Differential EquationsMTH 242

Lecture # 11

Dr. Manshoor Ahmed

Summary (Recall)

• Construction of 2nd solution from a known solution.

• Reduction of order.

• How to solve a homogeneous linear differential with constant coefficients?

• Auxiliary or characteristics equation.

• How to write the general of solution of linear differential with constant coefficients

Solution of non-homogeneous DE

Solution of non-homogeneous DE

The Method of Undetermined Coefficients

pcy

pby

pay ///

Caution!

• In addition to the form of the input function , the educated guess for must take into consideration the functions that make up the complementary function .

• No function in the assumed must be a solution of the associated homogeneous differential equation. This means that the assumed should not contain terms that duplicate terms in .

)(xg

cy

py

py

cy

Trial particular solutions

No Terms Choice for

1.

2.

3.

4.

py

xke xAe

( 0,1,2......)nkx n 20 1 2 ..... n

nA A x A x A x

cos

sin

k x

k x

cos sinA x B x

cos

sin

x

x

ke x

ke x

( cos sin )xe A x B x

Some functions and their particular solutions

1. Form of

2. 1(Any constant)

3.

4.

5.

6.

7.

8.

( )f x py

A

5 7x Ax B

3 1x x 3 2Ax Bx Cx D

sin 4x sin 4 cos 4A x B x

sin 4 cos 4A x B xcos 4x

5xe 5xAe

2 5xx e 2 5( ) xAx Bx C e

9.

10.

11.

12.

5(9 2) xx e 5( ) xAx B C e

3 sin 4xe x 3( cos 4 sin 4 ) xA x B x e

25 sin 4x x2 2( ) cos 4 ( )sin 4Ax Bx C x Ax Bx C x

3 cos 4xxe x3 3( ) cos 4 ( ) cos 4x xAx B e x Cx D e x

If equals a sum?

Suppose that • The input function consists of a sum of terms of the kind listed

in the above table i.e.

• The trial forms corresponding to be

.

Then, the particular solution of the given non-homogeneous

differential equation is

In other words, the form of is a linear combination of all the

Linearly independent functions generated by repeated differentiation

of the input function .

xg

xg

.21 xgxgxgxg m

xgxgxg m , , , 21

mppp yyy ,,,21

mpppp yyyy 21

py

)(xg

Example 1 Solve the non homogenous equation by using Undetermined

Coefficient Method.

Solution:

For complementary solution consider homogenous equation

Auxiliary equation is

24 8y y x

4 0y y

2

2

01 2

4 0

4

2

( cos 2 sin 2 )xc

m

m

m i

y e C x C x

For particular solution we assume

Substituting into the given differential equation

Equating the coefficients of and constant we have

Constant: Solving these equations we get the values of undetermined coefficients

2py Ax Bx C

2py Ax B 2py A

24 8p py y x 2 2

2 2

2 4( ) 8

2 4 4 4 8

A Ax Bx C x

A Ax Bx C x

2 ,x x2x : 4A 8

: 4 0x B 2 4 0A C

Thus

Hence the general solution is

2, 0, 1A B C

22 1py x

c py y y 2

1 2cos 2 sin 2 2 1y C x C x x

Example 2 Solve the non homogenous equation by using Undetermined Coefficient Method.

Solution:

siny y x

Duplication between and ?

• If a function in the assumed is also present in then this function is a solution of the associated homogeneous differential equation. In this case the obvious assumption for the form of

is not correct.

• In this case we suppose that the input function is made up of terms of kinds i.e.

and corresponding to this input function the assumed particular solution

is

py cy

py cypy

n

)()()()( 21 xgxgxgxg n

py

npppp yyyy 21

• If a contain terms that duplicate terms in , then that must be multiplied with , being the least positive integer that

eliminates the duplication.

Example 3Find a particular solution of the following non-homogeneous differential

Equation

Solution:

To find , we solve the associated homogeneous differential equation

We put in the given equation, so that the auxiliary equation is

ipyipy

cynx n

xeyyy 845 ///

cy

045 /// yyymxey

Thus

Since

Therefore

Substituting in the given non-homogeneous differential equation, we

obtain

or

Clearly we have made a wrong assumption for , as we did not

remove the duplication.

4 ,1 0452 mmm

xxc ececy 4

21

xexg 8)(

xp Aey

xexAexAexAe 845

xe80 py

Since is present in .Therefore, it is a solution of the associated

homogeneous differential equation

To avoid this we find a particular solution of the form

We notice that there is no duplication between and this new

Assumption for .

Now

Substituting in the given differential equation, we obtain

or

xAe cy

045 /// yyy

xp Axey

cy

py

xxxxp AeAxeAeAxey 2y , //

p/

.84552 xxxxxx eAxeAeAxeAeAxe

.3883 AeAe xx

So that a particular solution of the given equation is given by

Hence, the general solution of the given equation is

Example 4

Find a particular solution of

Solution:

Consider the associated homogeneous equation

Put

xp ey )38(

xxxc exececy )3/8(4

21

xeyyy /// 2

02 /// yyy

mxey

Then the auxiliary equation is

Roots of the auxiliary equation are real and equal. Therefore,

Since

Therefore, we assume that

This assumption fails because of duplication between and .

We multiply with .

Therefore, we now assume

1 ,1

0)1(12 22

m

mmm

xxc xececy 21

xexg )(

xp Aey

cy py

x

xp Axey

However, the duplication is still there. Therefore, we again multiply with

and assume

Since there is no duplication, this is acceptable form of the trial .

Example 5Solve the initial value problem

SolutionAssociated homogeneous DE

xx

p eAxy 2

py

xp exy 2

2

1

2)(y0,)y(

,sin104/

//

xxyy

Put

Then the auxiliary equation is

The roots of the auxiliary equation are complex. Therefore, the

complementary function is

Since

Therefore, we assume that

0// yy

mxey

imm 012

xcxcyc sincos 21

)()(sin104)( 21 xgxgxxxg

sincos C , 21

xDxyBAxy pp

So that

Clearly, there is duplication of the functions and .To remove this duplication we multiply with . Therefore,we assume that

So that Substituting into the given non-homogeneous differentialequation,we have

Equating constant terms and coefficients of , , we obtain

xcos xsin

2py x

.sincos xC xDxxBAxy p

2 sin cos 2 cos sinpy C x Cx x D x Dx x

xDxBAx cosx2sin C2yy p//

p

xxxDxBAx sin104cosx2sin C2

x xsin xx cos

xDxBAx sincos Cyp

So that

Thus

Hence the general solution of the differential equation is

We now apply the initial conditions to find and

Since

Therefore

02 ,102 ,4 ,0 DCAB

0 ,5 ,0 ,4 DCBA

xxxy p cos54

xxxcxcyyy pc cos5- x4sincos 21

1c 2c

0cos54sincos0)( 21 ccy

1cos,0sin

91 c

Now

Therefore

Hence the solution of the initial value problem is

Example 6

Solve the differential equation

xxxxcxy cos5sin54cossin9 2/

2cos5sin54cossin92)( 2/ cy

7.c2

.cos54sin7cos9 xxxxxy

122696 32/// xexyyy

Solution:The associated homogeneous differential equation is

Put

Then the auxiliary equation is

Thus the complementary function is

Since

We assume that

Corresponding to:

096 /// yyy

mxey

3 ,30962 mmm

xxc xececy 3

23

1

)()(12)2()( 2132 xgxgexxg x

2)( 21 xxg CBxAxy p 2

1

Corresponding to:

Thus the assumed form of the particular solution is

The function in is duplicated between and .

Multiplication with does not remove this duplication. However, if

we multiply with , this duplication is removed.

Thus the operative from of a particular solution is

Then

And

Substituting into the given differential equation and collecting like

term, we obtain

xexg 32 12)(

xp Dey 3

2

xp DeCBxAxy 32

xe32py cy py

x

2py 2x

xp eDxCBxAxy 322

xxp eDxDxeBAxy 323 322

xxxxp eDxDxeDxeDeAy 32333 96622

Equating constant terms and coefficients of and yields

Solving these equations, we have the values of the unknown

coefficients

Thus,

.

Hence, the general solution

xxppp exDeBAxBAAxy 3232/// 12262C962)912(9y6y

2, xx xe3

0912 2,C962 BABA

122 ,6 9 DA

-6D and 32,98,32 CBA

xp exxxy 322 6

3

2

9

8

3

2

.63

2

9

8

3

2yy 3223

23

1pcxxx exxxxececy

Higher –Order EquationThe method of undetermined coefficients can also be used for higher

order equations of the form

with constant coefficients. The only requirement is that consists

of the proper kinds of functions as discussed earlier.

Example 7Solve

Solution:To find the complementary function we solve the associated

homogeneous differential equation

)(............ 011

1

1 xgyadx

dya

dx

yda

dx

yda

n

n

nn

n

n

)(xg

xeyy x cos/////

0///// yy

Put

Then the auxiliary equation is

Or

The auxiliary equation has equal and distinct real roots. Therefore, the

complementary function is

Since

Therefore, we assume that

Clearly, there is no duplication of terms between and .

mxmxmx emymeyey 2,,

0 23 mm

1,0,00)1(2 mmm

xc ecxccy 321

xexg x cos)(

xBexAey xxp sincos

cy py

Summary (Recall)