CHAPTER 6
FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
Chapter 6
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FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
Outline: 6.1 6.1 6.2 6.2
First-Order First-Order Lin ear Dif f erent ial Equations Solut ion s to Fir st-Ord t-Ord er Linear Linear Dif ferenti al Equation s
Augustin-Louis Cauchy (August 21, 1789 – May 23, 1857) was a French mathematician who was an early pioneer of analysis. He started the project of formulating and proving the theorems of infinitesimal infinitesimal calculus in a rigorous manner, rejecting the heuristic principle of the generality of algebra exploited by earlier authors. He defined continuity in terms of infinitesimals and gave several important theorems in complex analysis and initiated the study of permutation groups in abstract algebra. A profound mathematician, mathematician, Cauchy exercised a great influence over his contemporaries and successors. His writings cover the entire range of mathematics and mathematical physics. Cauchy verified the existence of recurrent elliptic functions, gave the first impetus to the general theory of functions and laid the foundation foundation for the modern treatment of the convergence of infinite series. He perfected the method of integration of linear differential equations and invented the calculus of residues. www.wikipedia.com
Overview: A n o t h er t y p e o f d i f f e r e n t i a l eq eq u a t i o n i s t h e f i r st - or or d er er l i n e a r d i f f e r en t i a l eq eq u a t i o n . W i t h t h e a i d o f i n t e g r a t i n g f a c t o r , t h e so so l u t i o n o f t h i s d i f f e r en t i a l eq eq u a t i o n i s o b t a i n ed .
Objectives: Upon comp letion of th is chapt er, the stu stu dents wi ll be able to: 1 . D e f i n e f i r st -o -o r d e r l i n e a r d i f f e r e n t i a l e q u a t i o n . 2 . D e t er er m i n e f i r st - o r d e r l i n e a r d i f f e r e n t i a l eq eq u a t i o n s. s. 3 . So l v e f i r st - o r d e r l i n e a r d i f f er e n t i a l e q u a t i o n s.
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FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
CHAPTER 6
6 .1
Fi r st -O r d e r L i n e a r D i f f e r e n t i a l Eq u a t i o n s A first-order linear differential equation can be written in the form are all function of alone.
+ ( ) = ( ), where and
To solve for the general solution of the differential equation, we write the ( , ) + ( , ) = 0, thus, from equation in the form + ( ) = ( ), we
have,
Now, set
+ () = () [( ) − ( )] + = 0 = ( ) − () and = 1, thus,
= ( ) and = 0. The
differential equation is not exact, therefore, we can apply the concept of integrating factors. From, () = = ( )
we obtained ∫ () as the integrating factor. From the original differential equation,
() − ()] + = 0
[
Multiply both sides of the equation by the integrating factor thus, we have,
() − ()] + = 0} ∫ ()∫ − ()∫ + ∫ = 0 ( )
{[
( )
( )
( )
Since,
()∫ + ∫ = ∫ ( )
( )
( )
Then,
∫() − ( ) ∫() = 0 Integrating both sides of the equation, the general solution is,
∫ ∫ − ∫ () ∫ = ∫ 0 ∫ − ∫ () ∫ = ∫ = ∫() ∫ + ( ) = ∫ ( ) () + ( )
( )
( )
( )
where
= ∫ . ( )
( )
( )
or
∫ , ( )
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6 .2
FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
53
So l u t i o n s t o Fi r st -O r d e r L i n ea r D i f f e r en t i a l Eq u a t i o n s
A solution to a first-order linear differential equation can be obtained by ( ) ( ) the formula ( ) = + , which is the obtained solution to a first + ( ) = ( ). order linear differential equation of the form
∫
Example(a).
Obtain the general solution of ( + )
Solution:
− = 0. ( ) ( ) = , therefore, +
Write the equation in the form
− = Find
() and (), thus, () = − and () =
For integrating factor, ( )
By
= ∫ = ∫ = = = = ( ) = ∫ () () + , the general solution is, = ∫ + = ∫ + = + or = +
Example(b).
Obtain the general solution of Solution:
+ = 4 . ( ) ( ) = , therefore, +
Write the equation in the form
+ = 4 Find
() and (), thus, () = and ( ) = 4
For integrating factor, ( )
By
= ∫ = ∫ = ( ) = ∫ () ( ) + , the general solution is, = ∫ 4 + = 4 + = 4 +
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FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
CHAPTER 6
Example(c).
− + = 0.
Obtain the general soluti on of Solution:
Write the equation in the form
+ = Find
+ () = ( ), therefore,
( ) and (), thus, ( ) = and ( ) =
For integrating factor,
= ∫ () = ∫ = () = + 1 By ( ) = ∫ ( ) ( ) + , the general solution is, ( + 1) = ∫ ( + 1) + ( + 1) = ∫( + ) + For ∫ , use integration by parts, therefore, ∫ = − Substituting to the formula, the general solution is,
( + 1) = − + + ( + 1) = + or = Example(d).
Obtain the general soluti on of sec Solution:
+ ( csc − 1) = 0.
Write the equation in the form
+ cot = cos Find
() ( ) = , therefore, +
( ) and (), thus, ( ) = cot and ( ) = cos
For integrating factor,
= ∫ = ∫ = = sin ( ) = ∫ ( ) () + , the general solution is, sin = ∫ cos sin + sin = sin + = sin + csc ( )
By
(
)
CHAPTER 6
FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
Example(e).
− = ln .
Obtain the general solution of Solution:
For this example, we can also use the form
() () = and +
) = ∫() () + , therefore, − () and (), thus, () = − and () = ln
the formula will be ( = ln Find
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For integrating factor,
= ∫ () = ∫ = = = = By
() = ∫ () () + , the general solution is, = ∫ ln + = ln + or = ln +
Example(f).
Obtain the particular solution of condition, = 0.
= ( + sin ), that satisfies the
Solution:
Write the equation in the form
− = sin Find
( ) ( ) = , therefore, +
() and (), thus, () = − and () = sin
For integrating factor, ( )
.
= ∫ = ∫ = = = By ( ) = ∫ ( ) ( ) + , the general solution is, = ∫ sin + = − cos + = − cos + To find the particular solution, set, = and = 0, then solve for
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FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
When
CHAPTER 6
= and = 0, = 0.
Then, the particular solution is,
= − cos Example(g).
Obtain the particular solution of condition ( 0) = 1.
+ ( + 3) = 0, that satisfies the
Solution:
+ () = ( ), therefore,
Write the equation in the form
+ 3 = Find
( ) and (), thus,p ( ) = 3 and () = −
For integrating factor,
= ∫ = ∫ = By ( ) = ∫ ( ) ( ) + , the general solution is, = ∫ − ( ) + = − ∫ + = − + = − + To find the particular solution, set = 0 and = 1, then solve for ( )
. When
= 0 and = 1, = .
Then, the particular solution is,
= − + Su p p l em en t a r y Pr o b l em s I . O b t a i n t h e g en e r a l so l u t i o n o f t h e f o l l o w i n g d i f f e r en t i a l eq u a t i o n s .
+ = 2. + 2 = 1.
3. 4.
+ = sin + cos = 1
Ans: Ans: Ans: Ans:
= + = + = + sin = +
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+ 6 = 2 6. − (1 − 2) = 0 7. +3 = 1 8. = + 3 5.
= ( − 2) = − 2 cot2 10. 9.
FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
Ans: Ans: Ans: Ans: Ans: Ans:
= + = + = + + = 4 = 2 − 1 + 4 = 1 − 2 cot2 + csc2
I I . O b t a i n t h e p a r t i c u l a r so l u t i o n s sa t i s f y i n g t h e i n d i c a t ed c o n d i t i o n s.
+ = 6 , (1) = 2 2. + 2 = 1, (0) = −2 1.
− 2) + = 0,(−1) = 3 = ( − 4), (0) = − ( tan − sin2 ) + = 0, (0) = −1
Ans: Ans:
3. (
Ans:
4.
Ans:
5.
Ans:
( − ) = −1 = − 3 ( − ) = 4 4 = 1 − 2 = cos (1 − 2cos )
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FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
CHAPTER 6