DIFFEQU5_2 vsasda

7/29/2019 DIFFEQU5_2 vsasda

1/19

Second Order Homogeneous Linear Differential Equation with Constant Coefficients

The equation 0d

d

d

d2

2 cyx

yb

x

ya , where 0a , b and c are constants is called a second order

homogeneous linear differential equation with constant coefficients.

By a convenient change of notation, writing Dyx

y d

d, yDDyD

x

y

xx

y 22

2

.d

d

d

d

d

d , where xD dd is

an operator which act on y.

Thus, 0d

d

d

d2

2 cyx

yb

x

ya becomes 02

ycbDaD or 0yDF , where cbDaDDF 2 . DF is a polynomial of degree 2 in variable D is called the characteristics function and 0DF is

called the characteristics equation of 0d

d

d

d2

2 cyx

yb

x

ya .

If 21

DDaDF , then1

and2

are called the characteristic roots of the characteristic

equation 0DF .

1


2/19

The solution of the second order homogeneous linear differential equation with constant coefficients,

0d

d

d

d2

2 cyx

yb

x

ya or 0yDF , is as follows:(A) If

1and

2are real and

21 , then the general solution of the equation 0yDF is

xxeCeCy 21

21

.(B) If

1and

2are real and

21 , then the general solution of the equation 0yDF is x

exCCy 121

.

(C) If1

and2

are complex, ie i1

and i2

, then the general solution of the equation 0yDF is xCxCey x sincos21

.

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3/19

Example: Find the general solution each of the following differential equations.

(a) 09d

d2

2 yx

y(b) 04

d

d4

d

d2

2 yx

y

x

y

(c) 013d

d8

d

d4

2

2 yxy

x

y(d) 09'12''4 yyy

(e) 06'7''3 yyy (f) 07'4'' yyy Solution

(a) 09d

d2

2 yx

y, 03392 DDDDF 3,3 D

The solution isxx

eCeCy3

2

3

1

.(b) 04

d

d4

d

d2

2 yx

y

x

y, 0244

22 DDDDF 2,2 D The solution is xexCCy 221 .

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4/19

(c) 013d

d8

d

d4

2

2 yx

y

x

y, 09141184 22 DDDDF iD

231

The solution is xCxCey x 23sin23cos 21 .(d) 09'12''4 yyy , 0329124 22 DDDDF 23,23D

The solution is xexCCy 2321

.(e) 06'7''3 yyy , 0323673 2 DDDDDF

3,32D

The solution isxx

eCeCy3

232

1 .

(f) 07'4'' yyy , 03274 22 DDDDF iD 32 The solution is xCxCey x 3sin3cos

21

2 .

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5/19

Example: Find the particular solution of 04d

d4

d

d2

2 yx

y

x

y, given that 3)0( y and 4)0(' y .

Solution

04dd4

d

d2

2 yxy

xy , 0244

22 DDDDF 2,2D The general solution is xexCCy 2

21 .

3)0( y , 31C . xx eCexCCy 2

2

2

212'

4)0(' y , 42 21 CC 22 C .The particular solution is xexy 223 .

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6/19

Example: Solve the differential equation 04'2'' yyy , given that 1)0( y and 321)0(' y .Solution

04'2'' yyy , 03142 22 DDDDF iD 31 The general solution is xCxCey x 3sin3cos

21 .

1)0( y , 11C .

xCxCexCxCey xx 3cos33sin33sin3cos' 2121 321)0(' y , 3213 21 CC 22 C .

The particular solution is xxey x 3sin23cos .

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7/19

Example: Find the solution of 025'8'' yyy , given that 24

y and 34' y .Solution

025'8'' yyy , 094258 22 DDDDF iD 34 The general solution is xCxCey x 3sin3cos 214 .

24

y , 243sin43cos 21 CCe eCC 22121 21 eCC 22

21---------------(1) xCxCexCxCey xx 3cos33sin33sin3cos4'

21

4

21

4 3

4' y , 3232321214 2121 CCeCCe

eCC 23721

----------------(2)

(1)+(2) eC 262

eC6

22

eeCC 6 2112221 The particular solution is

xexeey x 3sin

6

23cos

6

2114.

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8/19

Second Order Non-Homogeneous Linear Differential Equation with Constant Coefficients

The equation )(d

d

d

d2

2

xQcyx

yb

x

ya , where 0a , b and c are constants and 0)( xQ is called a

second order non-homogeneous linear differential equation with constant coefficients.

The equation can be written as )()( xQyDF .The general solution of the equation is pc yyy , where cy is the general solution of thehomogeneous equation 0)( yDF , and py is any particular integral of )()( xQyDF .Using the method of finding general solution of 0)( yDF , cy can be obtained. Thus, our problem isreduces to one of finding py .

Solving Second Order Non-Homogeneous Linear Differential Equation with Constant CoefficientsUsing Undetermined Coefficients Method

For the equation )()( xQyDF , the general solution is pc yyy , where cy is the general solution of0)( yDF and )(..........)()()( 321 xpKxpCxpBxApy tp , where A, B, C, ...., Kare arbitrary

constants. The function )(.,,.........)(,)(,)( 321 xpxpxpxp t are the terms in )(xQ and the terms arising

from differention of the terms in )(xQ . The values of the constant A, B, C, ...., Kcan be obtained using

)()( xQyDF p .

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9/19

Example (How to define py ) ( Remember xc exCCy 121 or xxc eCeCy 21 21 or xCxCey xc

sincos21

a) If the equation is xxyDF 2)(3 , and the solutions of 0)( DF is 0D , then

DCxBxxAyp 23 .b) If the equation is

23

32)( xxeyDF

x , and the solutions of 0)(

DF is 0D and 3D , then

EDxCxBexeAy xxp 233 .c) If the equation is xeaxyDF

x 23sin)( , and the solutions of 0)( DF is aiD , 0D and 3D , then EDxCeaxBaxAy

xp 3cossin .

d) If the equation is xxyDF 2)(3 , and one of the solutions of 0)( DF is 0D , then

EDxCxBxxAyp 234 .e) If the equation is

2332)( xxeyDF

x , and one of the solutions of 0)( DF is 0D and theother solution of 0)( DF is 3D (i.e xc eCCy 321 ), then

ExDxCxBexeAyxx

p 2333 .f) If the equation is

2332)( xxeyDF

x , and the solutions of 0)( DF are 0D and 3D (i.e xc eCCy

321 ), then ExDxCxBxeexAy xxp 23332 .

g) If the equation is23

32)( xxeyDFx , and the solutions of 0)( DF are 0D and 0D

(i.e xCCyc 21 ), then 23433 ExDxCxBexeAy xxp .h) If the equation is

2332)( xxeyDF

x , and the solutions of 0)( DF are 3D and 3D (i.e xc exCCy 321 ), then EDxCxeBxexAy xxp 23233 .

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10/19

Example Find the solution ofx

xexxy22

632''3 .Solution

03)(2 DDF 0,0D

Then xCCyc 21 .Since

x

xexxxQ

22

632)( andox

e also a term of cy corresponding to the roots 0 withmultiplicity 2,

Letxx

p EeDxeCxBxAxy22234

The constants A, B, C, D and E can be determined usingx

p xexxy22''

6323 xxx

p EeDxeDeCxBxAxy22223'

22234 xxx

xxxx

p

EeDxeDeCBxAxEeDxeDeDeCBxAxy

2222

22222''

444261244222612

Substitute intox

pxexxy 22'' 6323 , xxxx xexxEeDxeDeCBxAx 222222 63244426123

xxx xexxeEDDxeCBxAx 22222 63212121261836 Equating the coefficients of like terms

181236 AA ; 61318 BB ; 166 CC ; 121112 DD

12101212 EED

xxp exexBxxy

22234

121

121

61

181

The general solution ofx

xexxy22

632''3 isxx

exexBxxxCCy22234

21 12112161181 10


11/19

Example Find the general solution of xxyx

y

x

ysin232

d

d

d

d2

2 .Solution 0122)( 2 DDDDDF 1,2 D

Thenxx

c eCeCy 221 .

Since xxxQ sin23)( and the terms in Q(x) are totally different from the terms in cy take xDxCBAxyp cossin

xDxCAyp sincos'

xDxCyp cossin''

Substitute into xxyyy ppp sin232''' xDxCBAxxDxCAxDxC cossin2sincoscossin xx sin23 xxBAAxxCDxCD sin2322cos3sin3 Equating the coefficients of like terms

2332 AA ;

4302 BBA ;

CDCD 3223 ; 53032303 CCCCD ; 51D Therefore xxxyp cos5

1sin53

43

23 .

The general solution of the equation is xxxeCeCyxx

cos51sin

53

43

23

22

1

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12/19

Example Find the general solution ofx

xexyx

y

x

y 22

2

232d

d3

d

d .Solution 01223)( 2 DDDDDF 1,2D

Then

xx

c eCeCy 22

1 .Since

xxexxQ

223)( , and the term xe2 in Q(x) also a term in cy corresponding to single root 2 ,

then

Let xxxp eDxCxBAxDxeeCxBAxy 22222 xxp eDxCxeDCxAy 222' 22 xeDxDCCxA 22 222

xxp eDxDCCxeDCCxy 222'' 2222224 xeDCxDCCx 22 42484 Substitute into

xppp xexyyy

2'''2323

xeDCxDCCx 22 42484 xeDxDCCxA 22 2223 xx xexeDxCxBAx 222 232 xx xexeDCCxABAx 22 2322322

Equating the coefficients of like terms

2332 AA ;

49032 BAB

122 CC ; ; 202 DDC Therefore xp exxxy 22 24923 .The general solution of the equation is xx exxCeCxy 2221 24923 12


13/19

Example: Solve xxyy 2sin4'' Solution: iDDDF 204)(

2 Thus xCxCyc 2sin2cos 21

xxxQ 2sin)( and x2sin also a term in cy corresponding to the a root i2 of multiplicity one.Let xDxxCxxBxxAxyp 2cos2sin2cos2sin

22 xCxBxxBxxAxxAxyp 2sin2sin22cos22cos22sin2

22' xDxDxCx 2sin22cos2cos2 xxCBxBxxAx 2cos)22(2sin22cos2

22 xCxDxxDA 2sin2cos2sin)22( xBxxBxxAxxAxyp 2cos42sin42sin42cos4

22'' xDAxxCBxCB 2sin)22(2sin)22(22cos)22(

xCxDxxDA 2cos22sin22cos)22(2 xxCBxBxxAx 2sin)48(2cos42sin4 22

xCBxDAxxBA 2cos)42(2sin)42(2cos)48( Substitute into xxyy pp 2sin4

'' xxxCBxDAxxAxxBx 2sin2cos)42(2sin)42(2cos82sin8

Equating the coefficients of like terms

18 B ; 08 A ; 042 DA ; 042 CB So 0A ,

81B ,

161C and 0D

Therefore xxxxyp 2sin1612cos

81 2

The general solution of the equation is pc yyy xxxxxCxCy 2sin

1612cos

812sin2cos

221

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14/19

Example: Find the particular solution ofx

exyyy 834'5'' given that 0)0('and1)0( yy .

Solution:

4,10)4)(1(45)(2 DDDDDDDF

Thus xxc eCeCy4

21 x

exQ 8)( and xe also a term ofc

y corresponding to the single root 1D .

We takex

p CxeBAxy xxx

p eCxCACxeCeAy )('

xxxx

p eCxCCxeCeCey )2(

''

Substitute into

xppp exyyy 8345

''' We get xxxx exCxeBAxeCxCAeCxC 834)(5)2( xxxx exCxeBAxeCxCAeCxC 834)(5)2(

xx exCeABAx 833544 4334 AA ;

1615054 BAB ;

3883 CC

Thereforex

p xexy 38

1615

43

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15/19

The general solution of the equation is pc yyy xxx

xexeCeCy38

1615

434

21 xxxx xeeeCeCy

38

38

434'

4

21

Given 1)0( y ,161

21 CC 0)0(' y ,

12234 21 CC

14489

2 C and 951 C The particular solution is

xxx xexeey38

1615

43

14489

95 4

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16/19

Example: Solve the equationx

xeyyy2

2'3'' , given that 2)0('and1)0( yy .Solution:

2,10)2)(1(23)(2 DDDDDDDF

Thusxx

c eCeCy2

21 xxexQ 2)( and xe2 also a term in cy corresponding to the single root 2D .

We takexxx

p eBxAxBxeeAxy22222

)( xxx

p eBBxAxAxeBxAxeBAxy22222'

)222()(2)2( xxp eBBxAxAxeBAAxy 222'' )222(2)224(

xeBABxAxAx

22)42484(

Substitute intox

ppp xeyyy2'''

23 We get

x

eBABxAxAx22

)42484( xxx

xeeBxAxeBBxAxAx22222

)(2)222(3 xx

xeeBAAx22

)22( Equating the coefficients of like terms

12 A 02 BA

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17/19

So

21A and 1B

Thereforexx

p xeexy222

21

The general solution of the equation isxxxx xeexeCeCy 2222

21 21

xxxxxx xeeexxeeCeCy 22222221

22' Given 1)0( y , 1

21CC

2)0(' y , 3221CC

22

C and 11

C The particular solution is

xxxx xeexeey 2222212

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18/19

Example: Find the general solution of xxeyyyx

sin232'''2 , given that

1)0('and1)0( yy .Solution: 1,20)1)(2(2)(

2 DDDDDDDF Thus

xxc eCeCy

221 xxexQ x sin23)( 2 , the term xe2 in Q(x) also a term in cy correspiding to single root

2D .

Let xDxCBxeeAxyxx

p cossin222

xDxCBxeBeeAxAxeyxxxx

p sincos22222222'

xDxCBxeBeBeeAxAxeAxeAeyxxxxxxx

p cossin422444222222222''

xDxCBxeBeeAxAxeAe xxxxx cossin44482 222222 Substitute into xxyyy ppp sin232

''' We find

xDxCBxeBeeAxAxeAexxxxx

cossin44482222222

)sincos222(22222

xDxCBxeBeeAxAxexxxx

xxexDxCBxeeAx xxx sin23cossin2 2222 xeAxBA 2632 + xxexCDxDC x sin23cos)3(sin)3( 2 Equating the coefficients of like terms

36 A ; 032 BA ; 23 DC ; 03 CD ;So

21A ,

31B ,

53C and

51D

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19/19

Therefore

xxxeexy xxp

cos51sin

53

31

21 222

The general solution of the equation is

pc yyy xxxeexeCeCy xxxx cos

51sin

53

31

21 222

2

2

1

xxxeeexxeeCeCy xxxxxx sin51cos

53

32

312' 22222

2

2

1

xxeexxeeCeC xxxxx sin51cos

53

31

312 2222

2

2

1

1)0( y , 5421 CC 1)0(' y ,

1512

21CC

4511

1C ,

95

2C

The particular solution of the equation is

xxxeexeey xxxx cos51sin

53

31

21

95

4511 2222

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