Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2008/2009

KNF1023Engineering

Mathematics II

Second Order ODEs

Learning Objectives

Explain about 2nd Order Linear ODEs

Discuss about General Solutions of

Homogeneous ODEs

Explain about Homogeneous ODEs with

constants coefficients

Second Order Linear ODEs

�A second-order ODEs is called linear if it can be written as

and nonlinear if it cannot be written in this form.

�Here p(x), q(x) and r(x) are given functions of x.

,, '( ) ( ) ( ) (1)y p x y q x y r x+ + = − − −

Second Order Linear ODEs

� called homogeneous. If , then equation (1) is called non-homogeneous.

� Below are some examples of 2nd order linear ODEs

1.

2.

3.

0)()( ',, =++ yxqyxpy

0)( ≠xr

xydx

dy

dx

yd854

2

2

=+−

092

2

=+ ydx

yd

2,,, )()1()(2)( xxyxxxyxy =+++

General Solutions Of Homogeneous ODEs

� Consider the 2nd order linear homogeneous ODE given by

�We have the following lemmas and theorem concerning the solutions of the ODE.

0)()( ,,, =++ yxgyxfy

Lemma #1, #2 and Theorem 2

Lemma #1

� If and are solutions of the above

ODE (over a certain interval), then where A and B are any arbitrary constants, is also a solution of the ODE (over the interval).

Lemma# 2

� If and are solutions of the above 2nd order linear homogeneous ODE, then

where D is a constant

( )xy1 ( )xy2

( ) ( )xByxAyxy 21)( +=

( )xy1( )xy 2

( )( )' '

2 1 1 2 expy y y y D f x dx− = −∫

Theorem 2

� If and are solutions of the above

2nd order linear homogeneous ODE (over a certain interval) and if and are linearly independent of each other (over the interval), then the general solution of the ODE is given by

where A and B are arbitrary constants.

( )xy1( )xy2

( )xy1 ( )xy2

( ) ( ) ( )xByxAyxy 21 +=

Homogeneous ODEs With Constant Coefficients

� A 2nd order linear homogeneous ODE with constant coefficients is one which can be written

in the form

� Here are given constants.

�According to Theorem #2, to construct the general solution of the ODE, we have to find any two linearly independent solutions of the ODE.

( ) ( ) ( )'' ' 0ay x by x cy x+ + =

candba ,0≠

Homogeneous ODEs With Constant Coefficients

� To look for a solution of the ODE, let us try

Where is a constant

� Differentiating, we obtain

( ) xy x e

λ=

λ

2

'( )

"( )

x

x x

y x e

y x e e

λ

λ λ

λ

λ λ λ

=

= ⋅ =

Homogeneous ODEs With Constant Coefficients

� Substituting into the ODE, we obtain

which will be true for all x if

� We can therefore determine the constant form this quadratic equations. We consider the following cases.

2

2

0

0

x x x

x

a e b e ce

e a b c

λ λ λ

λ

λ λ

λ λ

+ + =

+ + =

2 0a b cλ λ+ + =

Case (a):

Now if then the quadratic equation

has two distinct real solutions given by

Thus, we obtain two solutions for the ODE is

2 4 0b ac− >

2 4 0b ac− >

2

1

2

2

4

2

4

2

b b ac

a

b b ac

a

λ λ

λ λ

− + −= =

− − −= =

1 2

1 2,x x

y e and y eλ λ= =

Continue…

� Now since

Hence, the two solutions are linearly independent. For this case where

from Theorem #2, the general solution of the ODE (a≠0, b, and c are constants) is given by

where A and B are arbitrary constants and are the solutions of the quadratic equation

[ ]1

1 2

2

1

2

tan ,x

x

x

y ee cons t

y e

λλ λ

λ

−= = ≠ 1 2.λ λ≠

2 4 0b ac− >

" ' 0ay by cy+ + =

1 2x xy Ae Be

λ λ= +

1 2andλ λ

2 0a b cλ λ+ + =

Example 1

Solve the ODE subject to

and

*This is 2nd order linear homogeneous ODE with

constant coefficient. So, we use

Substituting into the ODE, we obtain

06'" =−+ yyy 1)0( =y

7)0(' =y

xxxeyandeyey

λλλ λλ 2"', ===

( )( )

2

2

2

6 0

6 0

6 0

3 2 0

3 2

x x x

x

e e e

e

λ λ λ

λ

λ λ

λ λ

λ λ

λ λ

λ λ

+ − =

⇒ + − =

⇒ + − =

⇒ + − =

⇒ = − =

Continue…

The two linearly independent solutions are

The general solution is

where A and B are arbitrary constants.

We will now use to work

out A and B. Differentiating the general solution,

we have

xx eyandey 2

2

3

1 == −

xxBeAey

23 += −

7)0('1)0( == yandy

xxBeAey

23 23' +−= −

Continue…

So,

Solving for A and B, we obtain

Thus, the required particular solution of the ODE

is

723;7)0('

1;1)0(

=+−=

=+=

BAy

BAy

21 =−= BandA

xxeey

23 2+−= −

Case b:

� In this case, the quadratic equation has only one real solution given

by . Hence, in trying we have succeeded in finding only one solution for the ODE

� To construct the general solution of the ODE, we need another solution, which is linearly independent to the one we have already found. To look for another solution, let us try the substitution

2 4 0b ac− =

2 0a b cλ λ+ + =

2

b

aλ = − x

y eλ=

/(2 )bx ay e

−=

( ) xy u x e

λ= ⋅

Continue…

where is a function to be determined.

Differentiating, we obtain

Substituting into the ODE, we have

( )u x

2

' ( ) e '( )

'' ( ) e 2 '( ) "( )

x x

x x x

y u x u x e

y u x u x e u x e

λ λ

λ λ λ

λ

λ λ

= ⋅ + ⋅

= ⋅ + ⋅ +

( ) ( )2 ( ) e 2 '( ) "( ) ( ) e '( ) ( ) 0x x x x x xa u x u x e u x e b u x u x e cu x e

λ λ λ λ λ λλ λ λ⋅ + ⋅ + + ⋅ + ⋅ + =

Continue…

Since and , the equation

above reduces to

Since and , we find that

A solution for this simple ODE is (We

do not have to look for the general solution of this

simple ODE, we are just interested in finding two

linearly independent solutions of the ODE )

2 0a b cλ λ+ + =2

b

aλ = −

"( ) 0xau x e

λ =

0a ≠ 0xe

λ ≠

"( ) 0u x =

( )u x x=

"( ) '( ) ( ) 0ay x by x cy x+ + =

Continue…

To summarise, for this case where

, two particular solutions of the ODE (a≠0, b and c are constants) are given by

Now, since , the two solutions above

are linearly independent and hence from Theorem

#2, the general solution of the ODE is

2 4 0b ac− =

"( ) '( ) ( ) 0ay x by x cy x+ + =

21 e

bx

ay

−

= 22 e

bx

ay x

−

= ⋅

1

2

1(co tan )

yns t

y x= ≠

2 2

bx bx

a ay Ae Bxe

− −

= +

Example 2:

Solve the ODE subject to

This is a 2nd order linear homogeneous ODE with

constant coefficients. So let us try

Substituting into the ODE, we obtain

xxxeyandeyey

λλλ λλ 2"', ===

09'6" =++ yyy

1)0(')0( == yy

3

0)3(

096

2

2

−=

=+

=++

λ

λ

λλ

Continue…

Since is the only possible solution of the

quadratic equation, the general solution of

the ODE is given by

Differentiating the general solution, we obtain

Using the given conditions, we have

3−=λ

xxBxeAey

33 −− +=

413;1)0('

1;1)0(

==+−=

==

BorBy

Ay

' 3 3 33 3x x xy Ae Bxe Be

− − −= − − +

Continue…

Hence, the required particular solution is

xxxeey

33 4 −− +=

Case C:

In this case, the quadratic equation

does not have any real solutions. It

has two distinct complex solutions given by

Where

If we simply ignore the fact that and

are complex and proceed as in case (a) above, the

general solution of the ODE is given by

2 4 0b ac− <

2 0a b cλ λ+ + =

2

1

2

2

4

2 2

4

2 2

b acbi

a a

b acbi

a a

λ λ

λ λ

−= = − +

−= = − −

1i = −

1λ 2λ

1 2x xy Ae Be

λ λ= +

Continue…

Theory which will be useful to us in dealing

with when is complex.

1. If z and w are any numbers (either real orcomplex) then

2. If x is any real number then

xe

λλ

z w z we e e

+ = ⋅

cos( ) sin( )ixe x i x

± = ±

Example 3

Solve the ODE subject to . What is the value of y at

?

This is 2nd order linear homogeneous ODE with

constant coefficients. Let us try

Substituting into the ODE, we obtain

013'4'' =+− yyy2)0('1)0( == yandy

2

π=x

xxxeyandeyey

λλλ λλ 2"', ===

ii 3232

01342

−=+=

=+−

λλ

λλ

Continue…

Thus, the general solution is given by

Before we proceed any further, it is useful to

Rewrite the general solution as

xixiBeAey

)32()32( −+ +=

[ ] [ ]( )

[ ] [ ]( )

(2 3 ) (2 3 )

2 (3 ) (3 )

2

2

( )

cos(3 ) sin(3 ) cos(3 ) sin(3 )

cos(3 ) sin(3 )

i x i x

x i x i x

x

x

y Ae Be

y e Ae Be

y e A x i x B x i x

y e A B x i A B x

+ −

−

= +

= +

= + + −

= + + −

Continue…

Thus, we can rewrite the general solution as

where C=A+B and D=i[A-B] are arbitrary

constants.

Differentiating, we have

Using the given conditions, we find that

[ ]2 cos(3 ) sin(3 )xy e C x D x= +

[ ] [ ])3sin()3cos(2)3cos(3)3sin(3' 22 xDxCexDxCey xx +++−=

0223;2)0('

1;1)0(

==+=

==

DorCDy

Cy

Continue…

Thus, the required particular solution is

The value of y at is

)3cos(2xey

x=

2

π=x ( ) ( ) 0

23cos

2== ππ π

ey

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2007/2008