Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2007/2008

KNF1023Engineering

Mathematics II

Second Order ODEs

Learning Objectives

Explain about Euler-Cauchy ODEs

Discuss about Second order

inhomogeneous ODEs

Explain about Particular Solution by

Guesswork

Euler-Cauchy ODEs

� An Euler-Cauchy ODE is one of the form

Where are given constants.

� To look for linearly independent solutions of this

ODE, try

where λ is a constant yet to be determined.

� Differentiating, we have

2 "( ) '( ) ( ) 0ax y x bxy x cy x+ + =

candba ,0≠

y xλ=

( )1 2' " 1y x and y xλ λλ λ λ− −= = −

Euler-Cauchy ODEs

Substituting into the ODE gives

Hence, the value of the constant λ can be

determined from the quadratic equation above. We

consider the following cases.

( )

( )( )

( )

( )

2 2 1

2

1 0

1 0

1 0

0

ax x bx x cx

x a b c

a b c

a b a c

λ λ λ

λ

λ λ λ

λ λ λ

λ λ λ

λ λ

− −⋅ − + ⋅ + =

⇒ − + + =

⇒ − + + =

⇒ + − + =

Case (a):

In this case, we can find two distinct real values for

as given by

Thus, we have two solutions for the Euler-Cauchy

ODE, namely

[ ]2

4 0b a ac− − >

[ ] [ ]2

1

4

2

b a b a ac

aλ λ

− − + − −= =

[ ] [ ]2

2

4

2

b a b a ac

aλ λ

− − − − −= =

1 2

1 2,y x y xλ λ= =

Continue…

These solutions are linearly independent, since

For this particular case where and

are real, the general solution of the Euler-Cauchy

ODE is given by

where A and B are arbitrary constants.

1

1 2

2

11 2

2

( tan ) ( )y x

x cons t asy x

λλ λ

λλ λ−= = ≠ ≠

1 2λ λ≠1 2andλ λ

1 2y Ax Bxλ λ= +

Example 1

Solve the ODE subject to

This is an Euler-Cauchy ODE. So try

Substituting into the ODE, we obtain

( ) 21 1"', −− −=== λλλ λλλ xyandxyxy

02'4"2 =++ yxyyx

1)1(')1( == yy

( )

( )( )

2,1

021

023

0241

2

−=−=

=++

=++

=++−

λλ

λλ

λλ

λλλ

2 2 1( ( 1) ) 4 ( ) 2 0x x x x xλ λ λλ λ λ− −− + + =

0]24)1([ =++− λλλλx

Continue…

y = Ax-1 + Bx-2

Where A and B are arbitrary constant.

Differentiating the general solution gives

Putting the given conditions into the general

solution, we have

32 2' −− −−= BxAxy

12;1)1('

1;1)1(

=−−=

=+=

BAy

BAy

Continue…

Solving for A and B, we obtain A=3 and B=-2.

The required particular solution is

21 23 −− −= xxy

Case (b):

In this case, the quadratic equation

has only one solution.

Thus, in trying we manage to find only

one solution for the Euler-Cauchy ODE. We need

two linearly independent solutions to construct the

general solution of the ODE. To find another

solution, let us try

[ ]2

4 0b a ac− − =

2 ( ) 0a b a cλ λ+ − + =

[ ]1

2

b a

aλ λ

− −= =

y xλ=

Continue…

� Here, is a function yet to be determined.

� Substitution into the ODE, we obtain

( )

1

1 1

1 1 1

1

1

2 1

1 1 1

( ) ( )

'( ) ( ) '( )

"( ) 1 ( ) 2 '( ) "( )

y x x u x

y x x u x x u x

y x x u x x u x x u x

λ

λ λ

λ λ λ

λ

λ λ λ

−

− −

= ⋅

= ⋅ + ⋅

= − ⋅ + ⋅ + ⋅

( )u x

( ) 1 1 1

1 1 1

2 12

1 1 1

1

1

1 ( ) 2 '( ) "( )

( ) '( ) ( ) 0

ax x u x x u x x u x

bx x u x x u x cx u x

λ λ λ

λ λ λ

λ λ λ

λ

− −

−

− ⋅ + ⋅ + ⋅

+ ⋅ + ⋅ + ⋅ =

Continue…

Since and this further

reduces to

To solve for let so that

This leads to

( )1 0a b cλ λ λ− + + =[ ]

12

b a

aλ

−= −

'( ) "( ) 0u x xu x+ =

( )u x ( ) '( )v x u x= 0dv

v xdx

+ =

1ln( ) ln( ) ln

1

1

ln( )

dv dx

v x

v xx

vx

du

dx x

u x

= −

⇒ = − =

⇒ =

⇒ =

⇒ =

∫ ∫

Continue…

The solution of the Euler-Cauchy ODE which

we are looking for is therefore given by

The solutions and are linearly

independent to each other.

where A and B are arbitrary constants.

1 ln( )y x xλ= ⋅

1y xλ= 1 ln( )y x x

λ= ⋅

1 1 ln( )y Ax Bx xλ λ= +

Example 2

Solve the ODE subject to

This is an Euler-CODE. So let us try

Substitution into the ODE gives

( )1 2, ' , " 1y x y x y xλ λ λλ λ λ− −= = = −

2 " 3 ' 0x y xy y+ + =

(1) 1 ( ) 2y and y e= =

( )

( )

2

2

1 3 1 0

2 1 0

1 0

1 is the only solution

λ λ λ

λ λ

λ

λ

− + + =

+ + =

+ =

= −

0]13)1([ =++− λλλλx

Continue…

Thus, general solution of the ODE is

Where A and B are arbitrary constants.

Using the given conditions, we have

The required particular solution is therefore

1 1 ln( )y Ax Bx x− −= +

(1) 1; 1

1 ln( )( ) 2; 2; 2 1

y A

B ey e B e

e e

= =

= + = = −

( )1 12 1 ln( )y x e x x− −= + −

Case C:

In this case, the quadratic equation has two complex solutions given

by.

If we proceed on, without bothering about the facts

that and are complex, we can construct the

general solution of the Euler-Cauchy ODE as

[ ]2

4 0b a ac− − <

2 ( ) 0a b a cλ λ+ − + =

[ ] [ ]

[ ] [ ]

2

1

2

2

4

2 2

4

2 2

i b a acb a

a a

i b a acb a

a a

λ λ

λ λ

− −−= = − +

− −−= = − −

1λ 2λ

1 2y A x B xλ λ= +

Continue…

Our only problem is to interpret what

means when is complex and also to find a way to

calculate this complex power. Once again we can

resort to the theory of complex functions to resolve

this problem. We can use the following results

1 are any numbers (real or complex) and x is a real number

2 are real numbers

3 is any real numbers

xλ

λ

z w z wx x x if z and w

+ = ⋅

ln( ) 0iy iy xx e if x and y= >

cos( ) sin( )ixe x i x if x

± = ±

Example 3

Solve the ODE

subject to

2 "( ) 3 '( ) 2 ( ) 0x y x xy x y x+ + =

(1) '(1) 1y y= =

( )1 2, ' , " 1y x y x y xλ λ λλ λ λ− −= = = −

( )2

1 3 2 0

2 2 0

1 , 1i i

λ λ λ

λ λ

λ

− + + =

+ + =

= − + − −

1 1i iy Ax Bx

− + − −= +

( ) 0]231[ =++− λλλλx

Continue…

( )( )( )

[ ]

1 1

1

1 ln( ) ln( )

1

1

cos ln( ) sin(ln( ) (cos(ln( )) sin(ln( ))

cos(ln( )) sin(ln( ))

i i

i i

i x i x

y Ax Bx

x Ax Bx

x Ae Be

x A x i x B x i x

x C x D x

− + − −

− −

− −

−

−

= +

= +

= +

= + + −

= +

Continue…

where are arbitrary

constants.

Differentiating, we have

Using the given conditions, we have

( )C A B and D i A B= + = −

( )( ) ( )( ) ( )( ) ( )( )1 1 1 2' sin ln cos ln cos ln sin lny x Cx x Dx x x C x D x− − − − = − + − +

(1) 1; 1

'(1) 1; 1; 2

y C

y D C D

= =

= − = =

Continue…

Thus, the required particular solution is

( )( ) ( )( )1 cos ln 2sin lny x x x− = +

THEORY FOR INHOMOGENEOUS ODEs

Consider the 2nd order linear inhomogeneous

ODE

Let be any particular solution of

the inhomogeneous ODE. Then

To solve the inhomogeneous ODE, let us make the

substitution

" ( ) ' ( ) ( )y f x y g x y r x+ + =

( )p py y x=

" '( ) ( ) ( )p p py f x y g x y r x+ + =

( ) ( )p

y y x Y x= +

Continue…

On substituting into the ODE, we obtain

Thus, we obtain a 2nd order linear

homogeneous ODE in

" '

" '

( ) "( ) ( ) ( ) '( ) ( ) ( ) ( ) ( )

( ) ( ) " ( ) ' ( ) ( )

( ) " ( ) ' ( ) ( )

" ( ) ' ( ) 0

p p p

p p p

y x Y x f x y x Y x g x y x Y x r x

y f x y g x y Y f x Y g x Y r x

r x Y f x Y g x Y r x

Y f x Y g x Y

+ + + + + =

+ + + + + =

+ + + =

+ + =

( )Y Y x=

Continue…

To summarise, assuming that we can find a

particular solution for the 2nd order linear

inhomogeneous ODE and also that we can solve the

homogeneous ODE in for its general solution of the

inhomogeneous ODE is given by

{ {

homhom

( ) ( )p

general solution of thea particular solutioncorresponding ogeneous ODEof in ogeneous ODE

y y x Y x= +

PARTICULAR SOLUTIONS BY GUESSWORK: Example

Solve the ODE subject to

Firstly let us solve the corresponding homogeneous

ODE

This is a 2nd order linear ODE with constant

coefficients. Let us try

" 3 ' 2 0Y Y Y+ + =

"( ) 3 ' 2 2exp(5 )y x y y x+ + =

(0) '(0) 0y y= =

2, ' , "x x xY e Y e Y e

λ λ λλ λ= = =

Continue…

Then, substitution into the homogeneous ODE

Gives

The general solution of the homogeneous ODE is

( )( )

2 3 2 0

1 2 0

1, 2

λ λ

λ λ

λ

+ + =

+ + =

= − −

2x xY Ae Be

− −= +

Continue…

Now, to construct the general solution of the

inhomogeneous ODE we

need to find one particular solution of the ODE. We

will try to look for one by guesswork. The right

hand side of the ODE, namely the term

suggests that we try a particular solution of the

Form

where the constant is to be selected to satisfy the

ODE

exp(5 )py xα=

"( ) 3 ' 2 2exp(5 )y x y y x+ + =

2exp(5 )x

α

Continue…

Differentiating, we have

Substituting into the ODE, we obtain

Hence, is a particular solution of the

inhomogeneous ODE.

"' 5 exp(5 ), 25 exp(5 )p py x y xα α= =

25 exp(5 ) 15 exp(5 ) 2 exp(5 ) 2exp(5 )

42 exp(5 ) 2exp(5 )

1

21

x x x x

x x

α α α

α

α

+ + =

⇒ =

⇒ =

1exp(5 )

21p

y x=

Continue…

The general solution of the inhomogeneous

ODE is

1(0) 0;

21

5'(0) 0; 2

21

y A B

y A B

= + = −

= + =

2 1

7 3B and A= = −

1 1 2exp(5 ) exp( ) exp( 2 )

21 3 7y x x x= − − + −

( ) xxBeAexy

25exp21

1 −− ++=

( ) )2exp(2)exp(5exp21

5'xBxAxy −−−−=

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2008/2009