ¸ π ° ¸

° ¸ 2t π n n n 0 n 0 2 2 π πn πn π πn 2) sin( 2) u 2vcos( 2) sin(πn2) 1 u 2vcos( t ba s

π n n n n 1 uπ2vcos(πn2) 2 sin(πn2) t ba s

n n π ba 0 2 1 2 1 2 b b bR bV b π 2 9π 4 25π ba


Page

178

Solution:



6.)

Page

179

Solution:



7.)

Page

180

Solution:



aì

1 π π Í Î 2 2π 2

aa


aì 0

1 s tcos nt dt

tcos nt dtt aa π Q

1 ° tsin nt 0 cos nt 0 tsin nt π cos nt π q ° ¸ ¸ ° ¸ r

° ¸

π n π n π n 0 n 0 ( ) ( ) 1 πsin πn cos 0 cos πn πsin πn cos πn cos 0 aa sE F E F

t π n n n n n n 1 2πsin πn 2cos πn 2cos 0 r aa q

π n n n 4 4 4 a a 0 aR aV 0 a π 9π 25π aa

ba

ba


1 s tsin nt dt tsin nt dtt π Q

1 tcos nt 0 sin nt 0 tcos nt π sin nt π s ° ¸ ° ¸ π ° ¸ ° ¸ t π n π n n 0 n 0 2 1 πcos(πn) õö÷(ø÷) (π)cos(πn) sin(πn) ba s

t ¿ π n ÷ n n 2 ba 0 πcos(πn)2 π ba 2cos(πn)

b 2

b 0

bR

2 3

bV 0

b

2 5

Page

π 1 1 1 1 f(t) Scos t cos 3t cos 5t ò T 2(sin t sin 3t sin 5t ò 4 9 25 3 5

181

ba



9.) t

πÈ È p

Solution:

aì

1 . f(t)dt 2π P

1 s aì t dt

t dtt 2π Q

aì

tR π 1 tR 0 Ís t

s t Î 2π 3 π 3 0

aa

1 aa s t cos nt dt t cos nt dtt π Q

7 úa(Qa) a

a 4

C A0

ì(Qa)

aa

a 1

a7

C A0

úa(Qa) aZ

C

7 úa(a)

1 4πcos πn 4sin πn q r πn n nR 4 1 aR aV 9 4

a

ì a

a

a7

4 25

úa a aZ

{


182

1 t° sin nt 0 2tcos nt 0 2sin nt 0 t° sin nt π 2tcos nt π 2sin nt π ° s ù ¸ ¸ ù ° ¸ ° ¸ t

°

R R R π n π n π n π n 0 n 0 n 0

aa zA0


Page

aa

1 2πR s t 2π 3 π aì 3

aì


ba f(t)


1 s t sin nt dt t sin nt dtt π Q

ba 0

π 1 1 1 1 4 Scos t cos 2t cos 3t cos 4t cos 5t ò T 3 16 25 4 9

0 È t È 2π

Solution:

aì

1 . f(t)dt 2π P

1 s aì t dtt 2π

aì

aì aa

1 tR π s t 2π 3 0

8πR 4π 6π 3


nt 2π 2sin nt 2π 1 t° sin nt 2π 2tcos s ù ¸ ¸ t

° ° R π n 0 n 0 n 0 1 4π sin(2πn) 4πcos(2π) 2sin(2πn) t aa s

π n n nR 4 1 4 a 4 a 1 aR aV a 9 4 25 aa

ba


183

1 aa s t cos nt dtt π

Page

10.) t

ba



ba f(t)

1 s t sin nt dtt π ba 0

4π 1 1 1 1

4 Scos t cos 2t cos 3t cos 4t cos 5t ò T 3 16 25 4 9

Solution:

f(t) Ò

k k

2 È tÈ 0 û p4 0ÈtÈ2


Page

Find the Fourier series expansion of the function whose period is p4

184

Example 3.2

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS p 2L

4 2L L2

Since the graph of the function is symmetrical to the origin, it is an odd function ï a 0 and aa 0

1 . πn ba f(t) sin dt L ð L

1 πn πn ba s (k) sin dt k sin dtt 2 Q 2 2

2 k πn 2 πn ba s °S cos T¸ °S cos T¸ t πn 2 Q πn 2 2 nπt nπt k ° scos ¸ °cos ¸ t πn 2 Q 2

k 0cos 0 cos(πn) cos πn cos 02 πn since, cos(nπ) cos nπ

k 02 2 cos πn2 πn

ba

n 1; b

4k π

n 3 ; bR

4k 3π

n 2; b 0

2k 01 cos πn2 πn

n 2; b 0

n 3; bR

n 4; bV 0

n 5; b

f(t)

4k 3π

4k 5π

4k π 1 3π 1 5π Ssin t sin t sin t ò T π 2 3 2 5 2 Final answers are shaded in green.

185

ba

Page

ba

ba


Drill Problem 3.2 Direction: Find the Fourier series expansion of the following periodic functions. Terms involving cos 5t and sin 5t must be included.

1 1.) f(t) Ò 1 Solution:

2ÈtÈ0 û p4 0ÈtÈ2

Since the graph is symmetrical to the origin, it is an odd function therefore aa 0 and a 0 L

1 nπt f(t) sin ba dt L L QL

p 2L

4 2L

L2

1 nπt nπt ba x sin

sin dty 2 2 2 Q

Q

1 2 nπt 0 nπt 0 S T q – cos T

S cos T 2 nπ 2 2 2 2 2nπ 2nπ 1 0)2 0( cos 0 cos ) cos

cos ba nπ 2 2 2 ba 0 1 cos nπ2 nπ bR

4 π

4 3π

b 0

b

4 5π

bV 0

186

b

Page

ba



0 4

Solution:

4 πt 1 3πt 1 5πt (sin sin

sin

ò) π 2 3 2 5 2


a

1 L f(t)dt 2L QL

p 2L

4 2L a

L2

1 s 0dt 4dtt 2L Q

a

1 0 4t 2 2L

1 a 082 4 a 2

nπt 1 L aa f(t) cos dt L QL L nπt 1 aa 4 cos dt 2 2 2 nπt r aa 2 q sin nπ 2

aa

4 0sin nπ sin 02 nπ

ba

1 L nπt f(t) sin dt L QL L

aa 0

187

2.) f(t) Ò

f(t)

Page

Therefore,


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 nπt 4 sin dt 2 Q 2

ba

ba 2 q

ba

Therefore,

3.) f(t) t

Solution:

b

2 nπt r cos nπ 2

4 0cos nπ cos 02 nπ

ba

4 0cos nπ 12 nπ

8 8 8 ; b 0; bR ; bV 0; b π 3π 5π

8 π 1 3π 1 5π f(t) 2 Ssin t sin t sin t ò T π 2 3 2 5 2

1 È t 1, p 2

Since the graph is symmetrical to the y-axis, it is an even function ï ba 0

1 L f(t) dt a 2L QL p 2L

2 2L L1

1 a t dt 2 Q 1 tR a s t 2 3 Q

Page

188

1 1 1 a q r 2 3 3 Final answers are shaded in green.

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS a

1 3

1 L nπt dt aa f(t) cos L QL L

aa t cos nπt dt Q

Using integration by parts

t 2t 2 aa ° sin πtù ° cos nπt¸ ° R R sin nπt¸ nπ n π n π Q Q Q

aa

a

Therefore,

f(t) Z

Solution:

0cos(nπ) cos(nπ)2 aa

4

n π

cos nπ

4 1 4 1 4 ; a ; aR ; aV ; a π π 9π 4π 25π

1 4 1 1 1 1 Scos πt cos 2πt cos 3πt cos 4πt cos 5πt ò T 3 π 4 9 16 25 1 È t È 1, p 2

Since the graph is symmetrical to the origin, it is an odd function ï aa 0 and a 0

ba

1 L nπt f(t) sin dt L QL L p 2L

2 2L L1

π L R nπt t sin dt 2 QL L

189

ba

Page

4.) f(t)

2

n π


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Using integration by parts

t 2t 6t 4 ° ° ° ° ba cos πtù sin nπtù R R cos nπt¸ V V sin nπt¸ nπ n π n π n π Q Q Q Q

π 12 2 q cos nπ R R cos nπr 2 nπ n π 1 6 ba cos nπ R cos nπ n n π

ba

b S1 f(t) qS1

6 1 3 1 2 1 3 1 6 T ; b S T ; bR S T ; bV S T ; b S T π 2 4π 3 9π 4 32π 5 125π Therefore,

6 1 3 1 2 1 3 1 6 T sin πt S T sin 2πt S T sin 3πt S T sin 4πt S T sin 5πt ò r π 2 4π 3 9π 4 32π 5 125π

5.) f(t) sin πt 0 È t È 1, p 1

1 L f(t) dt 2L QL p 2L 1 2L 1 L 2 1 sin πt dt a 1 2 A2C 1 a 0cos πt2 π 1 a 0cos π 12 π 2 a π 1 L nπt aa f(t) cos dt L QL L 1 aa sin πt cos 2nπt dt 1 2 a

190

aa 2 sin πt cos 2nπt dt

Page

Solution:



Using integration by parts 2 2 ° ° aa sin πt sin 2nπt¸

cos πt cos 2nπt¸ π(2n 1) n(2n 1)π 2 0cos 2πn cos 02 aa n(2n 1)π 4 4 4 a 0; a ; aR 0; aV ; a 0; a~ 3π 15π 35π f(t)

2 4 1 1 1 S cos 2πt cos 4πt cos 6πt ò T π π 3 15 35

6.) f(t) cos πt È È , p 1 Solution:

1 L f(t) dt a 2L QL p 2L

1 2L L

1 2

1 cos πt dt a 1 2 A2C Q

a

1 0sin πt2 π Q

1 01 12 π 2 a π

a


1 L nπt aa f(t) cos dt L QL L

Page

191

1 aa cos πt cos 2nπt dt 1 Q 2 Final answers are shaded in green.


Using integration by parts

cos πt sin 2nπt 2 0cos πt cos 2nπt2

π(2n 1) n(2n 1)π Q 2 0cos πn cos(πn)2 aa 2n(2n 1)π 4 aa cos πn n(2n 1)π 4 4 4 a 0; a ; aR 0; aV ; a 0; a~ 3π 15π 35π

aa

f(t)

2 4 1 1 1

S cos 2πt cos 4πt cos 6πt ò T π π 3 15 35

7.) f(t) |t| 1 È t È 1, p 2 Solution:


a

1 L f(t) dt 2L QL

p 2L

2 2L L1

1 a |t| dt 2 Q 1 |t| a s t 2 2 Q

Page

192

1 1 1 a q r 2 2 2 1 a 2


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 L nπt aa f(t) cos dt L L QL

aa |t| cos nπt dt Q

a

1 t 1t

Solution:

4 4 4 ; a 0; aR ; aV 0; a π 9π 25π 1 4 1 1 Scos πt cos 3πt cos 5πt ò T 2 π 9 25


Since the graph is symmetrical to the y-axis, it is an even function a

ba 0

1 L f(t) dt 2L QL p 2L 2 2L L1

1 ( ) a s 1 t dt (1 t) dtt 2 Q

1 t t a «°t ù °t ù ¬ 2 2 Q 2 1 1 1 a q1 1 r 2 2 2 1 a 2

193

Page

8.) f(t) Ò

f(t)

t 2 aa ° sin nπt cos nπt¸ nπ n π Q 2 aa ° cos nπt¸ n π Q 2 aa (1 cos nπ) n π



1 L nπt aa f(t) cos dt L L QL

aa (1 t) cos nπt dt (1 t) cos nπt dt Q

aa cos nπt dt t cos nπt dt cos nπt dt t cos nπt dt Q

Q

1 t 1 1 t 1 aa ° sin nπt¸ q sin nπt cos nπtr ° sin nπt¸ q sin nπt cos nπtr nπ nπ n π nπ nπ n π Q Q

aa

a

f(t)

2

n π

cos nπ

aa

2

n π

cos nπ

(1 cos nπ)

2

n π

4 4 5 ; a 0; aR ; aV 0; a π 9π 25π

1 4 1 1

Scos πt cos 3πt cos 5πt ò T 2 π 9 25

9.) f(t) 1 t 1 È t È 1, p 2 Solution:

2

n π

a


1 a (1 t ) dt 2 Q

1 tR a st t 2 3 Q 1 1 1 a S1 1 T 2 3 3 2 a 3

Since the graph is symmetrical to the y-axis, it is an even function

Page

ba 0

194


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 L nπt aa f(t) cos dt L L QL

aa (1 t ) cos nπt dt Q

aa cos nπt dt t cos nπt dt Q

Q

1 t 2t 2 ° ° ° ° aa sin nπt¸ sin πtù cos nπt¸ R R sin nπt¸ nπ nπ n π n π Q Q Q Q 2 aa R R 0cos nπ cos(nπ)2 n π

Solution:

4

n πR

cos πn

4 1 4 1 4 ; a ; aR ; aV ; a π π 9π 4π 25π

2 4 1 1 1 1

Scos πt cos 2πt cos 3πt cos 4πt

òT 3 π 4 9 16 25 cos 5πt 2 È t È 0 û p4 0ÈtÈ2

a


1 a s 0 dt t dtt 4 Q

1 t a s t 4 2 1 a 2

1 nπt aa t cos dt 2 2 1 2t 4 aa q sin πt cos nπtr 2 nπ n π

195

0 10.) f(t) Ò t

aa

Page

f(t)

a



a

aa

2

n π

(cos 2nπ 1)

4 4 4 ; a 0; aR ; aV 0; a π 9π 25π 1 L nπt ba f(t) sin dt L L QL 1 nπt ba t sin dt 2 2 1 2t 2 ba s° cos nπt¸ sin πtt 2 nπ n π ba

2 cos nπt nπ

1 4 πt 1 3πt 1 5πt Scos cos

cos

òT 2 π 2 9 2 25 2 2 πt 1 2πt 1 3πt 1 4πt 1 5πt

Ssin sin

sin sin

sin

òT π 2 2 2 3 2 4 2 5 2 f(t)

b

2 1 2 1 2 ; b ; bR ; bV ; b π π 3π 2π 5π

Example 3.3 Find the Fourier series expansion of the function

2 È t È 1 1 È t È 1ô p 4 1ÈtÈ2

4 2L L2

Since the graph of the function is symmetrical to the y-axis, it is an even function Final answers are shaded in green.

196

p 2L

Page

Solution:

0 f(t) ók 0

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS ï ba 0

aa

1 . f(t) dt 2L ð

Q 1 s 0 dt k dtt 2(2) Q Q

aa

1 aa 0°kt|Q 2 4

k aa 01 12 4 aa

aa

2k 4 k 2

nπt 1 . aa f(t) cos dt L L ð

1 nπt dt 0t aa s0 k cos 2 2 Q

nπt k t aa s cos 2 Q 2

k °2 nπt s ¸ t aa sin 2 nπ 2 Q aa

(πn) k πn ssin t

sin nπ 2 2

n 1; b

2k π

n 2; b 0

aa

(πn) πn sin 2 2 2k πn zsin { nπ 2

n 3; bR

2k 3π

n 4; bV 0


197

since, sin

Page

aa

k ° nπt ssin ¸ t nπ 2 Q

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS n 5; b

2k 5π f(t)

k 2k π 1 3π 1 5π

Scos t cos t cos t ò T 2 π 2 3 2 5 2

Drill Problem 3.3

Direction: Determine whether the following functions are even or odd then find its Fourier series expansion. Terms involving cos 5t and sin 5t must be included.

1.) f(t) π |t| Solution:

πÈ t È p

Since the graph is symmetrical to the y-axis, it is an even function Then, solve for a

ba 0

a

1 (π |t|) dt 2π Q

|t| 1 sπt t a 2π 2 Q

a

1 π π (π π 2π 2 2

aa

1 (π |t|) cos nt dt π Q

π 2π π a 2

1 s π cos nt dt t cos nt dtt π Q Q

1 ° t 1 sπ sin nt|Q E° sin nt¸ ° cos nt¸ Ft π nπ n Q Q


198

a

Page

aa

aa

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2 (1 cos nπ) nπ 4 4 4 a ; a 0; aR ; aV 0; a π 9π 25π aa

Therefore,

2.) f(t) 2t|t|

f(t)

π 4 1 1

Scos t cos 3t cos 5t ò T 2 π 9 25

1 È t È 1

Solution:

Since the graph is symmetrical to the origin, it is an odd function ï aa 0 and a 0

1 L nπt ba f(t) sin dt L QL L L1

ba 2 q° a cos nπt± 7

ba 2 z0

Q

ì(Qa) a

Q

°a77 sin nπt±

aZZ

cos 0

Q

°

°aZZ cos nπt± r 2 q° a cos nπt± °a77 sin nπt±

aZZ

aZZ

cos nπt± r

Q

cos(nπ){ 2 z

7

a

cos nπ

aZZ

cos nπ

aZZ

cos 0{

b

ba 4 q

4 cos nπ 4 cos π R R R Rr n n π n π

4 2 4 1 4 (π 4); b ; bR (9π 4); bV ; b (25π 4) R R π π 27π π 25πR Final answers are shaded in green.

199

2 cos(nπ) 4 4 2 4 4 R R cos 0 R R cos(nπ) cos nπ R R cos(nπ) R R cos 0 nπ n π n π nπ n π n π

Page

ba

ba 2 t sin nπtdt 2 t sin nπtdt

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS f(t) Z z(π 4) sin πt (9π 4) sin 3πt (25π 4) sin 5πt ò { V

t

3.) f(t) ó πt Solution:

z Asin 2πt sin 4πt sin 6πt ò C{ R

ÈtÈ

ÈtÈ

Rô

Since the graph is symmetrical to the origin, it is an odd function ï a 0 and aa 0

ba

1 . f(t) sin nt dt π ð R

1 Q ba x t sin nt dt (π t) sin nt dty π

R

R

Q 1 t cos nt 1 cos nt t cos nt 1 ¸ q ba q

sin ntr °

sin ntr π n n n n n

R

Q 1 1 1 ba x° sin nt¸ ° sin nt¸ y π n n

1 πn 3πn q3 sin r sin n π 2 2

f(t)

4 1 1 Ssin t sin 3t cos 5t ò T π 3 25

200

4 4 4 ; b 0; bR ; bV 0; b π 9π 25π

Page

b

ba


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 4.) f(t) Ëπe πe

Q

Solution:

π È t È 0Ì 0ÈtÈπ


1 . f(t) dt a 2π ð

1 Q s πe dt πe dtt a 2π Q

1 a 0e e e e 2 2 2 a 0e 12 2 a 0e 12

π Q aa e cos nt dt e cos nt dt π Q

aa °

neQ eQ eQ eQ cos ntù ° cos ntù °

cos ntù ° cos ntù n 1 n(n 1) n 1 n(n 1) Q Q

aa a(a7 P) 01 e cos(nπ)2 a(aP) 01 e cos(nπ)2 aP 0e cos nπ 12 a(a7 P) 0e cos nπ

2 (e cos nπ 1)

1)

n(n

1 2 1 2 (e 1); a (e 1) a (e 1); a (e 1); aR ; aV 5 5(e 1) 17 26

2 1 2 1 f(t) q(e 1) (e 1) cos t (e 1) cos 2t (e 1) cos 3t (e 1) cos 4t (e 1) ò r 5 4 17 26


201

aa

12

a

Page

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2 5.) f(t) Ò 0

Solution:

2ÈtÈ0 û 0ÈtÈ2 a

1 . f(t) dt 2L ð p4 L2

1 1 a 2 dt 0 dt 4 Q 4 1 a ° 2t¸ 4 Q

1 a 00 22 2 a 1

aa 0

Since it is odd function

1 nπ ba f(t) sin t dt L ð L nπ 1 ba 2 sin t dt 2 2 Q

ba

2 nπ zcos t{ nπ 2 Q

ba

2 01 cos nπ2 nπ

4 4 4 b ; b 0; bR ; bV 0; b π 3π 5π 4 πt 1 3πt 1 5πt f(t) 1 Ssin sin

sin

òT π 2 3 2 5 2


202

2 0cos 0 cos (nπ)2 nπ

Page

ba


Example 3.4

Find the Fourier transform of f(t) 1 if |x| È 1 and f(t) 0 otherwise. Solution:

F(ω)

F(ω)

F(ω)

1

√2π 1

f(t)eQY¯ dt

√2π 1

Q

s eQY¯ dtt

1 QY¯ ∞ ^ T \e 1 √2π jω S

1 F(ω) q (eY¯ eQY¯ )r jω

since 2j sin ω eY¯ eQY¯ Therefore,

F(ω)

2j sin ω q r jω √2π 1

2 sin § F(ω) | S T π §

Drill Problem 3.4

Direction; Find the Fourier transform of the following functions using the Fourier transform integral.

|t| ü a

Î otherwise

F(ω)

1

√2π

f(t)eQY¯ dt

203

Solution:

||

Q

Page

1. f(t) Í 1 0


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS F(ω)

For |t|eQYý

F(ω)

1

1

√2π

E1 Q

|t| QY¯ Fe dt a

1 s eQY¯ dt |t|eQY¯ dtt a Q √2π Q

Performing integration by parts u |t|

du dt

|t|eQY¯ dt Q

dv eQY¯

|t|

jω

|t|eQY¯ dt Q

1 QY¯

e dt jω Q

1 QY¯ e jω

|t| QY¯ 1 e eQY¯ j ω jω

|t|eQY¯ dt Q

e

QY¯

v

|t| QY¯ 1 e

eQY¯ jω ω

|t| 1 |t|eQY¯ dt ° eQY¯ ù ° eQY¯ ¸ jω ω Q

1

√2π

√2π

F(ω) 2.) f(t) eQ|| Solution:

q

s°

1 QY¯ ° |t| QY¯ 1 ¸ e ù ° eQY¯ ¸ t e jω ω jω

1 QY¯ 1 a 1 1 e

eQY¯ eQY¯ r jω jω ω ω jω

1 a QY¯ 1 ueQY¯ 1v s S t Te

jω jω ω √2π 1

F(ω)

F(ω)

1

√2π 1

√2π

f(t)eQY¯ dt Q

204

F(ω)

1

eQ|| eQY¯ dt Q

Page

F(ω)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS F(ω) 1

F(ω) ° F(ω)

Solution:

S

√2π

1 È t È 0û otherwise

F(ω)

F(ω)

Performing integration by parts

S

F(ω)

F(ω) °

F(ω)

1 T 0eQ eQ 2 a jω

F(ω) 0

√2π 1

√2π 1

√2π

ut

du dt

1 T eQ(PY¯)|| ¸ a jω Q

1

F(ω)

Q

f(t)eQY¯ dt Q

teQ eQY¯ dt Q

teQ(QY¯) dt Q

dv eQ(QY¯) dt

v

1 eQ(QY¯) 1 jω

t 1 eQ(QY¯) eQ(QY¯) dt j jω 1

jω Q

t 1 Q(QY¯) ¸ Q(QY¯) ¸ ° e e (1 jω) j jω Q Q

1 1 1 ePY¯ eQ(QY¯)

(1 jω) (1 jω) 1 jω

F(ω)

ePY¯ 1 u1 ePY¯ v

1 jω (1 jω)

F(ω)

1

ePY¯ 1 ePY¯ t

√2π 1 jω (1 jω) s

205

Q

1

√2π

eQ(PY¯)|| dt

Page

3.) f(t) Ò te 0

√2π

1



F(ω)

Performing integration by parts

1

√2π

|t|eQY¯ dt Q

ut

du dt

dv eQY¯ dt

F(ω)

|t|

jω

F(ω)

eQY¯

|t|

jω

F(ω)

eQY¯

Q

1 QY¯ e dt jω

v

1 QY¯ e jω

1 QY¯ e dt jω Q

|t| QY¯ 1 e eQY¯ jω j ω

|t| QY¯ 1 QY¯ ° ° ( ) ù e ¸ F ω e ω jω Q Q

F(ω)

1 QY¯ 1 Y¯ 1 1 e

e eQY¯ eY¯ jω jω ω ω

since: sin θ

eY eQY ; 2j sin θ eY eQY 2j

2 sin θ

F(ω)

1

eY eQY j

2j sin jω 2 q sin jω r ω √2π ω

2 sin jω j sin jω r F(ω) | q π ω ω

206

Solution:

1 ü t ü 1û otherwise

Page

|| 4.) f(t) Ò t 0


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1

0ütü

5.) f(t) Ê1 ü t ü 0þ 1

F(ω)

F(ω)

1

√2π

« e Q

Q

QY¯

dt eQY¯ dt¬

1 1 QY¯ ° x e ¸ ° eQY¯ ¸ y F(ω) jω √2π jω Q

F(ω)

1

√2π

f(t)eQY¯ dt

1 1 Y¯ 1 QY¯ 1 q e e r jω jω √2π jω jω 1

F(ω)

2 1 Y¯ 1 QY¯ e e r jω √2π jω jω 1

q

eY eQY since: cos θ 2 2 cos θ eY eQY

2 cos

F(ω)

Y¯ Y¯ jωa e eQ 2

2 jωa q 2 cos r 2 √2π jω

F(ω)

1

1

1 jωa q cos r 2 √2π jω

207

Solution:

otherwise

Page

0



Example 3.6 If the function f(t) has the Fourier Transform (ω), use the linearity and shifting in the frequency domain properties of the Fourier Transform to find 0f(t) cos at2 Solution:

since, cos at

0f(t) cos at2

eQ e , use the linearity theorem: 2

0af(t) bg(t)2 a (ω) bG(ω)

0f(t) cos at2 sf(t)

By shifting in the frequency domain, Therefore, And

\eY f(t)^ (ω a)

s

eQ (ω a) f(t)t 2 2

e (ω a) s f(t)t 2 2 (ω)

Example 3.7

1 ü t ü 1 û plot f(2t) and f A C and find their Fourier Transform. Use the results of otherwise

Solution:

(ω)

1

√2π

208

Example 3.4

(ω a) (ω a)

2 2

f(t)eQY¯ dt

Page

1 If f(t) Ò 0

eQ e t sf(t) t 2 2

Q



(ω)

(ω)

(ω)

1

√2π

1

√2π

1

√2π

eQY¯ dt

q

q

Q

1 r \eQY¯ ^Q jω

1 r \eQY¯ eY¯ ^ jω

since, eY¯ eQY¯ 2j sin ω (ω)

1

2j sin ω r jω √2π q

2 sin ω r (ω) | q π ω

Using the property of time scaling, We found out, and

0f(ct)2

ω AcC c

ω 2 sin 2 y 0f(2t)2 | x π ω

t 2 sin 2ω r qf S Tr | q 2 π ω

Example 3.8

1 1ütü1 û using the results in Example 3.4 and the Find the Fourier Transform of f(t) Ò 0 otherwise Differentiation Property of the Fourier Transform 1

√2π

f(t)eQY¯ dt

209

(ω)

Q

Page

Solution:



(ω)

1

√2π

Performing integration by parts,

teQY¯ dt Q

ut

du dt

dv eQY¯ dt

v

1 QY¯ e jω

1 QY¯ t QY¯ ° ° ¸ e ¸ (ω) e ω jω Q Q

(ω)

1 QY¯ 1 Y¯ 1 1 e e eQY¯ eY¯ jω jω ω ω

(ω)

1 1 QY¯ \e

eY¯ ^ \eQY¯ eY¯ ^ jω ω

since, eQY¯ cos ω j sin ω

√2π

q

1 1 (cos ω j sin ω cos ω j sin ω) (cos ω j sin ω cos ω j sin ω)r jω ω (ω)

(ω)

1

√2π 1

√2π

q

q

2 2j (cos ω) (sin ω)r jω ω

2ω 1 (cos ω) 2j(sin ω)r j ω

Applying the Differential Property of Fourier Transform, (ω)

1

0tf(t)2 j ½ (ω)

√2π

q

2 1 (j sin ω) 2j ω cos ωr jω ω

2 ω cos ω sin ω r (ω) | q π ω

210

1

Page

(ω)

eY¯ cos ω j sin ω



Drill Problem 3.5 Prove rules 3.1.1 through 3.1.10 using the Integral for the Fourier Transform and its properties. 3.1.1

Ò1 0

b È t È bû otherwise

(ω)

1

√2π

(ω)

f(t)eQY¯ dt Q

1

eQY¯ dt

√2π

Q

eQY¯ s t (ω) √2π jω Q

(ω)

1

1

jω√2π

\eQY¯ eY¯ ^

since, eY¯ eQY¯ 2j sin bω 2j sin bω

(ω)

b È t È cû otherwise

(ω)

1

√2π

(ω)

(ω)

f(t)eQY¯ dt

1

Q

√2π

(ω)

(ω)

1

eQY¯ dt

eQY¯ t √2π jω 1

s

jω√2π

1

jω√2π

\eQY¯ ^

\eQY¯ eQY¯ ^

211

Ò1 0

2 sin bω r (ω) | q π ω

Page

3.1.2

jω√2π



eQY¯ eQY¯

(ω) Q

t É 0û a É 0 otherwise

(ω)

(ω)

(ω)

1

√2π

(ω)

3.1.6

Òe 0

b È t È cû otherwise

1

√2π 1

√2π

s°S

1

(ω)

√2π

(ω)

(ω)

(ω)

1

eQ(PY¯) dt

1 Q(PY¯) ¸ t Te a jω

1

1

1

f(t)eQY¯ dt

√2π

1

1 T (0 1) a jω

√2π(a jω)

1

(ω)

Q

S

√2π

(ω)

f(t)eQY¯ dt

√2π

Q

e eQY¯ dt

e(QY¯) dt

1 T °e(QY¯) a jω √2π S

1 T \e(QY¯) e(QY¯) ^ a jω √2π S

(ω)

e(QY¯) e(QY¯) √2π(a jω)

212

Òe 0

Page

3.1.5

jω√2π


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 3.1.7

Ëe 0

Y

b È t È bÌ otherwise

(ω)

(ω)

1

√2π 1

(ω) 1

Q

eY eQY¯ dt

√2π

1

1

Q

e(Q¯)Y dt

√2π

Q

1 T °e(Q¯)Y Q √2π j(a ω)

(ω)

(ω)

f(t)eQY¯ dt

S

1 T \e(Q¯)Y eQ(Q¯)Y ^ ( √2π j a ω) S

By shifting in frequency domain as eY¯ eY¯ 2j sin ωb where, eY¯ 0f(t)2 (ω a)

Ëe 0

Y

2j sin b(ω a) t j(ω a) √2π 1

s

π sin b(ω a) t (ω) } s 2 ωa

b È t È cÌ otherwise

(ω)

(ω)

√2π

(ω)

(ω)

(ω) Ë

1

1

1

1

√2π 1

f(t)eQY¯ dt Q

eY eQY¯ dt

√2π

e(Q¯)Y dt

1 T °e(Q¯)Y √2π j(a ω) S

1 j T \e(Q¯)Y e(Q¯)Y ^Ì j √2π j(a ω) S


213

3.1.8

(ω)

Page

Then,

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Multiplying the whole equation by Y (ω)

Where j 1

3.1.9 Q¢

7

( É 0)

Y

1 S T \e(Q¯)Y e(Q¯)Y ^ √2π j (a ω) 1

(ω)

j

eY(Q¯) eY(Q¯) t aω √2π

(ω) (ω)

(ω)

(ω)

1

√2π

s

1

√2π 1

√2π 1

√2π

eQ eQY¯ dt

eQ(

(ω)

(ω)

7 QY¯)

dt

dt

√2π

1

Q¯7 V

e

7

¯7 ¯7 QSP 7 T P 7 dt e

úa ¢

(ω)

1

√2π

214

¢

sin at QY¯ e dt t

Page

3.1.10

√2π

eQ(

7 QY¯)

Y¯ Y7 ¯7 Y7 ¯7 Q(7 Q P 7 Q 7 ) e dt

1

7


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1

(ω)

(ω)

√2π

(ω)

e

(QY¯)

s

e

e 2t

Q QY¯

t s

2t

eQ(PY¯) t 2t

t¬

1 ∞ 1 ∞ «q e(QY¯) r q e(QY¯) r ¬ 0 a jω 0 √2π a jω π } (ω) 2 0

.

|ω| È a °

|ω| É a

215

1

√2π

e e 2t

QY¯

«s 1

Page

(ω)

1

√2π

e eQ QY¯ ue dtv 2t



UNIT 4

Page

216

POWER SERIES AND POWER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS



Example 4.1

1.) Find the sum of the geometric series a ar ar ar R … ar aQ … aQ ∑ .Determine the necessary conditions for this series to converge and the value to a ar which this series converges. Solution:

Sn ar aQ aQ

Sn a ar ar … ar aQ

rSn ar ar ar R ar aQ … ar a Sn rSn a ar a

Sn (1 r) a(1 r a ) Sn

a(1 1 r a ) (1 r)

r2

a1

Sn 1 2 4 8 ò lim Sn

a

r 0.5

1(1 2a ) 12 a1

Sn 1 0.5 0.5R 0.5V ò lim Sn

a

1(1 0.5a ) 1 0.5

1 2 1 0.5 rÈ1

lim Sn

aQ

ar aQ

aQ

The nth partial sum is Sn

(Q w ) Q

converges to Sn

a 1r

a(1 r N ) 1r

r Û 1. The series converges when r È 1 and this

Q

, otherwise the series diverges.


217

N

ar aQ

Page

a

a 1r

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 2.) Using the results above, determine the sum of the geometric series whose first term is 1/9 and a common ratio 1/3. Solution:

a

Sn

1 9

r 1/3

Sn a ar ar ò ò

1 1 1 1 1 aQ

… S TS T 9 27 81 9 3 Sn

Sn

a 1r 1 9

1 1 3

1 Sn 9 2 3 Sn

1 6

3.) Using again the results of the first problem, determine the sum ∑

a

(1)a 5 5 5 5 1 a 5

…

5 S T 4 16 64 4 4a a5

Sn

r 14

a 5 1 1r a 4

218

V

Sn 4

Page

Solution:

.

(Q)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 4.) Find the sum of the series ∑ a a(aP) and ∑a (aQ)(aP) .

Solution:

~

a

1 n(n 1)

1 A B q r n(n 1)

n(n 1) n n 1 1 A(n 1) Bn n0 n1 A 1 B 1

1 1 1 1 1 1 1 1 1

S T S1 T S T S T … S T … 2 2 3 3 4 n n 1 n n 1

a

Sn 1

1 1 1 ∞ 1 n 1

lim Sn 1

a

a

aQ

a

aQ

1 1 1 1 1 1 1 1 1 T 3 qS1 T S T S T … S T …r 2n 1 2n 1 3 3 5 5 7 2n 1 2n 1 Sn 3 S1

a

1 T 2n 1

lim Sn 3

a

6 3 (2n 1)(2n 1)

219

3 S

6 1 1

3 S T (2n 1)(2n 1) 2n 1 2n 1

Page

1 1 n 1



Example 4.2

Direction: Determine the following series if it diverges using the nth term test for divergence. 1.) ∑ a n

n 01 4 9 … n . . 2

a

Since the nth term of the series is approaching infinity (© ∞), 2.) ∑ a

Therefore the series diverges.

aP a

a

3 4 5 6 n 1 n 1 r q2 … .

2 3 4 5 n n

Since the nth term is approaching to 1 A aP 3.) ∑ a (1)

aP a

Therefore the series diverges

1C,

(1)aP 01 1 1 1 1 1 … . (1)aP 2

a

The nth term doesn’t exist , therefore the series diverges.

Example 4.3 Direction: Test the following series for convergence using the integral test. 1. ∑ a a.

220

aa f(x)

Page

Solution:



f(x)dx converges N

Then

aa converges

aa

a

1 ; n

1 1 f(x) a x

1 f(x)dx ax (lnx) diverges x

a

1 diverges n

ï the series diverges

2.) ∑ a a7 .

a

aa

1 1 f(x) n x

f(x)dx

1 dx converges 1 x

1 f(x)dx q r

221

1 converges n

Page

Solution:



f(x)dx 00 12 1

ï the series converges

Example 4.4 Direction: Investigate the convergence of the following series, and if the series is convergent, find its sum. w P Rw

Solution:

aa

2a 5 3a

P

aaP aa

2aP 5 aP 3a 2 5 3a

p

2a · 2 5 3a · 2a 5 3a · 3

aaP

2aP 5 3aP

2aP 5 3a p · a 3aP 2 5

1 2a · 2 5 2n p ·ã ä 1 3(2a 5) 2n

Since p È 1,

2 52a

3A1 52a C

p

2 3

222

p lim

Page

1.) ∑ a



w P Rw

Solution:

a

aa

aa 1

(2n)! n! n!

(2n)! n! n!

(2n 2)! (n 1)! (n 1)!

(2n 2)! aa 1 n! n! lim a aa (n 1)! (n 1)! (2n)!

p lim

(n 1)! (n 1)(n)(n 1)(n 2) … (2)(1) (n 1)! (n 1)(n!) (2n 2)! (2n 2)(2n 1)(2n)!

(2n 2)(2n 1)(2n)! n! n! . a (n 1)n! (n 1)n! (2n)!

p lim

(2n 2)(2n 1) a (n 1)

p lim

2(2n 1) a n 1

p lim

1 4n 2 n p . n 1 1 n

p4

223

Applying L’ Hospital’s Rule

2 4n n p 1 1 n

Page

2.) ∑ a

Therefore, the series converges.


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS p lim

a

4n 2 n 1

p4

p É 1,

ï the series diverges Vw a!a! (a)!

Solution:

aaP p p

aaP aa

aa

4a n! n! (2n)!

4a (n 1)! (n 1)! (2n 2)!

4a (n 1)! (n 1)! (2n 2)! 4a n! n! (2n)!

4a (n 1)! (n 1)! (2n)! · a (2n 2)! 4 n! n!

since n! n(n 1)! (n 1)! (n 1)! n! (2n 2)! (2n 2)(2n 1)(2n)!

4aP (n 1)! n! (n 1)! (2n)! · (2n 2)(2n 1)(2n)! 4a n! n! 4aP (n 1)(n 1) p a 4 (n 1)(2n 1) p

p

4aP (n 1) 24a (2n 1)

4a · 4(n 1) 24a (2n 1)

224

p

Page

3.) ∑ a


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS p

2(n 1) 2n 1

1 2n 1 n lim · a 2n 1 1 n 2 2 p1 p

Using the nth term test

a

8 16 128 4a n! n! 4a n! n! r q2

….

(2n)! (2n)! 3 5 35

since n ∞, an 0, therefore the series diverges Since p1, the test is inconclusive.

Example 4.5 Investigate the convergence of the following series using root test. 1. ∑ a w a7

a

aa

n 2a

n 2a

p lim wfaa a

p lim

a

225

w n p lim | a a 2

√n 1 2 2

w

Page

Solution:



p lim √n 1 a

w

pÈ1

ï the series converges 2.) ∑ a a7

w

Solution: aa

2a n

w 2a p lim w√an lim | a a n

p lim

Since p É 1,

Pa

C

a

1 a aa S T 1 n

p lim wfaa |S a

Since p È 1

w

1 a 1 1 T 1 n 1 n n

p0

226

The series will diverge

The series will converge.

Page

3.) ∑ aP A

2

a na



Example 4.6

Direction: Investigate the convergence of the following series. aP 1.) ∑ a (1) a7

Solution:

(1)aP

a

aa 1

1 1 1

ò ò n 4 9

The series converges absolutely.

Q 2.) ∑ /© (1)

Solution:

(1)aP 1 1 1 1 1

ò

4 9 16 n n

(1)Q /© s1

By absolute convergence test

1 1 1 (1)Q t

2 3 4 ©

| | 0 1

1 1 1 1 2 3 4 ©

Page

For pÉ0, the series converge, for pÉ1, the corresponding series of absolute values ∑| |,converges, therefore the series converge absolutely. For 0Èpü 1, since the series that converge but does not converge absolutely, therefore it converges absolutely.

227

.



Drill Problem 4.1

A. Which of the following sequences converges? Find the limit of each convergent sequence 1.) aa 2 (0.1)a Solution:

Evaluating the limit

lim aa 2 (0.1)

a

Qa

As you approach n to infinity, the value will be two. Therefore the sequence converges.

Pa

lim aa S

a

1 2n 1n T 1 2n 1n

1 2 n lim aa E F 1 n 2n a lim aa a

2 2

lim aa 1 a

The sequence converges

228

Solution:

Page

2.) aa

lim aa 2

a



3.) aa A1 aC Solution:

a

lima aa A1 aC

lim aa a

7 ∞

lim aa 0 a

The sequence converges

B. Find the sum of the following series.

aQ

4.) ∑ a Vw q~ ~V ~ ò ~ A C Solution:

V

a 0, r

Sa

1 4

a for |r| È 1 1r

1 16 Sa 1 14 1 16 Sa 1 14 Sa

5.) ∑ a (VaQR)(VaP) Solution:

r

1 1


1 A B

(4n 3)(4n 1) 4n 3 4n 1 Final answers are shaded in green.

229

Page

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 A(4n 1) B(4n 3) n:

S

a

0 4A 4B

n : 4 A 3B A 1, B 1

1 1 1 1 1 1 1 1 T q1

òr 4n 3 4n 1 5 5 9 9 13 13 Sa lim q1 a

Sa 1

1 r 4n 1

C. Investigate the convergence or divergence of the following series. 6.) ∑ a A C

√

a

Solution:

Using root test:

1 a 1 1 | P lim faa lim S T lim a a √2 a 2 √2

Using Ratio test:

P

a a

Since P È 1, the series converges. aaP aa

1 aP 1 a 1 S T S T T √2 1 √2 a √2 1 1 a √2 S T S T √2 √2 S

since P È 1, then the series converges.

Solution:

Using ratio test:

Let aa

a a a

and aaP

a(aP) aP


230

a

Page

7.) ∑ a

w

w


aaP P a a

ln(n 1) n 1 Sln n ln 1T A n C ln n n 1 ln n n lim z

a

n 1n { n 1 1n

lim s

a

1 t 1 1n

since P 1, the series is inconclusive. By nth term test:

a

8.) ∑ a

ln n ln n r q0.353 0.352 0.351 0.32 …

n n

since n ∞, and aa ∞ then the series converges.

a

a7P

Solution:

By nth term test:

a

n

n 1 2 3 4 n r q

ò 2 5 10 17

1 n 1

since n ∞, and aa 0 then the series converges. n P lim faa lim } lim a a n 1 a

a

w

na

1 n

1a

since P 0, then the series converges.

1 1 1

Page

9.) ∑ a A1 RaC

w

231

By root test



Solution:

By root test

P lim wfaa lim |S1 a

a

The test is inconclusive, so By nth term test

S1

a

w

1 a 1 a T lim S1 T 1 a 3n 3n

2 25 519 1 a 1 a T s

ò S1 T t 3n 3 36 729 3n

since n ∞, and aa not approaches to ∞, then the series diverges.

10.) ∑ a Aa

Solution:

C 7

a

a

By nth term test:

1 1 a 1 8 80 1 1 a

S T s0

ò S T t n n 16 729 65536 n n

a

since n ∞, and aa 0 then the series converges.

By root test

w 1 1 a 1 1 a P lim wfaa lim |S T lim S T 0 a a a n n n n

since P È 0 then the series converges.

11.) ∑ a a(aP)!

232

a! a a

Page

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS By nth term test:

a

n! ln n n! ln n r q0 0.17 0.44 1.38 ò

n(n 2)! n(n 2)!

since n ∞, and aa not approaches to ∞, then the series diverges. Using ratio test:

Let aa a(aP) and aaP a! a a

aaP P a a

Ps

Ps

(n 1)! ln(n 1) (n 1)(n 3) n! ln n n(n 2)

(n 1)! (n 1)(n)!

ln(n 1) ln n ln 1

(n 1)n! (ln n ln 1) n(n 2) ts t (n 1)(n 3) n! ln n lim s

lim s

a

a!

(aP)(aPR)!

(n 1)! ln(n 1) n(n 2) ts t (n 1)(n 3) n! ln n

a

12.) ∑ a

(aP)! a(aP)

n(n 2) 1n t (n 3) 1n

n 2 t∞ 1 3n

since P ∞, then the series diverges.

a

Let aa

a! a

and aaP

(aP)! (aP)


Page

Using ratio test:

233

Solution:

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS (n 1)! aaP (n 1) P a n! a n Pq

(n 1)! n rz { (n 1) n!

lim n ∞

a

since P ∞, then the series diverges.

By nth term test:

a

n! n! q1 1 2 6 ò r n n

as n ∞, and aa not approaches to ∞, then the series diverges. aP A C 13.) ∑ a (1) a

Solution:

a

By nth term test:

(1)aP A

a

n a 1 1 27 16 n a C q

ò (1)aP A C r 10 10 10 25 1000 625

By alternating series convergence test:

|a | (1)aP A

a

n a 1 1 27 16 C q

r

10 10 25 1000 625

aP ∑ ln A1 C a (1)

Solution:

a

Page

14.)

234

since the series converge but doesn½ t converge absolutely, then the series converges conditionally.


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS By nth term test:

1 1

(1)aP ln S1 T q0.69 0.41 0.29 0.22 0.18 ò (1)aP ln S1 Tr n n

a

since the corresponding series of absolute values,

|aa | converges, therfore the series converges

a

absolutely

series absolute convergence test:

1

|aa | (1)aP ln S1 T 00.69 0.41 0.29 0.22 0.18 ò 2 n

By using alternating

a

Qa 15.) ∑ a n e

Solution:

Using integral test:

f(x) x eQ

Performing integral by parts

Let u x

du 2xdx

dv eQ

v eQ

x eQ 2 ªQ xeQ dx

du dx

dv eQ

v eQ

235

Let u x

xeQ 2 sxeQ eQ dxt

Page



x eQ 2xeQ 2 eQ dx

x eQ 2xeQ 2eQ

0x eQ 2

∞ ∞ ∞ 02xeQ 2 02eQ 2 1 1 1

∞eQ eQ 2∞eQ 2eQ 2eQ 2eQ f(x) eQ 2eQ 2eQ f(x) 5eQ f(x) 1.85

since f(x) 1.85 x eQ dx converges and

Q

also the n eQa also converges. a

Example 4.7 Direction: For what values of x do the following power series converge? Solution:

w a

(1)aQ

a

1 1 1 xa xa 0x x x R x V (1)aQ n 2 3 4 n

|aa | sx

aQ

xa aa n

p

x xR xV xa

òt 2 3 4 n aaP

aa 1 x aP n · aa n 1 xa

x aP n 1

236

aQ ∑ a (1)

Page

1.)


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS xn a n 1

p lim

By L’Hospital’s Rule

px

|x| È 1 1 È |x| È 1

Solution:

By ratio test

7wÐ6 aQ

(1)aQ

a

x aQ xR x sx …t 2n 1 3 5

|aa | 0x

a

x aa 2n 1

by L’ Hospital’s rule

wÐ6

xR x x aPQ

…

5 2n 2 1 3 aaP

x aPQ 2n 2 1

x aPQ 21 p 2n wÐ6 x 2n 1 aP x 2n 1 p · aQ 2n 1 x x (2n 1) p 2n 1

p x 1 ü x ü 1 1 1 1 as aa q1 … r 3 5 n

237

aQ ∑ a (1)

Page

2.)

1 1 1 xa

|aa | 1 (1)aQ 2 3 4 n 1 È |x| ü 1



w a!

x xR xa xa

s1 x ò t 2! 3! n! n!

a

By ratio test

aa

xa n!

aaP

x aP (n 1)! p xa n! x lim a n 1 p0

4.)

a ∑ a n! x

Solution:

By ratio test

By L’ Hospital’s rule

x aP (n 1)!

since p È 1 , for all values of x.

n! x a 01 x 2! x n! x a

a

aa n! x a aaÓ6 (n 1)! x aP p (n 1)! x aP /n! x a p lim x(n 1) a

px

all values of x except x 0

238

∑ a

Example 4.9

Page

3.)

1 1 1

aa q1 … r 3 5 n 1 ü x ü 1


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS Directions: Obtain the first and second derivative of the function.

1 f(x) 1 x x x R . . x a … x a 1x

for 1 È x È 1

a

f

1 f ½ (x) 0 1 2x 3x 4x R ò nx a Q nx aQ 2 (1 x) a

2 x) 2 6x 12x ò n(n 1)x aQ n(n 1)x aQ (1 x)R

½½ (

a

all for 1 È x È 1

Drill Problem 4.2 A. For the following series, find the series’ radius and interval of convergence.

a 1.) ∑ a (x 5)

Solution:

(x 5)a 0(x 5) (x 5) (x 5)R ò (x 5)a 2

a

aa (x 5)a P

aaP (x 5)aP

aaP (x 5)aP (x 5)a (x 5) aa (x 5)a (x 5)a P lim (x 5) a

P (x 5)

aa 10 100 1000 ò 10a

ï the series½ radius is 5 and the interval of convergence is 5 ü x ü 5 Final answers are shaded in green.

Page

5 ü x ü 5

239

r5


a a 2.) ∑ a (1) (4x 1)

Solution:

(1)a (4x 1)a 0(4x 1) (4x 1) (4x 1)R ò (1)a (4x 1)a 2 By alternating convergence test

|aa | 0(4x 1) (4x 1) (4x 1)R ò (4x 1)a 2

By ratio test

aa (4x 1)a P

aaP (4x 1)aP

aaP (4x 1)aP (4x 1)a (4x 1) aa (4x 1)a (4x 1)a P lim (4x 1) a

P (4x 1) r

1 4

1 1 üxü 4 4

aa 10 100 1000 ò 10a

R 1 1 1

aa Íq4 S T 1r q4 S T 1r q4 S T 1r ò Î 4 4 4

ï the series½ radius is

1 1 1 and the interval of convergence is ü x ü 4 4 4

240

aa 2 4 8 ò 2a

Page

a


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 3.) ∑ a

(RQ)w a

Solution:

3x 2 R 3x 2 a (3x 2)a 3x 2

T S T òS T t s(3x 2) S 3 n n 2

a

By ratio test

aa S

3x 2 a T n

aaP S

3x 2 aP T n 1

3x 2 aP aaP A n 1 C 3x 2 aP na P S T . aa 3x 2 a n 1 (3x 2)a A C n P

By L-Hospitals’ rule

(3x 2)a (3x 2) . n (n 1)(3x 2)a (3x 2) . n a (n 1)

P lim

P (3x 2) r

2 3

2 2 üxü 3 3

R 1 2 3A C 2 4 A4C 1 (4)a 2

aa Êq3 S T 2r x 3 y x y ò þ 2 3 3 n


4.) ∑ a aP aw

aa 4 8 ò

65 3

2 2 2 and the interval of convergence is ü x ü 3 3 3

Page

241

Solution:



a

By ratio test

aa

nx a x x 3x R nx a t s

ò n 2 5 n 2 3 2

nx a n 2

P

By L-Hospital’s rule

aaP

(n 1)x (aP) (n 3)

aaP (n 1)x (aP) n 2 . (n 3) aa nx a P

(n 1)x a x n 2 . (n 3) nx a

P

x(n 1)(n 2) n(n 3)

x(n 3n 2) x(2n 3) n 3 2n P 2x

ï the series½ radius is 0 and the interval of convergence is at all values of 2x except x 0 (Q)w (P)w

Solution:

a

(1)a (x 2)a (x 2) (x 2)R (1)a (x 2)a )

t s(x 2

ò n 2 3 n

By ratio test aa

(1)a (x 2)a n P

aaP

(1)aP (x 2)aP n 1

aaP (1)aP (x 2)aP n . a ( ) aa n 1 1 (x 2)a


242

a

Page

5.) ∑ a


P

(1)a (1)(x 2)a (x 2) . n (n 1)(1)a (x 2)a P

n(x 2) (n 1)

P lim

a


n(x 2) (n 1)

P (x 2)

ï the series½ radius is 2 and the interval of convergence is 2 ü x ü 2 B. For the following series, find the series’ interval of convergence and within this interval, the sum of the series as a function of x. (Q)7w Va

a

(x 1) (x 1)V (x 1)~ (x 1)a (x 1)a t

ò s 8 12 4n 4n 4

By ratio test aa

(x 1)a 4n P

aaP

(x 1)(aP) 4(n 1)

aaP (x 1)(aP) 4n . aa 4(n 1) (x 1)a (x 1)a (x 1) . 4n P 4(n 1) (x 1)a P

n(x 1) (n 1)

243

Solution:

Page

6.) ∑ a


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS n(x 1) a (n 1)

P lim


P (x 1) (x 1)(x 1) °x È 1|x È 1 1 È x È 1

ï the series½ radius is 1 and the interval of convergence is 1 ü x ü 1 (P)7w µw

Solution:

a

(x 1)a (x 1)a (x 1)Ra (x 1)a s1 t

ò 9a 9 81 9a

By ratio test aa

(x 1)a 9a P

aaP

(x 1)(aP) 9aP

aaP (x 1)(aP) 9a . aa 9aP (x 1)a P

(x 1)a (x 1) . 9a 4(n 1) (x 1)a

P

n(x 1) 9a . 9 (x 1)a (x 1) a 9


1 1 ÈxÈ 9 9

1 1 1 and the interval of convergence is È x È 9 9 9 Final answers are shaded in green.

244

P lim

Page

7.) ∑ a


√ 8.) ∑ a A 1C

Solution:

a

a

R

V

a

√x √x √x √x √x √x

E 1F «E 1F E 1F E 1F E 1F ò E 1F ¬ 2 2 2 2 2 2

a

By Root test:

lim wfaa p

a

√x p lim |E 1F a 2 w

a

√x P lim E 1F a 2 P

√x 1Sa 2

a1

√x E 1F 2 1

√x 1 2 1

4 È x È 4Sa

2

√x

The series of interval of convergence is: 4 È x È 4 the radius is 4 and Sa

√

a 9.) ∑ a (lnx)

(lnx)a 01 lnx 2lnx 3lnx ò 2 Sa

a1

Sa

a 1r

r lnx

245

a

1 1 lnx

Page

Solution:



the series of interval or convergence is: eQ È x È e and the Sa R

Solution:

a

xR 1 xR 1 xR 1 xR 1

E F «E F E F E F ¬ 3 3 3 3

a

a

a1

Sa Sa

1 xR 1 1 3

xR 1 3

1 3 xR 1 3

Sa By root test

r

R

3 4 xR

p lim wfaa a

xR 1 p lim |E F a 3 w

a

xR 1 a 3

p lim

1 È x ü 1

the series of interval is 1 È x ü 1 and the Sa

3 4 xR

246

C

ZP

Page

10.) ∑ a A

1 1 lnx

Example 4.10 Final answers are shaded in green.


Directions: Investigate the following series . 1.) f(x) x Solution:

Z

[

ò

f(x)

1 x

a2

f ¼ (2) ck k! 1 1 f(2) 1 k 0 f(x) f(2) x 2 0! 2 · 0! 1 ½() 1 f(2) 1 k 1 f(x) f x 4 1! 4 · 1! 1 2 ½½() 1 f¾¾(2) k 2 f(x) R f 4 2! 4 · 2! x 6 ½½½() 3 f¾¾¾(2) 1 k 3 f(x) V f x 8 3! 8 · 3! f ¼ (2) 1 c¼ (1)¼ · ¼P k! 2

f(x) ca (x a)a a

f

)

(x 2)a 1 (x 2) (x 2 a (x) s ( ) t

1 2 2 2R 2aP for 0 È x È 4

2.) Find the Taylor series expansion, and the Tyalor polynomials generated by f(x) e at x 0.

f(x) ca x a

aa c¼

f ¼ (0) k!

f(0) 1 0! f(0) k 1 f(x) e f¾(0) 1 1 1! f(0) 1 k 2 f(x) e f¾¾(0) 1 2! 2! k 0 f(x) e f(0) 1


247

a

Page

Solution:


(x)

a

e¼ f

1 k!

1 a x xR xa x 01 x ò 2 n! 2! 3! n! all values of x

3.) Find the Taylor series and Taylor polynomials of ()¨£ () ∑ ! . The Taylor polynomials generated by the function is

ò. 2 ©!

() 1

f(x) cos x

x0

a

ca c¼

f ¼ (0) k!

k 0 f(x) cosx f(0) 1

k 1 f ½ (x) sinx f ½ (0) 0

k 2 f ½½ (x) cosx f ½½ (0) 1

f(0) 1 0! f ½ (0) 0 1! f ½ (0) 1 2! 2 Final answers are shaded in green.

248

f(x) ca x a

Page

Solution:


aa

f ¼() xa a k! 2 k!

f(x) ca x a 01

a

x xV xa

ò a 4 48 2 n!

Drill Problem 4.3 A. Find the Maclaurin series (Taylor series at x 0) of the following functions 1.) f(x) eQ x 0 Solution:

f(x) ca x a a

ca c¼

f ¼ (0) k!

f(0) 1 0! f ½ (0) 1 k 1 f ½ (x) eQ f ½ (0) 1 1! 1! k 0 f(x) eQ f(0) 1 k 2 f"(x)e-x f(0) 1

Therefore the series will be,

f(x) Ca x a 0 1 x

a

x xR xa ò (1)a 2 2! 3! n! Final answers are shaded in green.

249

f ¼ (0) 1 (1)a k! k!

Page

ca

f"(0) 1 2! 3!


2.) f(x) e7

Solution:

f(x) Ca x a a

ca c¼

f ¼ (0) k!

k 0 f(x) e f(0) 1

f(0) 1 0! 1!

x x f(0) x k 1 f ½ (x) e f ½ (0) 2 2 1! 2.1!

k 2 f"(x)

x2 x x f(0) x e2 f"(0) 4 4 2! 4.2!

ca

Therefore, the series will be,

f ¼ (0) xa a k! 2 k!

f(x) Ca x a s1

a

Ca C¼

f ¼ (0) k!

1 (1 x)

f(0) 1

k 0 f(x)

k 1 f(x)

1 1 x

f(0) 1

f(0) 1 0!

f(0) 1 1! Final answers are shaded in green.

250

f(x) Ca x a

Page

3.) f(x) P

a

x xV x a

ò a t 4 48 2 n!

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS k 2 f"(x)

2(1 x) (1 x)R

Ca

Therefore, the series will be

f(0) 2

fk(0) n! (1)a k! n!

f(0) 1 2!

f(x) Ca x a 01 x x x R ò (1)a x a 2 a

4.) f(x) Q

Solution:

f(x) Ca x a a

Ca C¼

f ¼ (0) k!

k 1 f¾(x)

1 (1 x)

f(0) 1

f(0) 1 1!

k 3 f¾¾¾(x)

6 (1 x)V

f(0) 6

f(0) 1 3!

2 (1 x)R

Ca


f(0) 2

f(0) 1 2!

f ¼ (0) n! 1a k! n!

251

k 2 f"(x)

1 f(0) f(0) 1 1 1x 0!

Page

k 0 f(x)



5.) sin3x

k0

k1

k2

(x) Ca x a a

f(x) sin3x

f(0) 0

f(0) 3

f ½ (x) 3cos3x

f"(x) 9sin3x

k 3 f ½½½() 27cos3x

f(0) 27


f(0) 0 1!

f(0) 3 2! 2!

f(0) 0

f(0) 0 3!

(1)a 27 f(0) Ca . (3)a 3! n! 4!

f(x) Ca x a s a

(1)a 3a a 3x 27x R ò x t 2 2 n!

f(x) Ca x a a

Ca C¼

f ¼ (0) k!

k 0 f(x) coshx f(0) 1

k1

f ½ (x) sinhx

k 2f ½½ (x) coshx

f¾(0) 0

f ½½ (0) 1

Solution:

f(0) 1 0! 0!

f(0) ∞ 1!

f(0) 1 2! 2!


252

a

Page

6.) coshx

f(x) Ca x a (1 x x x R ò x a )

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS k3

f ½½½ (x) sinhx f ½½½ (0) 0

k 4 f ½½½½ (x) coshx Ca


f ½½½½ (0) 1

f ¼ (0) . 1 k! n!

f(x) Ca x a s1

a

f(0) ∞ 3!

f(0) 1 4! 4!

x xV xa

ò t 2 24 n!

7.) sinhx

(x) Ca x a

Ca C¼

f ¼ (0) k!

k 0 f(x) sinhx f(0) 0

k 1 f¾(x) coshx f¾(0) 1

k 2 f ½ ¾(x) sinhx f¾¾(0) 0

k 3 f¾¾¾(x) coshx f¾¾¾(0) 1

k 4 f¾¾¾¾(x) sinhx f¾¾¾¾(0) 0 Therefore the series will be

Ca C¼

1 n!

f(0) ∞ 0!

f¾(0) 1 1! 1!

f¾¾(0) ∞ 2!

f¾¾¾(0) 1 3! 3!

f¾¾¾¾(0) ∞ 4!

253

a

Page

Solution:



f(x) Ca x a sx

a

xR x 1

ò t 6 120 n!

8.) V 2 R 5 4 Solution:

(x) aa x a a

Ca C¼

f ¼ (0) k!

k 0; f(x) x V 2x R 5x 4 f(0) 4;

f(0) 4 0! 0!

k 1; f ½ (x) 4x R 6x 5 f ½ (0) 5;

f¾(0) 5 1! 1!

k 2; f ½½ (x) 12x 12x f ½½ (0) 0;

f¾¾(0) 0 2!

k 3; f ½½½ (x) 24x 12 f ½½½ (0) 12;

f¾¾¾(0) 12 3! 3!

k 4; f ½½½½ (x) 24

f ½½½½ (0) 24;

f¾¾¾¾(0) 25 4! 4!

Ca x a 04 5x 0 2x R x V ò 2

Page

a

254

Therefore, the sequence will be


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 9.) (x 1)

Solution:

f(x) aa x a a

Ca C¼

f ¼ (0) k!

k 0; f(x) 2(x 1) f(0) 2;

f(0) 2 0!

k 1; f ½ (x) 2x

f ½ (0) 0;

f¾(0) 0 1!

k 2; f ½½ (x) 2

f ½½ (0) 2;

f¾¾(0) 1 2!

k 3; f ½½½ (x) 0

f ½½½ (0) 0; Therefore, the series will be

Ca C¼

Ca x a q2 x ò

a

f¾¾¾(0) 0 3! 2 n!

2 a x r except odd numbers n!

10.) sin

255

(x) aa x a a

Page

Solution:



Ca C¼

f ¼ (0) k!

k 0; f(x) sin f(0) 0;

x 2

f(0) 0 0!

1 x k 1; f ½ (x) cos 2 2

1 f¾(0) 1 f ½ (0) ; 2 1! 2 · 1! k 2; f ½½ (x) f ½½ (0) 0;

1 x 42

f¾¾(0) 0 2!

1 x k 3; f ½½½ (x) cos 8 2

1 f¾¾¾(0) 1 f ½½½ (0) ; 3! 8 · 3! 8


(1)aQ a 1 xR

Ca x a s x

x t 2 12 2n · n!

a

11.) f(x) x 2x 4; a 2

f ¼ (2) C¼ k!

k 0; f(x) x R 2x 4 f(2) 8;

8 8 0!

f(2) 10;

256

k 1; f ½ (x) 3x 2 10 10 1!

Page

Solution:


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS k 2; f ½½ (x) 6x

f(2) 12;

12 6 2!

k 3; f ½½½ (x) 6 f(2) 6;


6 1 3!

f(x) Ca (x a)a s8(x 2) 10(x 2) 6(x 2)R (x 2)V ò a

f a (2) (x 2)a t n!

12.) f(x) 2x R 2x 3x 8 Solution:

C¼

f ¼ (1) k!

k 0; f(x) 2x R 2x 3x 8 f(1) 1;

f(1) 1 0! 0!

k 1; f ½ (x) 6x 4x 3 f ½ (1) 13;

f¾(1) 13 1! 1!

k 2; f ½½ (x) 12x 4

f ½½ (1) 16;

f ½½ (1) 16 2! 2!

k 3; f ½½½ (x) 12

257


f ½½½ (1) 12 3! 3!

Page

f ½½½ (1) 12;



f(x) Ca (x a)a s(x 1) 13(x 1) 8(x 1)R 2(x 1)V ò a

13.) f(x) x V x 1; a 2 Solution:

f a (1) (x 1)a t n!

f ¼ (2) C¼ k!

k 0; f(x) x V x 1 f(2) 21;

f(2) 21 0!

k 1; f ½ (x) 4x R 2x

f ½ (2) 36; f

f ½ (2) 36 1! 1!

k 2; f(x) 12x 2

½½(Q)

f ½½ (2) 26 26; 2! 2!


f(x) Ca (x a)a s21(x 2) 36 (x 2) 26(x 2)R ò a

f a (2) (x 2)a t n!

C¼

f ¼ (1) k!

k 0; f(x) 3x x V 2x R x 2 f(1) 7;

f(1) 7 0! 0!

Page

Solution:

258

14.) f(x) 3x x V 2x R x 2; a 1


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS k 1; f ½ (x) 15x V 4x 6x x f ½ (1) 24;

f ½ (1) 24 1! 1!

k 2; f ½½ (x) 60x R 12x 12x 1 f ½½(Q) 83;


f ½½ (1) 83 2! 2!

f(x) Ca (x a)a 07(x 1) 28(x 1) a

15.) f(x) 7 ; a 1

Solution:

k 0; f(x)

1 x

k 1; f(x)

2 xR

k 2 f(x)

6 xV

83 f ¼ (1) (x 1)R ò (x 1)a 2 2 n!

f(1) 1;

f(1) 1 0!

f¾(1) 2

f¾(1) 2 1!

f¾¾(1) 6

f¾¾(1) 3 2!


f(x) Ca (x a)a 0(x 1) 2(x 1) 3(x 1) ò n(x )a 2 a

Page

16.) f(x) Q a 0

259



Solution:

Ca

f ¼ (0) n!

k 0; f(x)

x 1x

f(0) 0;

f(0) 0 0!

f ½ (0) 1;

f¾(0) 1 1!

k 1; f ½ (x)

1 (x 2)

k 2; f ½½ (x)

f¾¾(0) 1 2!

f ½½ (0) 2;

k 3; f ½½½ (x) f ½½½ (0) 6; Ca


2 (1 x)R

6 (1 x)V

f¾¾¾(0) 1 3!

n (1)aQ n!

f(x) Ca (x a)a x x x R x V ò a

Ca

f ¼ (2) n!

k 0; f(x) e

f(2) e 0!

260

f(2) e ;

Page

17.) f(x) e ; a 2

n (1)aQ n!


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS k 1; f ½ (x) e

f¾(2) e ;

f¾(2) e 1!

k 2; f ½½ (x) e

f ½½ (2) e ;

f ½½ (2) e 2! 2!

C¼

f(x)


a

f(x) e e (x 2)

e n!

e (x 2)a n!

e (x 2) ò e n! (x 2)a 2

18.) f(x) 2 ; a 1 Solution:

k 0; f(x) 2x

f(0) 1;

k 1; f ½ (x)

2x ln 2

k 2; f ½½ (x)

2x ln 2

k 3; f ½½½ (x)

2x lnR 2

1 f ½ (0) 1 ; ln 2 1! 1! ln 2

f ½½ (0)

1 f ½½ (0) 1 ; 2 ln 2 2! 2! 2 ln 2

f ½½½ (0)

1 f ½½½ (0) 1 ; 3 ln 2 3! 3! 3 ln 2

261

f 0)

Page

½(

f(0) 1 0! 0!


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 1 n! n ln 2

Ca

Ca (x a)a q1

a

1 1 1 1 (x 1)

(x 1)

(x 1)R ò (x 1)a r ln 2 4 ln 2 36 ln 2 n! ln 2

Example 4.11 1.) Solve the equation y ½½ 4y 0 near the ordinary point x 0 Solution:

y ½½ 4y 0

y aa x a

n(n 1)aa x

a

n(n 1)aa x

aQ

aQ

4 aa x a 0 a

4 aaQ x aQ 0 a

a

n(n 1)aa 4aaQ 0 aa

4aaQ n(n 1)

for n 2

262

0n(n 1)aa 4aaQ 2x aQ 0

Page

a

a



aV a¼

a a¼P

4a 3·2

4aR 5·4

4a¼Q (2k 1)2k

(1)¼ 4¼ a a ò a¼Q (2k)!

(1)¼ 4¼ a (2k)!

for k 1

1 y a cos 2x a sin 2x 2

for k

y aa x a

y a a¼ x ¼

¼

a

a x a¼P x ¼P ¼

(1)¼ 4¼ x ¼P (1)¼ 4¼ x ¼ ¬ a «x

¬ y a «1

(2k)! (2k 1)! ¼

¼

(1)¼ (2x)¼ (1)¼ (2x)¼P 1 ¬ a «2x

¬ y a «1

(2k)! (2k 1)! 2

¼

¼


263

Page

a¼P

(1)¼ 4¼ a (2k 1)! 1

a¼

4a 4·3

4a¼Q 2k(2k 1)

aR

a aV ò a¼

4a 2·1


2.) Solve the equation (1 x )y ½½ 6xy ½ 4y 0near the ordinary point x 0 Solution:

(1 x )y ½½ 6xy ½ 4y 0

a

a

n(n 1)aa x

a

n(n 1)aa x

a

n(n 1)aa x

a

n 2:

aQ

aQ

aQ

6 aV a 4 8 a~ aV 6

a

a

a

(n 5n 4)aa x a 0 a

(n 1)(n 4)aa x a 0 a

(n 1)(n 2)aaQ x aQ 0 a

n 0:

n 1:

0 · a 0, 0 · a 0,

n(n 1)aa (n 1)(n 2)aaQ 0 n(n 1) Û 0

n 2: 4 a a 2

n(n 1)aa x 6naa x 4aa x a 0 a

aa

n 2 aaQ n

a¼

2k 2 a¼Q 2k

5 aR a 3


264

n(n 1)aa x

aQ

a

Page

y aa x a

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 7 a aR 5 9 a a 7 k 1:

a aV a~ ò a¼ k 1:

k 1:

a¼P

a¼ (k 1)a

5 · 7 · 9 ò (2k 3) a 3 · 5 · 7 ò (2k 1)

¼

a¼P

a

¼ ¬

y a «1 (k 1)x ¼

y

«a x a¼P x ¼P ¬

¼ ¬

y a (k 1)x ¼

2k 3 a 3

y aa x a

y «a a¼ x

2k 3 a 2k 1 ¼Q

4 · 6 · 8 ò (2k 2) a a aV ò a¼Q 2 · 4 · 6 ò (2k)

a¼P

k 1:

¼

¼

a «x

a

¼

¼

2k 3 ¼P ¬ x 3

2k 3 ¼P x 3

a a (3x x R )

(1 x ) 3(1 x )

Drill Problem 4.4

Find the general solution of the following differential equations near the origin. 1. y ½½ 3xy ½ 3y 0

Page

265

Solution:



n(n 1)aa x

aQ

n(n 1)aa x

a

n(n 1)aa x

a

a

a

aQ

aQ

a

(3n 3)aaQ x aQ 0 a

0n(n 1)aa (3n 3)aaQ 2x aQ 0

n(n 1)aa (3n 3)aaQ 0 aa

9 a a 2

(3n 3)aaQ n(n 1)

for n 2 aR 2a

a

5 aV a 4

6k 3 a 2k(2k 1) ¼Q

a¼P

a

(3n 3)aa x a 0

a

a¼

3naa x 3 aa x a 0 a

6k 6 a¼Q (2k 1)(2k)

¼

y aa x a

y a a¼ x

a¼P

¼

a

a¼

9 a 10 R

(3)¼ x ¼ a 2¼ k!

(3)¼ x ¼P a 3 · 5 · 7 ò (2k 1)

a x a¼P x ¼P ¼

(3)¼ x ¼ (3)¼ x ¼P ¬ a «x

¬ ; valid for all finite y a «1

2¼ k! 3 · 5 · 7 ò (2k 1)

¼

¼

Page

2. (1 4x )y ½½ 8y 0

266

a

y aa x a



Solution:

y aa x a

n(n 1)aa x

n(n 1)aa x

a

aQ

n(n 1)aa x

a

n(n 1)aa x

a

a

(4n 4n 8)aa x a 0

aQ

aQ

a

a

(n 2)(n 1)aa x a 0 a


n(n 1)aa (n 2)(n 1)aaQ 0

5 a4 a 6

(n 2)(n 1) aaQ n(n 1)

a

(2k 2)(2k 1) a 2¼ k!

y aa x a

y a a¼ x a

y a (1 4x

a «

3.) (1 x )y ½½ 10xy ½ 20y 0

a¼

)

2 aR a 3

¼

a

¼

9 a 10 R

2¼ x ¼P a 4k 1

a x a¼P x ¼P ¼

(1)¼P 2¼ x ¼P 1 ¬ ; valid for|x| È 4k 1 2

267

aa

a 0

a¼

4 n(n 1)aa x 8aa x a 0 a

Page

a

aQ

a



n(n 1)aa x

a

aQ

a

n(n 1)aa x

a

aQ

n(n 1)aa x

a

n(n 1)aa x

n(n 1)aa x

a

a )¼ (

a¼ (1

aV

21 a 2 15 a 4

a

a

a

aQ

aQ

(n 9n 20)aa x a 0 a

(n 4)(n 5)aa x a 0 a


n(n 1)aa (n 4)(n 5)aaQ 0 aa

(n 4)(n 5) aa n(n 1)

k 1)(2k 1)(2k 3)x a ¼

a

aR

16 a 3

a 3aR

a¼P (1)¼ (k 1)(k 2)(2k 3)x ¼P a

y aa x a

y a a¼ x

a

(n n 10n 20)aa x a 0

aQ

a

n(n 1)aa x 10naa x 20aa x a 0 a

a

¼

a x a¼P x ¼P ¼

a a y (1)¼ (k 1)(2k 1)(2k 3)x ¼ (1)¼ (k 1)(k 2)(2k 3)x ¼P 3 6 ¼

¼


268

Page

Solution:

SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 4. )(x 9)y ½½ 3xy ½ 3y 0 Solution:

n(n 1)aa x 9n(n 1)aa x

a

9n(n 1)aa x a

aQ

9n(n 1)aa x a

9n(n 1)aa x a

9n(n 1)aa x

a¼

aV

3naa x 3aa x a 0 a

a

a

(n n 3n 3)aa x a 0

aQ

a

(n 2n 3)aa x a 0 a

(n 1)(n 3)aa x a 0 a


9n(n 1)aa (n 1)(n 3)aaQ 0 aa

a

(n 1)(n 3) aaQ 9n(n 1)

2 aR a 9

5 a 18

7 a 36

a

3 · 5 · 7 ò (2k 1)x ¼ a 18¼ (2k 1)k!

a¼P

y aa x a

y a a¼ x a

¼

a

8 a 45 R

2k 4 a 9(2k 2)

a x a¼P x ¼P ¼


269

a

aQ

aQ

aQ

Page

a

a


y a «1

¼

3 · 5 · 7 ò (2k 1)x ¼ ¬ a x; valid for|x| È 3 18¼ (2k 1)k!

5. )(x 4)y ½½ 6xy ½ 4y 0 Solution:

n(n 1)aa x 4 n(n 1)aa x

a

4 n(n 1)aa x a

aQ

4n(n 1)aa x

a

4n(n 1)aa x

a

4n(n 1)aa x

a

6naa x 4aa x a 0 a

a

a

(n n 6n 4)aa x a 0

aQ

aQ

aQ

aQ

a

(n 5n 4)aa x a 0 a

(n 1)(n 4)aa x a 0 a


4n(n 1)aa (n 1)(n 4)aaQ aa

(n 1)(n 4) aaQ 4n(n 1) 9 a a 4

270

a

a

Page



(1)¼ (k 1)x ¼ a 2¼ a

27 a 40 R

(1)¼ (2k 3)x ¼P a 3 · 2¼

271

a¼P

7 aR a 6

Page

a¼

5 aV a 6



y aa x a

y a a¼ x y a «1

¼

6.) y ½½ x y 0

a x a¼P x ¼P ¼

(1)¼ (k 1)x ¼ (1)¼ (2k 3)x ¼P ¬ sx t

a

2¼ 3 · 2¼

n(n 1)aa x

a

n(n 1)aa x

a

aV

aQ

aQ

aa x aP a

aaQV x aQ a V

n(n 1)aa aaQV aa

1 a 12

aaQV n(n 1)

1 a 30

(1)¼ x V¼ ¼ a 2 k! · 3 · 7 · 11 ò (4k 1)

y a a¼ x ¼

a¼P

y aa x a ¼

a

a

a

1 a 20

1 a 42 R

(1)¼ x V¼P ¼ a 2 k! · 5 · 9 · 13 ò (4k 1)

a x a¼P x ¼P ¼

y a z1 7Ô ¼!·R··ò(V¼Q){ a zx 7Ô ¼!··µ·Rò(V¼P){ ; valid for all finite x (Q)Ô cÔ

(Q)Ô cÔÓ6

Page

a¼

a~

272

a

¼

a


SOLUTIONS MANUAL IN ADVANCED ENGINEERING MATHEMATICS 7.) (1 2x )y ½½ 3xy ½ 3y 0 a

a

n(n 1)aa x

a

n(n 1)aa x

a

n(n 1)aa x

n(n 1)aa x

a

a

a

a

(2n 2n 3n 3)aa x a 0

aQ

a

2n(n 1)aa x 3naa x 3aa x a 0 a

a

aQ

(2n n 3)aa x a 0 a

3

(n 1) Sn T aa x a 0 2

aQ

aQ

a

3

(n 1) Sn T aaQ x aQ 0 2 a

3 n(n 1)aa (n 1) Sn T aaQ 2 3 An C 2 aa aaQ n 7 a a 4

a¼

aV

11 a 8

3 A2k C 2 a 2k 3 aR a 2

a

a¼P

13 a 10 R

5 A2k C 2 a 2k 1

273

n(n 1)aa x

aQ

Page



y aa x a

y aì a¼ x

y a x a «1

¼Q

¼

¼

a

a x a¼P x ¼P ¼

(1)¼P 3 · 5 · 7 ò (4k 1)x ¼ 1 ¬ ; valid for |x| È ¼ 2 (2k 1)k! √2


Solution Manual for Advance Math

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