week 3 [compatibility mode]

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2008/2009

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

� If a function has continuous partial derivatives, its total or exact differential is

� From this it follows that if =c=const, then

� A first-order differential equation of the form

Is called exact if its left side is the total or exact differential

dyy

udx

x

udu

∂

∂+

∂

∂=

),( yxu

),( yxu0=du

( , ) ( , ) 0 (1)M x y dx N x y dy+ = − −− →

Exact Differential Equation

of some function . Then the differential equation (1) can be written

By integration we immediately obtain the general solution of (1) in the form

(2)u u

du dx dyx y

∂ ∂= + − −− →

∂ ∂

),( yxu

0=du

( , ) (3)u x y C= − −− →

Exact Differential Equation

� Comparing (1) and (2), we see that (1) is exact if there is some function such that

By the assumption of continuity the two second

derivatives are equal. Thus

)u

a Mx

∂=

∂

),( yxu

) (4)u

b Ny

∂=

∂

2M u

y y x

∂ ∂=

∂ ∂ ∂

2N u

x x y

∂ ∂=

∂ ∂ ∂

(5)M N

y x

∂ ∂= − −− →

∂ ∂

Exact Differential Equation

� This condition is not only necessary but also sufficient for to be an exact differential.

� If (1) is exact, the function can be found by guessing or in the following systematic way. From 4(a) we have by integration with respect to x

NdyMdx +

),( yxu

( )( ) 6u Mdx k y= + − −− →∫

Exact Differential Equation

� in this integration, y is to be regarded as a constant, and k(y) plays the role of a “constant” of integration. To determine k(y), we derive from (6), use (4b) to get and integrate to get k.

� Formula (6) was obtained from (4a). Instead of 4(a) we may equally well use (4b). Then instead of (6) we first have

/u y∂ ∂/k y∂ ∂ /k y∂ ∂

( )( ) *6u Ndy l x= + − −− →∫

Exact Differential Equation

� To determine l(x) we derive from (6*), use (4a) to get and integrate it to get .

/u x∂ ∂

/l x∂ ∂ l

Example 1: An exact equation

Solve

Solution:

1st step: Test for exactness.

Our equation is of the form (1) with

exact solution

3 2 2 3( 3 ) (3 ) 0 (7)x xy dx x y y dy+ + + =

3 2

2 3

3

3

6

6

M x xy

N x y y

Mxy

y

Nxy

x

= +

= +

∂=

∂

∂=

∂

Continue…

2nd Step: Implicit solution.

From (6) we obtain

To find k(y), we differentiate this formula

with respect to y and use formula (4b),

obtaining

( )3 2

4 2 2

( ) 3 ( )

1 3( ) (8)

4 2

u Mdx k y x xy dx k y

x x y k y

= + = + +

= + +

∫ ∫

2 2 33 3

u dkx y N x y y

y dy

∂= + = = +

∂

Continue…

� Hence , so that . Inserting this into (8) we get the answer

3rd step: Checking.

For checking, we can differentiate

implicitly and see whether this leads to

or , the given equation.

3ydy

dk= *

4

4c

yk +

=

4 2 2 41( , ) ( 6 ) (9)

4u x y x x y y c= + + =

cyxu =),(

NM

dxdy

−=

0=+ NdyMdx

Continue…

� In the present case, differentiating (9) implicitly with respect to x, we obtain

� Collecting terms, we see that this equals M + Ny’=0 with M and N as in (7); thus .This completes the check.

3 2 2 31(4 12 12 ' 4 ') 0

4x xy x yy y y+ + + =

0=+ NdyMdx

Example 2

� Solve

Solution:

Thus

exact

)cos(23

)cos(

2 yxyyN

yxM

+++=

+=

( ) ( )( ) 0cos23cos2 =+++++ dyyxyydxyx

sin( )

sin( )

Mx y

y

Nx y

x

∂= − +

∂

∂= − +

∂

Continue…

Step 2: Implicit general solution.

To find k(y), we differentiate this formula

with respect to y and obtain

Hence

By integration, . Inserting this result

( ) cos( ) ( ) sin( ) ( )u Mdx k y x y dx k y x y k y= + = + + = + +∫ ∫

2cos( ) 3 2 cos( )

u dkx y N y y x y

y dy

∂= + + = = + + +

∂

2/ 3 2dk dy y y= +

*23cyyk ++=

3 2( , ) sin( )u x y x y y y c= + + + =

Continue…

Step 3: Checking an implicit solution.

We can check by differentiating the implicit solution

u(x,y) = c implicitly and see whether this leads to

the given ODE:

This completes the check.

2cos( ) (cos( ) 3 2 ) 0

u udu dx dy x y dx x y y y dy

x y

∂ ∂= + = + + + + + =

∂ ∂

Homogeneous Equation

Identify whether the ODE is homogeneous or not???

�Differential equation is call Homogeneous equation if

for every real value of

Example 3: Identify whether is

homogeneous or not.

( ),dy

f x ydx

=

( ) ( ), ,f x y f x yλ λ =

λ

dy y x

dx y x

−=

+

Continue…

This equation is homogeneous

( )

( )

,

,

y xf x y

y x

y xf x y

y x

λ λλ λ

λ λ

−=

+

−=

+

y x

y x

−=

+

( ),f x y=

Example 4

� Check whether the equation given is homogeneous or not?

This equation is not homogeneous.

dyx y

dx= −

( )

( )

,

,

f x y x y

f x y x yλ λ λ λ

= −

= −

( )

( ),

x y

f x y

λ

λ

= −

=

Example 5

Solve

Solution:

The above equation is a homogeneous

x

yx

dx

dy

2

3+=

dy duu x

dx dx= +

y ux=

3( ) (1 3 ) 1 3

2 2 2

du x ux x u uu x

dx x x

+ + ++ = = =

Continue…

1 3

2

dv uu x

dx

++ =

1 3 1 3 2 1

2 2 2

du u u u ux u

dx

+ + − += − = =

1 1(1 )( )

2 2

du uu

dx x x

+= = +

1 2

du dx

u x=

+

1

1 2

du dx

u x=

+∫ ∫

Continue…

( )1

ln 1 ln ln2

u x c+ = +

( )1

2ln 1 ln lnu x c+ = +

( )1

2ln 1 ln( )u x c+ =

121u x c+ =

Continue…

cxxy 23

=+

cxx

y2

1

1 =+

Example 6

Solve subject y(0)=2

Let , rewrite this equation become

22 yx

xy

dx

dy

+=

y xu=

( )

( )22

21

x xuduu x

dx x xu

u

u

+ =+

=+

21

du ux u

dx u= −

+

3

21

u

u= −

+

Continue…

2

3

1 u dxdu

u x

+= −

3

1 1 dxdu

u u x

+ = −

3

1 1 dxdu

u u x

+ = −

∫ ∫

2

1ln ln

2u x c

u− + = − +

Continue…

Substitute

2

1ln ln

2u x c

u+ = +

2

1ln

2xu c

u= +

yu

x=

2

2

2ln

y

xcy +=

2

2

2 y

x

Aey =

Continue…

2)0( =y

2

2

2

0

2

2

y

x

ey

Ae

=

=

Summary on solving Homogeneous equations

1. Identify whether the equation is homogeneous or not.

2. Use substitution of and

in the original equation.

3. Separate the variable x and u in 2.

4. Integral both sides with respect to the related variables and put only one constant, say A.

5. Substitute back .

6. If the equation has subject to any value, substitute it to get the constant value A.

y xu= dy duu x

dx dx= +

yu

x=

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2008/2009