Nonhomogeneous Linear D.Es y a y x y z 0 for all x in n n...

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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.

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

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

1 xgyxadx

gaaa n ,,...,, 10

,pc yyy

.0)( Iinxallforxan

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

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 xgyadx

naaa ,...,, 10

,sincos xx

.0... 01

amamama n

The following table demonstrates the form of

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

with constant coefficients.

xDCxxBAxxx

eCBxAxexx

eBAxxe

DCxBxAxxx

CBxAxx

yofFormxg

sin)(cos)(cos

sincossin3

Example 1. Solve the following D. equation:

Solution. The associated Hom. E. is

hence the aux. eq. is

Therefore .2

c ececy

xBxAxx

xDxCxBxAxx

xeDCxxeBAxxxe

xBexAexe

xFExDxxCBxAxxxx

yofFormxg

sincoscos9sin4

3sin3cossincos3cos4sin3

sin)(cos)(sin

sincoscos4

sin)(cos)(cos)3(

)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

12)( xg Ay p py

)1(.46'5'' xyyy

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

hence and the general solution is

c ececy

,06'5'' yyy

py ,4)( xxg

,BAxy p

.0'',' yAyBAxy p

)2(.2)56(6

,,056,26185

31 BAABA

31 xy p

1 xececyyy xx

Example 3. Solve

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

c ececy

)1(,62'3'' xeyyy

py ,6)( xexg

.'',' xxx

p AeyAeyAey

,166 AA

,02'3'' yyy

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

pc eececyyy

Example 4. Solve

For we have where

Hence is on the form

Which implies

Therefore the general solution is

c ececy

)1(.22'3'' 3xeyyy

py ),()()( 21 xgxgxg

ppp BeAyyy

.'',' xx BeyBey

.2,22661

xx eyBAeBeA

,02'3'' yyy

xxx eececy

Beyexg

Example 5. Solve

For we have where

Hence is on the form

which implies

Using these values in Eq.(1) we obtain

c ececy

)1(.sin322'3'' xxyyy

,sincos21

xDxCBAxyyy ppp

,02'3'' yyy

.sincossin3)(

xDxCyxxg

BAxyxxg

.sincos',cossin' xDxCyxDxCAy pp

which implies

Therefore and the

general solution is

Example 6.

For we have , therefore

is on the form which implies

c ececy

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

,sin32sin)3(cos)3(322 xxxCDxDCABAx

,02'3'' yyy

pyxexxg )23()(

,33,03,032,22 CDDCABA

.,,,110

923 DCBA

,sincos103

23 xxxy p

.sincos103

1 xxxececyyy xx

p eBAxy

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 xyBAxBAAx p

1 xececyyy xx

.)(2'',)(' xx

p eBAxAeyeBAxAey

Example 7. Solve

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

c xececy

ppp eCxBAxyxyy

,0'2'' yyy

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

)1(.4'2'' xexyyy

eCyexg

BAxyxxg

2py 2x

Which implies

therefore and the he general

solution is

Example 8. Find the form of the particular solution

for each of the following differential equations

Solution. The auxiliary equation is

,22 2 x

p exxy

,2,242,02,1

CBCABA

exCeABAx xx

xxx exxececy

p eCxCxeCeyeCxCxeAy 22 42'',2'

.cos532''')5( xxexyy x

Which implies

therefore and the he general

solution is

Example 8. Find the form of the particular solution of

the following differential equation

,22 2 x

p exxy

,2,242,02,1

CBCABA

exCeABAx xx

xxx exxececy

p eCxCxeCeyeCxCxeAy 22 42'',2'

.cos532''')5( xxexyy x

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

c ececxcxccy

.1,1,0,0,0035 mmm

),()()()( 321 xgxgxgxg

.sin)(cos)(cos5)(

xGFxxEDxyxxxg

eCyexg

BAxyxxg

1py 3x

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

Example 8. Find the form of the particular solution of

the following differential equation

Now where

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

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

.sincossincos 654321 xxcxxcxcxcxccyc

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

.3sin3cos3sin7)(

,sincoscos)(

xHxGyxxg

xFxEyxxg

eDyexg

CBxAxyxxg

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

.sin3cos)sincos()( 322 xHxxGxxFxExCeCBxAxxy x

Nonhomogeneous Linear D.Es y a y x y z 0 for all x in n n...

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