Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2007/2008

KNF1023Engineering

Mathematics II

First Order ODEs

Learning Objectives

Demonstrate the solution of

inhomogeneous 1st order ODE in linear form

Demonstrate the solution of

Homogeneous 1st order ODE in linear form

Demonstrate how to find integrating

factor for non-exact differential equation

Integrating Factor

� The idea of the method in this section is quite

simple. If given an equation

that is not exact, but if multiply it by a suitable

function , the new equation

is exact, so that it can be solved, the function

is then called an integrating factor of equation (1).

( , ) ( , ) 0 (1)P x y dx Q x y dy+ = −−−

),( yxF

0 (2)FPdx FQdy+ = − − −

),( yxF

How to Find Integrating Factors

� Equation (2) with M=FP, N=FQ is exact by the definition of an integrating factor. Hence

now is

That is (subscripts denoting partial derivatives) which is complicated and useless. So we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one—the result may be useful (and may also help you later on). Hence we look for an integrating factor depending only on one variable;

x

N

y

M

∂

∂=

∂

∂( ) ( ) (3)FP FQ

y x

∂ ∂= −−−

∂ ∂

xxyy FQQFFPPF +=+

How to Find Integrating Factors

� fortunately, in many practical cases, there are such factors, as we shall see. Thus, let

Then and so that (3) becomes

Dividing by and reshuffling terms, we have

).(xFF =

0=yFdx

dFFF x == '

xy FQQFFP += '

F

xy FQQFFP += '

F

FQ

F

QFP x

y +='

How to Find Integrating Factors

xy QQdx

dF

FP +=

1

Qdx

dF

FQP xy

1=−

dx

dF

FQP

Qxy

1)(

1=−

∂

∂−

∂

∂=

x

Q

y

P

Qdx

dF

F

11

How to Find Integrating Factors

� Thus, we can write it as

This prove the following theorem.

Rdx

dF

F=

1 1(4)

P QR

Q y x

∂ ∂= − −−−

∂ ∂

How to Find Integrating Factors

* If we assume Then and so that (3) becomes

Dividing by and reshuffling terms, we have

).(yFF = 0=xFdy

dFFF y == '

xy FQFPPF =+'

F

How to Find Integrating Factors

( )

( )

'

1

1(4*)

1 1

1 1,

y

x

x y

x y

x y

x y

FPF PQ

F F

dFQ P P

F dy

dFQ P P

F dy

dFQ P

P F dy

dFR R Q P

F dy P

= +

= +

− = − − −

− =

= = −

Theorem 1 (Integrating factor F(x))

� If (1) is such that the right side of (4), call it R depends only on x, then (1) has an integrating factor F=F(x), which is obtained by integrating (4) and taking exponents on both sides,

Similarly, if F=F(y), then instead of (4) we get

Here And have the companion

( )( ) exp ( ) 5F x R x dx= − − −∫

( )1 1

6dF Q P

F dy P x y

∂ ∂= − − − −

∂ ∂

∂

∂−

∂

∂=

y

P

x

Q

pR

1

THEOREM 2 [Integrating factor F(y)]

� If (1) is such that the right side R of (6) depends only on y, then (1) has an integrating factor F=F(y), which is obtained from (4*) in the form

( )( ) exp ( ) 7F y R y dy= − − −∫

Example 1 (Integrating factor F(x))

� Solve by Theorem 1.

We have hence (4) on the right,

And thus

0)cos()sin(2 22 =+ dyyxydxy

)cos(),sin(2 22yxyQyP ==

[ ]x

yyyyyxy

R3

)cos()cos(4)cos(

1 22

2=−=

( ) 3/3exp)( xdxxxF == ∫

Example 2

xyP 2=

22 (4 3 ) 0,

(0.2) 1.5

xydx y x dy

y

+ + =

= −

234 xyQ +=

2)(

2)26(

2

1

yyThusF

yxx

xyR

=

=−=

� Solve the initial value problem

Here , , the equation is not exact, the right side of (4) depends on both x and y (verify!), but the right side of (6) is

Continue...

2y

� Is an integrating factor by (7). Multiplication by gives the exact equation

Which we can write as

0)34(2 2233 =++ dyyxydxxy

223

3

34

2

yxyN

xyM

+=

=

Continue...

� As we know that

So to get , we use

( ) cyxu =,

( ) cyxu =, ( )∫ += ykMdxu

( )ykdxxyu += ∫32

( )ykyxu += 32

Continue...

� To get , we differential with respect to ,from there we get

( )yk uy

3

22322

4

343

ydy

dkso

yxyNdy

dkyx

y

u

=

+==+=∂

∂

*4cyk +=

cyyxu =+= 432

ccyyxu =++= *432

5.1,2.0 −== yx

9275.4324 =+ yxy

Continue...

1st Order ODEs In Linear Form

� A first order differential equation is linear if it has the form

� If the right side r(x) is zero for all x in the interval in which we consider the equation (written r(x)≡0), the equation is said to be homogeneous other it is said to be nonhomogeous.

( ) ( )dy

f x y r xdx

+ =

Homogeneous 1st Order ODEs In Linear Form

� Linear ODE is said to be homogeneous if the function r(x) is given by r(x)=0 for all x. That is, a homogeneous 1st order ODE is given by

( ) 0dy

f x ydx

+ =

yxfdx

dy)(−=

dxxfdyy

)(1

−=

Homogeneous 1st Order ODEs In Linear Form

∫∫ −= dxxfdyy

)(1

ln( ) ( )y f x dx= −∫

Gives us the general solution of the homogeneous 1st order

ODE above.

exp( ( ) )y f x dx= −∫

Inhomogeneous 1st Order ODEs in Linear Form (Method 1:use of integrating factor)

1. is a general form of the

linear DE.

2. Here f and r are function of x or constants

3.

4. Integrating factor =I.f=

5. Solution is

( ) ( )xryxfdx

dy=+

∫ fdx

∫ fdx

e

( ) ( ) CdxfIrfIy += ∫ ..

Example

xey

dx

dy 25 =+

( ) ( )xryxfdx

dy=+

xerf

2,5 ==

xfdx

eefI5. =∫=

The above DE is of the form

( ) ( ) CdxfIrfIy += ∫ ..

Continue...

( )

xx

xx

xx

xxx

Ceey

Ce

ye

Cdxeye

Cdxeeey

52

75

75

525

7

1

7

−+=

+=

+=

+=

∫

∫

� In the 1st step, we solve the corresponding homogeneous ODE, i.e

� Let us say that we obtain as particular solution for the above homogeneous ODE. We will use it in the 2nd step below to construct a general solution for the original inhomogeneous ODE.

( ) 0dy

f x ydx

+ =

)(xyy h=

Method 2: Variation Parameter

Method 2: Variation Parameter

( )xyxy h=)(� In the 2nd step, for the general solution of the

inhomogeneous ODE, we let . v(x) and substitute it into ODE to obtain a 1st order separable ODE in v(x).

25

5 0

5

5

xdyy e

dx

dyy

dx

dyy

dx

dydx

y

+ =

+ =

= −

= −∫ ∫

Example

Continue...

( )

( )

5

5

5 ' 5

ln 5

.

. 5 .

x

x

x x

y x

y e

y e v x

dye v e v x

dx

−

−

− −

= −

=

=

= −

( )5 ' 5 5 2

5 ' 2

5 2

2

5

. 5 . 5. .x x x x

x x

x x

x

x

e v e v x e v e

e v e

dve e

dx

dv e

dx e

− − −

−

−

−

− + =

=

=

=

Continue…

7

7

75

25

7

( )7

7

x

x

xx

xx

dve

dx

ev c

ey e c

ey ce

−

−

=

= +

= +

= +

Continue…

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2007/2008