Nonhomogeneous Linear D.Es

Recall that a general order L.D.E. is on the form

where are continuous functions on some

interval I and

The general solution of Eq.(1) is on the form

where is the general solution of the associated Hom.

D.E.

and is a particular solution of Nonhom. E. Eq.(1).

)1(),()()(...)()( 011

1

1 xgyxadx

dyxa

dx

ydxa

dx

ydxa

n

n

nn

n

n

thn

gaaa n ,,...,, 10

cy

,pc yyy

py

.0)( Iinxallforxan

,0)()(...)()( 011

1

1

yxadx

dyxa

dx

ydxa

dx

ydxa

n

n

nn

n

n

Undetermined coefficients method

Consider an order L.D.E. with constant coefficients

where are constants.

We learned in Section 4.2 how can we determined

which is the general solution of the Hom. L.D. E.

associated with Eq.(1) using the auxiliary equation:

Now, if is one of the following types:

a constant, a polynomial, an exponential function on the

form , or finite sums and products of

these types, then has the same form as , but with

general unknown coefficients to be determined.

)1(),(... 011

1

1 xgyadx

dya

dx

yda

dx

yda

n

n

nn

n

n

thn

naaa ,...,, 10

)(xg

xe

py

,sincos xx

)(xg

cy

.0... 01

1

1

amamama n

n

n

n

The following table demonstrates the form of

depending upon the type of incase of L.D.Es

with constant coefficients.

)(xg

py

xDCxxBAxxx

xBxAx

eCBxAxexx

eBAxxe

Aee

DCxBxAxxx

CBxAxx

BAxx

BAxx

A

yofFormxg

xx

xx

xx

p

sin)(cos)(cos

sincossin3

)()6(

)(

7

2

12

95

3

)(

5252

55

33

233

22

Example 1. Solve the following D. equation:

Solution. The associated Hom. E. is

hence the aux. eq. is

Therefore .2

2

3

1

xx

c ececy

xBxAxx

xDxCxBxAxx

xeDCxxeBAxxxe

xBexAexe

xFExDxxCBxAxxxx

yofFormxg

xxx

xxx

p

sincoscos9sin4

3sin3cossincos3cos4sin3

sin)(cos)(sin

sincoscos4

sin)(cos)(cos)3(

)(

777

222

)1(.126'5'' yyy

,06'5'' yyy

.2,30)2)(3(0652 mmmmm

For we have

hence is on the form , where A

is a constant to be determined.

But Using these values in

Eq.(1) we get

hence the general solution is

Example 2. Solve the D. E.

,22126 pyAA

.0''' yyAy

.22

2

3

1

xx

pc

ecec

yyy

py

12)( xg Ay p py

)1(.46'5'' xyyy

Solution. The associated Hom. E. is

hence from Example 1 we have

For we have hence is on the form

where A and B are constants to be

determined. But

Using these values in Eq.(1) implies

Comparing coefficients on both sides of Eq.(2) we

get

hence and the general solution is

.2

2

3

1

xx

c ececy

,06'5'' yyy

py ,4)( xxg

,BAxy p

py

.0'',' yAyBAxy p

)2(.2)56(6

2)(65

xABAx

xBAxA

,,056,26185

31 BAABA

,185

31 xy p

.185

312

2

3

1 xececyyy xx

pc

Example 3. Solve

Solution. The associated Hom. E. is

hence the aux. equation and it’s roots are

therefore

For we have hence is on the form

where A is a constants to be

determined. But

Using these values in Eq.(1) we get

therefore , and the general solution is

.2

2

1

xx

c ececy

)1(,62'3'' xeyyy

py ,6)( xexg

,x

p Aey

.'',' xxx

p AeyAeyAey

,166 AA

,02'3'' yyy

,1,20)1)(2(0232 mmmmm

py

x

p ey

.2

2

1

xxx

pc eececyyy

Example 4. Solve

Solution. The associated Hom. E. is

hence from Example 3 we have

For we have where

Hence is on the form

Which implies

Using these values in Eq.(1) we get

Therefore the general solution is

.2

2

1

xx

c ececy

)1(.22'3'' 3xeyyy

py ),()()( 21 xgxgxg

,21

x

ppp BeAyyy

.'',' xx BeyBey

.2,22661

61 x

p

xx eyBAeBeA

,02'3'' yyy

py

.261

2

2

1

xxx eececy

,)(

2)(

2

1

2

1

x

p

x

p

Beyexg

Ayxg

Example 5. Solve

Solution. The associated Hom. E. is

hence from Example 3 we have

For we have where

Hence is on the form

which implies

Using these values in Eq.(1) we obtain

.2

2

1

xx

c ececy

)1(.sin322'3'' xxyyy

py ),()()( 21 xgxgxg

,sincos21

xDxCBAxyyy ppp

,02'3'' yyy

py

.sincossin3)(

2)(

2

1

2

1

xDxCyxxg

BAxyxxg

p

p

.sincos',cossin' xDxCyxDxCAy pp

which implies

hence

Therefore and the

general solution is

Example 6.

Solution. The associated Hom. E. is

hence from Example 3 we have

For we have , therefore

is on the form which implies

.2

2

1

xx

c ececy

)1(.)23(2'3'' xexyyy

py

,sin32sin)3(cos)3(322 xxxCDxDCABAx

,02'3'' yyy

pyxexxg )23()(

,33,03,032,22 CDDCABA

.,,,110

310

923 DCBA

,sincos103

109

23 xxxy p

.sincos103

109

23

2

2

1 xxxececyyy xx

pcp

,)( x

p eBAxy

Using these values in Eq.(1) we get

hence the general solution is

Remark.

Assume that the particular solution of a nonhom. L.D.E. is on the form

If there is a term in duplicates a term in , then this

must be multiplied by where s is the smallest positive integer that eliminates the duplication. In fact s is the multiplicity of the root of the associated auxiliary equation which causes the duplication.

.....1 kppp yyy

,,23656125

21

125

21 xyBAxBAAx p

ipy

.125

21

2

2

1 xececyyy xx

pcp

.)(2'',)(' xx

p

xx

p eBAxAeyeBAxAey

cy

ipy

,sx

Example 7. Solve

Solution. The associated Hom. E. is

hence the aux. equation and it’s roots are

therefore

For we have where

It is clear that the term in duplicates a term in

thus must be multiplied by to eliminate this

duplication. Hence

.21

xx

c xececy

py ),()()( 21 xgxgxg

,22

21

x

ppp eCxBAxyxyy

,0'2'' yyy

,1,10)1)(1(0122 mmmmm

2py

)1(.4'2'' xexyyy

.4)(

,2)(

2

1

2

1

x

p

x

p

eCyexg

BAxyxxg

,cy

2py 2x

Which implies

Using these values in Eq.(1) we get

therefore and the he general

solution is

Example 8. Find the form of the particular solution

for each of the following differential equations

(1)

Solution. The auxiliary equation is

,22 2 x

p exxy

,2,242,02,1

422

CBCABA

exCeABAx xx

.22 2

2

2

1

xxx exxececy

xxx

p

xx

p eCxCxeCeyeCxCxeAy 22 42'',2'

.cos532''')5( xxexyy x

Which implies

Using these values in Eq.(1) we get

therefore and the he general

solution is

Example 8. Find the form of the particular solution of

the following differential equation

Solution. The auxiliary equation is

,22 2 x

p exxy

,2,242,02,1

422

CBCABA

exCeABAx xx

.22 2

2

2

1

xxx exxececy

xxx

p

xx

p eCxCxeCeyeCxCxeAy 22 42'',2'

.cos532''')5( xxexyy x

Hence

Now where

It is clear that there are terms in duplicate terms

in therefore must be multiplied by to

eliminate this duplication. Also, the term in

duplicate a term in , therefore must be

multiplied by . Hence is on the form

.54

2

321

xx

c ececxcxccy

.1,1,0,0,0035 mmm

),()()()( 321 xgxgxgxg

.sin)(cos)(cos5)(

,3)(

,27)(

3

2

1

3

2

1

xGFxxEDxyxxxg

eCyexg

BAxyxxg

p

x

p

x

p

1py

cy

2py

1py 3x

cy2

py

x py

.sin)(cos)()(3 xGFxxEDxCxeBAxxy x

p

Example 8. Find the form of the particular solution of

the following differential equation

Solution. The auxiliary equation is

hence

Now where

.3sin7cos5''2 32)4()6( xxexyyy x

,,,0,00)1(02 222246 iimmmmmm

.sincossincos 654321 xxcxxcxcxcxccyc

)()()()()( 4321 xgxgxgxgxg

.3sin3cos3sin7)(

,sincoscos)(

,5)(

,)(

3

3

2

1

4

3

33

2

22

1

xHxGyxxg

xFxEyxxg

eDyexg

CBxAxyxxg

p

p

x

p

x

p

It is clear that there are terms in duplicate terms

in therefore must be multiplied by to

eliminate this duplication. Also, there are terms in

duplicate terms in , therefore must be

multiplied by . Hence is on the form

1pycy

3py

1py

2x

cy3

py

x py

.sin3cos)sincos()( 322 xHxxGxxFxExCeCBxAxxy x

p