ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I... · ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251)-P.VEERAIAH DEPARTMENT OF APPLIED MATHEMATICS

UNIT 2

ORDINARY DIFFERENTIALEQUATIONS

MATHEMATICS II(MA6251)UNIT 2

-P.VEERAIAHDEPARTMENT OF APPLIED

MATHEMATICS

12/23/2014

SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR

1

UNIT 2 SYLLABUS Higher order linear differential equations with

constant coefficients

Method of variation of parameters

Cauchy’s and Legendre’s linear equations

Simultaneous first order linear equations with

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 2

Simultaneous first order linear equations with constant coefficients

Second-order linear differentialequations

2

2Differential equations of the form ( )

are called second order linear differential equations.

d y dya b cy Q x

dx dx

When ( ) 0 then the equations are referred to as homogeneous,Q x When ( ) 0 then the equations are non-homogeneous.Q x


When ( ) 0 then the equations are non-homogeneous.Q x

Note that the general solution to such an equationmust include two arbitrary constants to becompletely general.


Theorem

If ( ) and ( ) are two solutions then so is ( ) ( )y f x y g x y f x g x 2

2we have 0d f df

a b cfdx dx

2

2and 0d g dg

a b cgdx dx

Adding 2 20

d f df d g dga b cf a b cg


Adding2 2 0

d f df d g dga b cf a b cg

dx dx dx dx

2 2

2 2 0d f d g df dg

a b c f gdx dx dx dx

And so ( ) ( ) is a solution to the differential equation. y f x g x


, for and , is a solution to the equation 0mxdy

y Ae A m b cydx

It is reasonable to consider it as a possible solution for

2

2 0d y dy

a b cydx dx


2dx dxmxy Ae mxdy Ame

dx

22

2mxd y Am e

dx

If is a solution it must satisfymxy Ae 2 0 mx mx mxaAm e bAme cAe assuming 0, then by division we getmxAe 2 0am bm c

The solutions to this quadratic will provide two values ofm which will make y = Aemx a solution.

Second-order linear differentialequations When the roots of the auxiliary equation are both

real and equal to m, then the solution would appear to be

y = Aemx + Bemx = (A+B)emx

A + B however is equivalent to a single constant


A + B however is equivalent to a single constant and second order equations need two

With a little further searching we find that y = Bxemx is a solution. So a general solution is

mx mxy Ae Bxe

Roots are complex conjugates

( ) ( )p iq x p iq xy Ae Be px iqx px iqxAe e Be e

px iqx iqxe Ae Be We know that cos sinie i

When the roots of the auxiliary equation are complex, they will be of the form m1 = p + iq and m2 = p – iq.

Hence the general equation will be


cos sin cos( ) sin( )pxe A qx i qx B qx i qx cos sin cos sinpxe A qx i qx B qx i qx

cos sinpxe A B qx A B i qx cos sinpxe C qx D qx

Where and ( )C A B D A B i

2

2 ( )d y dy

a b cy Q xdx dx

22 ( )

d g dga b cg Q x

dx dx

Non-homogeneousSecond-order lineardifferential equations


Non homogeneous equations take the form

Suppose g(x) is a particular solution to this equation. Then

2

2

( ) ( )( ) ( )

d g k d g ka b c g k Q x

dx dx

Now suppose that g(x) + k(x) is another solution. Then

Non homogeneoussecond order differential equations

Giving

2 2

2 2 ( )d g d k dg dk

a a b b cg ck Q xdx dx dx dx

2 2

2 2 ( )d g dg d k dk

a b cg a b ck Q xdx dx dx dx

2

2( ) ( )d k dk

Q x a b ck Q xdx dx

2d k dk


2

2 0d k dk

a b ckdx dx

From the work in previous exercises we know how to find k(x).

This function is referred to as the Complementary Function. (CF)

The function g(x) is referred to as the Particular Integral. (PI)

General Solution = CF + PI

22ndnd Order DEOrder DE –– Homogeneous LE withHomogeneous LE withConstant CoefficientsConstant Coefficients(2) If 1 and 2 are distinct real numbers (if b2 - 4ac > 0), then the general solution is:

xx ececy 21 21

(3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is:

xx xececy 11 21 xececy 21

(4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is:

xecxecy xx sincos 21 Where:

a

bac

a

b

2

4 and

2

2


22ndnd Order DEOrder DE –– Homogeneous LE withHomogeneous LE withConstant CoefficientsConstant CoefficientsHomogeneous Linear Equations with Constant Coefficients

A second order homogeneous equation with constant coefficients is written as: 0 0 acyybyawhere a, b and c are constant

The steps to follow in order to find the general solution is as follows: The steps to follow in order to find the general solution is as follows:

(1) Write down the characteristic equation

0 02 acba This is a quadratic. Let 1 and 2 be its roots we have

a

acbb

2

422,1


TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b

Solve the equation

coshx=1)y+2D-(D2

The given differential equation is ( =Coshx


The auxiliary equation ism2-2m + 1=0i.e. (m-1)2 =0 i.e. m = 1,1 . The roots are real and equal.The complementary function isCF =(A+Bx)ex

TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b

2122 12212

cosh

integralparticular thefind tohaveNow we

+ PI = PI)D+-(D

+ee =

)D+-(D

xPI =

-xx

twice)0r denominato themakes1= DSince( )e(x

= e

=Now PIx2x

1


twice)0r denominato themakes1= DSince( 4

= 1)+2D-2(D

=Now PI21

)-ting D = ( Substitu

e =

)D+-(D

e = PISimilarly

-x-x

1

8122 22

Now the general solution of the DE is GS = CF + PI1 + PI2 =

(A+Bx)ex +

is the solution of the given DE. 2. Solve (D2-2D+2) y = ex + 5 + e-2x

Solution: The given differential equation is(D2-2D+2) y = ex + 5 + e-2x

The auxiliary equation is


The auxiliary equation is m2-2m + 2= 0 Solving for m , we get m = i.e. m = 1 ± i . The roots are complex conjugates. The complementary function is CF = (Acos x + B sin x)ex

2

4.2)-(4±2

)1(12)+2D-(D

e2

x

1 DSincee

PIx

)uting D = (Substit = )D+-D

e = PISimilarly

x

02

5

22(

52

0

2


2) = Ding(Substitut/ 2

e=

2)+2D-(D

e=PI

2x

2

2x

3

DE.given theofsolution theis 2

e+

2

5+

1

e+x)esin B+x (Acos

= PI+ PI+ PI+CF=GS

solutiongeneralNow the

2xxx

321

Solve (D2-3D+2) y = 2cos(2x+3) Solution : The given differential equation is y = 2cos(2x+3) The auxiliary equation is m2- 3m + 2= 0 Solving for m , we get (m -1) (m-2) = 0

2)+3D-(D2


Solving for m , we get (m -1) (m-2) = 0 i.e. m = 1 ,2 The roots are real and distinct The complementary function is CF = (A ex + B e2x )

Now we have to find the particular integral

PI =

4)- = D(since

= 2)+3D-(-4

3)+cos(2x(2 =

2)+3D-(D

3)+cos(2x(2 =

2

2

)x+(D)+(-)x+(D+(-)x+( 32cos23232cos2)3232cos2


) ) D-

)x+(D)+(-

D) +D)(--(-

)x+(D+(- =

D)-(-

)x+(294(

32cos232

3232

32cos2)32

32

32cos2

= )9.(-4)-((4

3)+cos(2x3D)(2+(-2

40

3)+sin(2x12-3)+cos(2x(-4

10

] 3)+sin(2x3-3)+[-cos(2x


] 3)+sin(2x3-3)+[-cos(2x+ )e B+e(A= PI+CF=GS 2xx

Solve (D2+1)2y = 2sinx cos3x

Solution: The given differential equation is

y = 2sinx cos3x


(m2+1)2 =0 Solving for m , we get

22 1)+(D



(m2+1) (m2+1) = 0

i.e. m2 = -1 = i2 twice

Therefore m = ±i, ±i

The roots are pair of complex conjugates

The complementary function is

CF = (A+Bx)cos x +(C+Dx) sin x


2222 1

2sin4sin

1

)3cossin2 =

)+(D

xx- =

)+(D

xx(PI =


21

2222 11

+ PIPI

)+(D)+(D

)16-= D(since

225

sin4x =

1)+(-16

sin4x=

1)+(D

sin4x = PI

2

2221

)4-= D(since 9

sin4x=

1)+(-4

sin4x =

PI

22

2

DE.given theof solution theis

9

sin2x+

225

sin4x+x sinDx)+(C+x Bx)cos+(A=PI+ PI+CF=GS 21


42 )4( xyD

42 )4( xyD

xBxAFC 2sin2cos.. 11

24121

1641

4

1

2

4

42

x

xDD

Solve the DE


2

4

4 Dx

4

2

)4

1(

141

xD

41

2

)4

1(41

xD

23

341

1624

412

41

24

2

4

xx

xx

23

341

2sin2cos 24 xxxBxAPICFGS

2.Solve the DE x = y 2)+3D+(D 22

x =2)y +3D+(D

isequationaldifferentigivenThe:Solution22

distinct.andrealarerootsThe

. ,-21-=m i.e.

0= 2)+(m1)+(mget we,mfor Solving

0=2+3m+m

isequationauxiliary The2

)e B+e(A=CF

isfunctionary complementThe

.

2x-x-

x2 (9.2)]1

+6x)+(21

-[x1

=PI

)....]x)2

3D)+(D(+

2

3D)+(D-([1

2

1 =

2

2222


]2

3D)+(D+[1

2

1=

)2

3D)+(D+(1

2

1 =

3D)+D+(2

x=PI

(-1)2

1-2

2

2

]2

7)-6x-(x[

2

1=PI

(9)]2

1+3x)+(1-[x

2

1=PI

(9.2)]4

+6x)+(22

-[x2

=PI

2

2

2

7)-6x-(x

2

1+ )e B+e(A=

PI+CF=GS2

2x-x-

TYPE-4 PARTICULAR INTEGRALS

1.Solve (D2-2D+2) y = ex x2

xe=2)y +2D-(DisequationaldifferentigivenThe 2x2

2

4.2)-(4±(2=mget we,mfor Solving

0=2+2m-m

isequationauxiliary The2

x)esin B+x (Acos=CF

isfunctionary complementThex

1)+ D= D(Since

1)+2-2D-1+2D+(D

e =

2)+2D-(D

)x(e=PI


2

2x

2

2x x


.conjugatescomplex arerootsThe

. i ±1=m i.e.

2

=mget we,mfor Solving1)+ D= D(Since

12

xe=dx

3

xe=

3

x

D

1 e= dxx

D

1 e=

D

x e

4x

3x

3x2x

2

2x

equation.aldifferenti theofsolutionrequired theis 12

xe+x)esin B+x (Acos =PI+CF=GS

4xx

2. sinx e = y 3)+4D+(D -x2

0=3+4m+m

isequationauxiliary The

sinx e =3)y +4D+(D

isequationaldifferentigivenThe:Solution

2

x-2

)e B+e(A=CF



. ,-31-=m i.e.

0= 3)+(m1)+(mget we,mfor Solving

3x-x-

1)- D= D(since 3)+1)-4(D+1)-((D

sinx)e =

3)+4D+(D

sinx)(e= PI


2

(-x)

2

(-x)


3)+4-4D+1+2D-(D

sinx)(e =

1)- D= D(since 3)+1)-4(D+1)-((D

=

2

(-x)

2

-1)= DSince( 5

2cosx))-(-sinxe=

2D)sinx -(-1)4D-(1

e=

2D))-2D)(-1+((-1

2D)sinx-)(-1(e =

-1)= DSince( 2D)+(-1

sinx)(e=

2D)+(D

sinx)(e=

2(-x)

2

(-x)(-x)

2(-x)

2

(-x)

equation.aldifferenti theofsolutionrequired theis 5

2cosx))-(-sinx(e+)e B+e(A

= PI+CF=GS(-x)

3x-x-

Solve (D3-1)y=x sinx

Solution: The given differential equation is (D3-1)y = x sinx


m3-1= 0


m3-1= 0

i.e (m-1)(m2 +m+1)=0

Solving for m , we get m =

i.e. m = 1,

2

)4.1)-(1±(-1

2

)4.1)-(1±(-1

x/2-)e)x (3/2sin( B+)x (3/2(Acos(=CF


rootreala andconjugatescomplex ofpair a arerootsThe

sin4

13

2

sin1

sin1

13

1

sin1

sin11

13

11

sin1

13

1

sin

22

22

22

2

22

22

2

2

x)-D) (D(

)(-

x))-x(D-=

x )-D(

)-D) (D(

))-((D

x))(-x(D-=

x) (D+-D)(

)-D) (D(-

)))(D-(D+

x))(-x(D-=

)(-D-)D(

)-(-D-

xx


4

cos23

2

sincos4

cos2sinsin3

2

sincos

42

x)))(((-

x))x-(x(=

x))(-x-x+(-(-

x))x-(x( =

)(-

equation.aldifferenti theofsolutionrequired theis2

3cosx)-sinx)-(x(cosx+)ex 3/2sin B+x 3/2(Acos(

= PI+CF=GS

x/2-

)e CF =(A+Bx

nction ismentary fuThe comple

l.l and equats are rea . The roo,i.e. m =

=)i.e. (m-

=m + - m

on isary equatiThe auxili

x )y = x e D+-on is (Dial equati different The givenSolution:

x )y = x e D+Solve (D

x

x

x-

11

01

012

sin12

sin12

2

2

2

2


) xdxx(D

e =

x]) ([xD

e =

)-D+

x] [xe =

)D-

xxePI =

egralular the partice to find Now we hav

x

x

x

x

sin

sin

11(

sin

1(

sin

int

2

2

2

uationrential eq the diffeolution ofrequired sx) is the x+ (-x +e(A+Bx)e

F +PI = GS = C

x) x+ (-x e=

dx x)x+(x [-x e=

)dx xx+(-xe =

x)) x+ (-x(eD

=

xx

x

x

x

x

cos2sin

cos2sin

]sinsinsin

sincos

sincos1

equal.andrealarerootsThe.1,1=m i.e.

0=1)-(m i.e.

0=1+2m-m


x loge=1)y +2D-(DisequationaldifferentigivenThe:Solution

x loge=1)y +2D-(D1.

2

2

x2

x2

1)-(D

dx)elogx ee=

1-D1-D

logxe

=

1)-(D

logx)(e=PI

(-x)xx

x

2

x

dx)Xe(e=Xa-D

1 ax-ax


Bx)e+(A=CF


equal.andrealarerootsThe.1,1=m i.e.

x

dxx)]-[(xlogxe=

dx]ex)-(xlogx[ee=

1)-(D

x))-(xlogx(e=

x

(-x)xx

x

3]-[2logx4

ex=

]3x-logx[2x4

e=

]4

3x-logx

2

x[e=

]2

x-

4

x-logx

2

x[e=

]2

x-dx

x

2

1logx

2

x[e=

x2

22x

22x

222x

222x

x

)e B+e(A=CF



2,-1-=m i.e.

0=1)+2)(m+(mget we,mfor Solving

0=2+3m+ m


e=2)y +3D+(D

isequationaldifferentigivenThe :Solution

e=2)y +3D+2.(D

x-2x-

2

)(e2

)(e2

x

x

21

2

2

1

1

123

int

-PI=PI

eD+

eD+

D+ +D

ePI =


xx

x

ee

e

xt(-x)

xex-e1

xx

e= t wheredtee =

dxeee= e1D

1=PI

dx.e.eee=e2+D

1 =PI xxe2x-)(e2

xx


)(e(-x)t(-x)

xt(-x)

x

ee=ee=

e= t wheredtee =

1)-.(e.ee=

))e-.(e(ee=)e-(tee=

e= t wheredttee=2+D

x)(e2x-

)(ex)(e(-2x)tt(-2x)

xt(-2x)

x

xx

equation.aldifferenti theofsolutionrequired theistion)simplifica(after .ee+)e B+e(A=

1)-.(e.ee-ee+)e B+e(A

= PI+CF=GS

)(e(-2x)x-2x-

x)(e(-2x))(e(-x)x-2x -

x

xx

)


conjugatescomplex arerootsThe

2i ±=m i.e.

0=2i)-2i)(m+(mget we,mfor Solving

4i=4- =m 0,=4+m


cos2xx =4)y +(D

isequationaldifferentigivenThe:Solution

cos2xx =4)y + (Dequation theSolve

222

22

22

2ix

2

22ix

22

22ix

2

22ix

2

i2x2

2

2

1e

4iD)+(D

xe ofpart Real=

4)+4i+4iD+(D

xe ofpart Real=

4)+2i)+((D

xeofpart Real=

4)+(D

exofpart Real=

4)+(D

cos2xx=PI



Bsin2x)+cos2x (A=CF 2(-1)2ix

x)

4iD

+1

1(

4iD

e ofpart Real=

Contd.,

)dx8

1 -

2i

x-(x(

4i

e ofpart Real=

)8

1 -

2i

x-(x

4iD

eofpart Real=

)16i

2+

4i

2x-(x

4iD

eofpart Real=

x)16

D+

4i

D-(1

4iD

e ofpart Real=

22ix

22ix

2

22ix

2(-1)

2

22ix

i


equation.aldifferenti theofsolutionrequired theiscos2x 4

x - )

8

x -

3

x(

4

sin2x)( +

Bsin2x)+cos2x (A= PI+CF=GS

cos2x4

x - )

8

x -

3

x(

4

sin2x)( =

x/8) - 4i

x-

3

x(

4i

e ofpart Real=

)dx8

- 2i

-(x(4i

ofpart Real=

23

23

232ix

1. Solve (D2-2D+2) y = ex cosx Solution: The given differential equation is

(D2-2D+2) y = ex cos x



m2-2m + 2= 0

Solving for m , i.e. m = 1 ± i .

The roots are complex conjugates.


CF = (Acos x + B sin x)ex

EXTRA PROBLEMS)1(

2)1(2)1(2)+2D-(D

cose22

x

1 DDSince

DD

exPI

x

)22212(

cos21

DDD

xePI

x

2

)sin(x)esin Bx (Acos x

xxePIisCFsolutiongeneralThe

x

2

)sin( xxePI

x

)1(cos21

D

xePI

x


2. Solve (D2-3D+2) y = 2cos(2x+3) + 2 ex

Solution : The given differential equation is (D2-3D+2 ) y = 2cos(2x+3) + 2ex



The auxiliary equation is m2- 3m + 2= 0 Solving for m , we get (m -1) (m-2) = 0 i.e. m = 1 ,2 The roots are real and distinct The complementary function is CF = (A ex + B e2x )

Now we have to find the particular integralPI1 =

4)- = D(since

= 2)+3D-(-4

3)+cos(2x(2 =

2)+3D-(D

3)+cos(2x(2 =

2

2

) ) D-

)x+(D)+(-

D) +D)(--(-

)x+(D+(- =

D)-(-

)x+(294(

32cos232

3232

32cos2)32

32

32cos2


)9.(-4)-((4

3)+cos(2x3D)(2+(-2

40

3)+sin(2x12-3)+cos(2x(-4

10

] 3)+sin(2x3-3)+[-cos(2x


] 3)+sin(2x3-3)+[-cos(2x+ )e B+e(A=PI PI+CF=GS 2xx21

)()21(

2

)2)(1(

22 caseexception

xe

DD

ePI

xx

3. Solve the differential equation

(D2 +4D+3)y= 6e-2x sinx sin 2x

Solution The given DE is (D2 +4D+3)y= 6e-2x

sinx sin 2x

The AE is m2 +4m +3 =0 i.e. m=-1, -3


The AE is m2 +4m +3 =0 i.e. m=-1, -3

The complementary function is CF = Ae-x +Be-3x 1

)cos3(cos3

38444

)2sinsin2(3

3)2(4)2(

)2sinsin2(3

)34(

2sinsin62

2

2

2

2

2

2

2

D

xxe

DDD

xxe

DD

xxe

DD

xxePI

xxxx

11

cos3

19

3cos3

1

)cos3(cos3 22

2

2

xexe

D

xxePI

xxx

2

cos3

10

3cos3 BeAe

223x-x- xexePICFGS

xx

4.Solve the equation (D2 +5D+4)y = e-x sin2x

Solution: The differential equation is

(D2 +5D+4)y = e-x sin2x

The auxiliary equation is m2 + 5m +4 =0

Solving for m, we get m = -1, -4.




CF = Ae-x+Be-4x

D

xe

DD

xe

DDD

xe

DD

xe

DD

xePI

xxxxx

54

2sin

5

2sin

14552

2sin

4)1(5)1(

2sin

45

2sin2222

116

)2cos102sin4(

)4(2516

)2cos102sin4(

2516

2sin)54(

)54)(54(

2sin)54(2

xxexxe

D

xDe

DD

xDePI

xx

xx


116)4(2516

116

)2cos102sin4(4 xxeBeAePICFGSx

xx

EXTRA PROBLEMS

)e CF =(A+Bx

nction ismentary fuThe comple

l.l and equats are rea . The roo,i.e. m =

=)i.e. (m-

=m + - m

on isary equatiThe auxili

x x e )y = D+-on is (Dial equati different The givenSolution:

x x e )y = D+Solve (D

x

x

x-

11

01

012

sin812

sin812.5

2

2

2

2

) xdxx(D

e =

x]) ([xD

e =

)-D+

x] [xe =

)D-

xxePI =


x

x

x

x

sin8

sin8

11(

sin8

1(

sin8

int

2

2

2


uationrential eq the diffeolution ofrequired sx) is the x+ (-xe +(A+Bx)e

F +PI = GS = C

x) x+ (-xe =

dx x)x+(x [-xe =

)dx xx+(-xe =

x)) x+ (-xe(D

=

xx

x

x

x

x

cos2sin8

cos2sin8

]sinsinsin8

sincos8

sincos81

6. Solve the differential equation (D2 -4D+3)y= ex cos 2xSolution The given DE is (D2 -4D+3)y= ex cos 2xThe AE is m2 +4m +3 =0 i.e. m=1, 3The complementary function is CF = Aex +Be3x


8

)2sin2(cos

32

)2sin42cos4(

)4(416

)2)(cos24(

)416(

)2)(cos24(

24

)2(cos

2

)2(cos22

xxe

xxexDe

D

xDe

D

xe

DD

xePI

x

xx

xxx

DD

xe

DDD

xe

DD

xe

DD

xePI

xxxx

2

2cos

34412

2cos

3)1(4)1(

2cos

34

)2(cos2222

8

)2sin2(cos BeAe 3xx

xxePICFGS

x

The complementary function is CF = Ae +Be


(D2 +3D+2)y= sin x + x2


(D +3D+2)y= sin x + x

Solution The given DE is (D2 +3D+2)y= sin x + x2

The AE is m2 +3m +2=0 i.e. m=-1, -2

The complementary function is CF = Ae-x +Be-2x

EXTRA PROBLEMS

19

sin)13(

)13)(13(

sin)13(

13

sin

231

sin

23

sin221

D

xD

DD

xD

D

x

D

x

DD

xPI

10

sincos3

1)1(9

sin)13(1

xxxD

PI

)2

3D)+(D+(1

2

1 =

3D)+D+(2

x=PI

1-2

2

2

2

11

(9.2)]4

1+6x)+(2

2

1-[x

2

1=PI

)....]x)2

3D)+(D(+

2

3D)+(D-([1

2

1 =

2

2

2222


]2

3D)+(D+[1

2

1=

)2

+(12

=

(-1)2

]2

7)-6x-(x[

2

1=PI

(9)]2

1+3x)+(1-[x

2

1=PI

2

2

2

2

2

7)-6x-(x

2

1

10

sincos3+ )e B+e(A=

PI PI+CF=GS2

2x-x-

21

xx


(D2 +16)y = cos 3 x

Solution The given DE is (D2 +16)y= cos 3 x



The AE is m2 +16=0 i.e. m = ±4i

The complementary function is CF = Acos4x +Bsin4x

7

3cos

169

3cos

16

3cos

20

cos

)15(4

cos3

)161(4

cos3

)16(4

cos3

)16(4

3coscos3

)16(

cos

22

21

22

3

xx

D

xPI

xxx

D

xPI

D

xx

D

xPI


7

3cos

20

cos4sin4cos21

xxxBxAPIPICFGS

2/x-

3

)2

3sin

2

3(BcosAe CF

root.reala andconjugatescomplex arerootsThe

. 2

3i±11,-=m i.e.

-1m,2

4)-(1±1=mget we,mfor Solving

0=1m


xexCx

1.Solve the DE(D3+1)y =0


1243133

22)2(42423232

22

2

22

2

22

2

22

xexedxxex

D

e

dxxD

e

D

xe

D

xe

D

exPI

xxxx

xxxx

2.Find the particular integral of (D2-4D+4)y

=x2e2x

EXTRA PROBLEMS-PARTA

3.Find the particular integral of (D2+4)y =sin2x

)(4

2cos

)4(

2sin2

caseExceptionxx

D

xPI

4.Find the particular integral of )1(sin1sinsin

11

sin

)1(

sin 2222

Dcexe

D

xe

D

xe

D

exPI

xxxx


4.Find the particular integral of (D-1)2y = exsinx

111)1( 222 DDD

5.Find the particular integral of (D-1)2y = coshx

84)1(2)1(212

)1(

cosh 2

2222

xxxx

xx

eex

D

e

D

e

D

ee

D

xPI


6.Find the particular integral of (D2-4)y = cosh2x

7.Find the particular integral of

8. Find the particular integral of


integral of (D2-4)y =1

integral of (D-2)2y = 2x


(D+1)2y =e-xcosx


42sinh

88)2)(2(2)2)(2(242

)4(

2cosh.6

2222

2

22

2

xxxexe

DD

e

DD

e

D

ee

D

xPI

xxxx

xx

4

1

2,2,,04

4

1

44)4(

1.7

22

22

2

0

2

0

2

xx

xx

xx

BeAePICFGS

BeAeCF

mgetwemforsolvingmisAEThe

e

D

e

DPI


4 BeAePICFGS

)()22(log2)2(2

.8 log2

2log

2

2log

2

axxxxx

eaSincee

D

e

DPI

)1(sin1

coscos

)11(

cos

)1(

cos.9 2

222

ceDxe

D

xe

D

xe

D

xePI

xxxx

EXTRA PROBLEMSSolve the equation (D2+2D+5)y = ex cos3 x

Solve the equation (D2+2D-1)y = (x+ex )2

Find the particular integral of (D2+a2 )y = b cos ax + c sin ax


SOLUTIONS(PART-A & B)


UNIT 2

-P.VEERAIAH

DEPARTMENT OF APPLIED MATHEMATICS

12/23/2014


1

ORDINARY DIFFERENTIAL EQUATIONSMATHEMATICS II(MA6251)

1

UNIT 2 SYLLABUS

12/23/2014


2

Higher order linear differential equations with constant coefficients

Method of variation of parameters

Cauchy’s and Legendre’s linear equations

Simultaneous first order linear equations with constant coefficients

Second-order linear differential equations

12/23/2014


3

Note that the general solution to such an equation must include two arbitrary constants to be completely general.


12/23/2014


4

Adding


12/23/2014


5

The solutions to this quadratic will provide two values of m which will make y = Aemx a solution.


12/23/2014


6

When the roots of the auxiliary equation are both real and equal to m, then the solution would appear to be

y = Aemx + Bemx = (A+B)emx

A + B however is equivalent to a single constant and second order equations need two

With a little further searching we find that y = Bxemx is a solution. So a general solution is

Roots are complex conjugates

12/23/2014


7

When the roots of the auxiliary equation are complex, they will be of the form m1 = p + iq and m2 = p – iq.

Hence the general equation will be

12/23/2014


8

Non homogeneous equations take the form

Suppose g(x) is a particular solution to this equation. Then

Now suppose that g(x) + k(x) is another solution. Then

Non-homogeneous Second-order linear differential equations

Non homogeneoussecond order differential equations

12/23/2014


9

Giving

From the work in previous exercises we know how to find k(x).

This function is referred to as the Complementary Function. (CF)

The function g(x) is referred to as the Particular Integral. (PI)

General Solution = CF + PI

2nd Order DE – Homogeneous LE with Constant Coefficients

(2) If 1 and 2 are distinct real numbers (if b2 - 4ac > 0), then the general solution is:

(3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is:

(4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is:

Where:

12/23/2014


10

2nd Order DE – Homogeneous LE with Constant Coefficients

Homogeneous Linear Equations with Constant Coefficients

A second order homogeneous equation with constant coefficients is written as:

where a, b and c are constant

The steps to follow in order to find the general solution is as follows:

(1) Write down the characteristic equation

This is a quadratic. Let 1 and 2 be its roots we have

12/23/2014


11

Type-1 particular integrals f(x) = eax+b

12/23/2014


12

Solve the equation


m2-2m + 1=0

i.e. (m-1)2 =0

i.e. m = 1,1 . The roots are real and equal.


CF =(A+Bx)ex

The given differential equation is (

=Cosh x

Type-1 particular integrals f(x) = eax+b

12/23/2014


13

Now the general solution of the DE is GS = CF + PI1 + PI2 =

(A+Bx)ex +

Type-1 particular integrals

12/23/2014


14

is the solution of the given DE.

2. Solve (D2-2D+2) y = ex + 5 + e-2x


(D2-2D+2) y = ex + 5 + e-2x


m2-2m + 2= 0


i.e. m = 1 ± i .





12/23/2014


15

Type-2 particular integralsf(x) = cos(ax+b) or sin (ax+b)

12/23/2014


16

Solve (D2-3D+2) y = 2cos(2x+3)

Solution :

The given differential equation is

y = 2cos(2x+3)


m2- 3m + 2= 0

Solving for m , we get (m -1) (m-2) = 0

i.e. m = 1 ,2

The roots are real and distinct


CF = (A ex + B e2x )


12/23/2014


17


PI =


12/23/2014


18

Solve (D2+1)2y = 2sinx cos3x


y = 2sinx cos3x



(m2+1) (m2+1) = 0

i.e. m2 = -1 = i2 twice

Therefore m = ±i, ±i

The roots are pair of complex conjugates


12/23/2014


19


CF = (A+Bx)cos x +(C+Dx) sin x



12/23/2014


20

Type-3 particular integrals f(x) = xm ( m a positive integer)

12/23/2014


21

Solve the DE


12/23/2014


22

2.Solve the DE


12/23/2014


23

1.Solve (D2-2D+2) y = ex x2

Type-4 particular integrals f(x) = eaxV(x)

12/23/2014


24

2.

Type-5 particular integrals f(x) = xV(x)

12/23/2014


25

Solve (D3-1)y=x sinx

Solution: The given differential equation is (D3-1)y = x sinx


m3-1= 0

i.e (m-1)(m2 +m+1)=0


i.e. m = 1,


12/23/2014


26


12/23/2014


27


12/23/2014


28


12/23/2014


29

MISCELLANEOUS MODEL (using Euler’s formula)

12/23/2014


30

MISCELLANEOUS MODEL

12/23/2014


31

Contd.,

EXTRA PROBLEMS

12/23/2014


32

1. Solve (D2-2D+2) y = ex cosx


(D2-2D+2) y = ex cos x


m2-2m + 2= 0

Solving for m , i.e. m = 1 ± i .




EXTRA PROBLEMS

12/23/2014


33

EXTRA PROBLEMS

12/23/2014


34

2. Solve (D2-3D+2) y = 2cos(2x+3) + 2 ex

Solution :

The given differential equation is

(D2-3D+2 ) y = 2cos(2x+3) + 2ex


m2- 3m + 2= 0

Solving for m , we get (m -1) (m-2) = 0

i.e. m = 1 ,2

The roots are real and distinct


CF = (A ex + B e2x )

EXTRA PROBLEMS

12/23/2014


35


PI1 =

EXTRA PROBLEMS

12/23/2014


36


(D2 +4D+3)y= 6e-2x sinx sin 2x

Solution The given DE is (D2 +4D+3)y= 6e-2x sinx sin 2x

The AE is m2 +4m +3 =0 i.e. m=-1, -3


EXTRA PROBLEMS

12/23/2014


37

4.Solve the equation (D2 +5D+4)y = e-x sin2x

Solution: The differential equation is

(D2 +5D+4)y = e-x sin2x

The auxiliary equation is m2 + 5m +4 =0



CF = Ae-x+Be-4x

EXTRA PROBLEMS

12/23/2014


38

EXTRA PROBLEMS

12/23/2014


39

EXTRA PROBLEMS

12/23/2014


40


(D2 -4D+3)y= ex cos 2x

Solution The given DE is (D2 -4D+3)y= ex cos 2x

The AE is m2 +4m +3 =0 i.e. m=1, 3

The complementary function is CF = Aex +Be3x

EXTRA PROBLEMS

12/23/2014


41


(D2 +3D+2)y= sin x + x2

Solution The given DE is (D2 +3D+2)y= sin x + x2

The AE is m2 +3m +2=0 i.e. m=-1, -2


EXTRA PROBLEMS

12/23/2014


42

EXTRA PROBLEMS

12/23/2014


43


(D2 +16)y = cos 3 x


The AE is m2 +16=0 i.e. m = ±4i

The complementary function is CF = Acos4x +Bsin4x

EXTRA PROBLEMS

12/23/2014


44


12/23/2014


45

1.Solve the DE

(D3+1)y =0

2.Find the particular integral of (D2-4D+4)y =x2e2x


12/23/2014


46


(D2+4)y =sin2x


(D-1)2y = exsinx


(D-1)2y = coshx


12/23/2014


47


(D2-4)y = cosh2x


(D2-4)y =1

8. Find the particular integral of

(D-2)2y = 2x


(D+1)2y =e-xcosx


12/23/2014


48

EXTRA PROBLEMS

12/23/2014


49

Solve the equation (D2+2D+5)y = ex cos3 x

Solve the equation (D2+2D-1)y = (x+ex )2

Find the particular integral of (D2+a2 )y = b cos ax + c sin ax

SOLUTIONS(PART-A & B)

12/23/2014


50

2

2

Differential equations of the form ()

are called second order linear different

ial equations.

dydy

abcyQx

dxdx

++=

When ()0 then the equations are referred

to as homogeneous,

Qx

=

When ()0 then the equations are non-homo

geneous.

Qx

¹

Theorem

If () and () are two solutions then so i

s ()()

yfxygxyfxgx

===+

2

2

we have 0

dfdf

abcf

dxdx

++=

2

2

and 0

dgdg

abcg

dxdx

++=

22

22

0

dfdfdgdg

abcfabcg

dxdxdxdx

+++++=

(

)

22

22

0

dfdgdfdg

abcfg

dxdxdxdx

æö

æö

+++++=

ç÷

ç÷

èø

èø

And so ()() is a solution to the differe

ntial equation.

yfxgx

=+

, for and , is a solution to the equati

on0

mx

dy

yAeAmbcy

dx

=+=

2

0

ambmc

++=

It is reasonable to consider it as a pos

sible solution for

2

2

0

dydy

abcy

dxdx

++=

mx

yAe

=

mx

dy

Ame

dx

Þ=

2

2

2

mx

dy

Ame

dx

Þ=

If is a solution it must satisfy

mx

yAe

=

2

0

mxmxmx

aAmebAmecAe

++=

assuming 0, then by division we get

mx

Ae

¹

mxmx

yAeBxe

=+

Where and ()

CABDABi

=+=-

()()

piqxpiqx

yAeBe

+-

=+

pxiqxpxiqx

AeeBee

-

=+

(

)

pxiqxiqx

eAeBe

-

=+

We know that cossin

i

ei

q

qq

=+

(

)

(

)

(

)

cossincos()sin()

px

eAqxiqxBqxiqx

=++-+-

(

)

(

)

(

)

cossincossin

px

eAqxiqxBqxiqx

=++-

(

)

(

)

(

)

cossin

px

eABqxABiqx

=++-

(

)

cossin

px

eCqxDqx

=+

2

2

()

dydy

abcyQx

dxdx

++=

2

2

()

dgdg

abcgQx

dxdx

++=

2

2

()()

()()

dgkdgk

abcgkQx

dxdx

++

+++=

22

22

()

dgdkdgdk

aabbcgckQx

dxdxdxdx

+++++=

22

22

()

dgdgdkdk

abcgabckQx

dxdxdxdx

æöæö

Þ+++++=

ç÷ç÷

èøèø

2

2

()()

dkdk

QxabckQx

dxdx

æö

Þ+++=

ç÷

èø

2

2

0

dkdk

abck

dxdx

Þ++=

x

x

e

c

e

c

y

2

1

2

1

l

l

+

=

x

x

xe

c

e

c

y

1

1

2

1

l

l

+

=

(

)

(

)

x

e

c

x

e

c

y

x

x

b

b

a

a

sin

cos

2

1

+

=

a

b

ac

a

b

2

4

and

2

2

-

=

-

=

b

a

(

)

0

0

¹

=

+

¢

+

¢

¢

a

cy

y

b

y

a

(

)

0

0

2

¹

=

+

+

a

c

b

a

l

l

a

ac

b

b

2

4

2

2

,

1

-

±

-

=

l

coshx

=

1)y

+

2D

-

(D

2

2

1

2

2

1

2

2

1

2

cosh

integral

particular

the

find

to

have

Now we

+ PI

= PI

)

D+

-

(D

+e

e

=

)

D+

-

(D

x

PI =

-x

x

twice)

0

r

denominato

the

makes

1

=

D

Since

(

4

)

e

(x

=

1)

+

2D

-

2(D

e

=

Now PI

x

2

2

x

1

)

-

ting D =

( Substitu

e

=

)

D+

-

(D

e

=

PI

Similarly

-x

-x

1

8

1

2

2

2

2

2

4.2)

-

(4

±

2

)

1

(

1

2)

+

2D

-

(D

e

2

x

1

=

=

=

D

Since

e

PI

x

)

uting D =

(Substit

=

)

D+

-

D

e

=

PI

Similarly

x

0

2

5

2

2

(

5

2

0

2

2)

=

D

ing

(Substitut

/

2

e

=

2)

+

2D

-

(D

e

=

PI

2x

2

2x

3

DE.

given

the

of

solution

the

is

2

e

+

2

5

+

1

e

+

x)e

sin

B

+

x

(Acos

=

PI

+

PI

+

PI

+

CF

=

GS

solution

general

Now the

2x

x

x

3

2

1

2)

+

3D

-

(D

2

4)

-

=

D

(since

=

2)

+

3D

-

(-4

3)

+

cos(2x

(2

=

2)

+

3D

-

(D

3)

+

cos(2x

(2

=

2

2

) )

D

-

)

x+

(

D)

+

(-

D)

+

D)(-

-

(-

)

x+

(

D

+

(-

=

D)

-

(-

)

x+

(

2

9

4

(

3

2

cos

2

3

2

3

2

3

2

3

2

cos

2

)

3

2

3

2

3

2

cos

2

=

?)/10

]

3)

+

sin??(2x

3

-

3)

+

([-cos?(2x

=

?)/40

3)

+

sin??(2x

12

-

3)

+

cos?(2x

((-4

?)/10

]

3)

+

sin??(2x

3

-

3)

+

([-cos?(2x

=

?)/40

3)

+

sin??(2x

12

-

3)

+

cos?(2x

((-4

=

)

9.(-4)

-

?)/((4

3)

+

cos??(2x

3D)2

+

((-2

==

)

9.(-4)

-

?)/((4

3)

+

cos??(2x

((2

3D)

+

-2

=

)

9.(-4)

-

((4

3)

+

cos(2x

3D)(2

+

(-2

40

3)

+

sin(2x

12

-

3)

+

cos(2x

(-4

10

]

3)

+

sin(2x

3

-

3)

+

[-cos(2x

DE.

given

the

of

solution

the

is

10

]

3)

+

sin(2x

3

-

3)

+

[-cos(2x

+

)

e

B

+

e

(A

=

PI

+

CF

=

GS

2x

x

2

2

1)

+

(D

2

1

2

2

2

2

1

2

sin

4

sin

1

)

3

cos

sin

2

+ PI

PI

=

)

+

(D

x

x-

=

)

+

(D

x

x

(

PI =

)

16

-

=

D

(since

225

sin4x

=

1)

+

(-16

sin4x

=

1)

+

(D

sin4x

=

PI

2

2

2

2

1

)

4

-

=

D

(since

9

sin4x

=

1)

+

(-4

sin4x

=

PI

2

2

2

DE.

given

the

of

solution

the

is

9

sin2x

+

225

sin4x

+

x

sin

Dx)

+

(C

+

x

Bx)cos

+

(A

=

PI

+

PI

+

CF

=

GS

2

1

4

2

)

4

(

x

y

D

=

+

4

2

)

4

(

x

y

D

=

+

x

B

x

A

F

C

2

sin

2

cos

.

.

+

=

2

4

4

D

x

+

4

2

)

4

1

(

1

4

1

x

D

+

4

1

2

)

4

1

(

4

1

x

D

-

+

÷

ø

ö

ç

è

æ

+

-

=

÷

ø

ö

ç

è

æ

+

-

=

÷

ø

ö

ç

è

æ

+

-

=

2

3

3

4

1

16

24

4

12

4

1

16

4

1

4

1

2

4

2

4

4

4

2

x

x

x

x

x

D

D

÷

ø

ö

ç

è

æ

+

-

+

+

=

+

=

2

3

3

4

1

2

sin

2

cos

2

4

x

x

x

B

x

A

PI

CF

GS

x

=

y

2)

+

3D

+

(D

2

2

x

=

2)y

+

3D

+

(D

is

equation

al

differenti

given

The

:

Solution

2

2

distinct.

and

real

are

roots

The

.

,-2

1

-

=

m

i.e.

0

=

2)

+

(m

1)

+

(m

get

we

,

m

for

Solving

0

=

2

+

3m

+

m

is

equation

auxiliary

The

2

)

e

B

+

e

(A

=

CF

is

function

ary

complement

The

.

2x

-

x

-

]

2

3D)

+

(D

+

[1

2

1

=

)

2

3D)

+

(D

+

(1

2

1

=

3D)

+

D

+

(2

x

=

PI

(-1)

2

1

-

2

2

2

]

2

7)

-

6x

-

(x

[

2

1

=

PI

(9)]

2

1

+

3x)

+

(1

-

[x

2

1

=

PI

(9.2)]

4

1

+

6x)

+

(2

2

1

-

[x

2

1

=

PI

)

....]x

)

2

3D)

+

(D

(

+

2

3D)

+

(D

-

([1

2

1

=

2

2

2

2

2

2

2

+

2

7)

-

6x

-

(x

2

1

+

)

e

B

+

e

(A

=

PI

+

CF

=

GS

2

2x

-

x

-

x

e

=

2)y

+

2D

-

(D

is

equation

al

differenti

given

The

2

x

2

.

conjugates

complex

are

roots

The

.

i

±

1

=

m

i.e.

2

4.2)

-

(4

±

(2

=

m

get

we

,

m

for

Solving

0

=

2

+

2m

-

m

is

equation

auxiliary

The

2

x)e

sin

B

+

x

(Acos

=

CF

is

function

ary

complement

The

x

1)

+

D

=

D

(Since

1)

+

2

-

2D

-

1

+

2D

+

(D

e

=

2)

+

2D

-

(D

)

x

(e

=

PI

integral

particular

the

find

to

have

Now we

2

2

x

2

2

x

x

12

x

e

=

dx

3

x

e

=

3

x

D

1

e

=

dx

x

D

1

e

=

D

x

e

4

x

3

x

3

x

2

x

2

2

x

ò

ò

equation.

al

differenti

the

of

solution

required

the

is

12

x

e

+

x)e

sin

B

+

x

(Acos

=

PI

+

CF

=

GS

4

x

x

sinx

e

=

y

3)

+

4D

+

(D

-x

2

0

=

3

+

4m

+

m

is

equation

auxiliary

The

sinx

e

=

3)y

+

4D

+

(D

is

equation

al

differenti

given

The

:

Solution

2

x

-

2

)

e

B

+

e

(A

=

CF

is

function

ary

complement

The

distinct.

and

real

are

roots

The

.

,-3

1

-

=

m

i.e.

0

=

3)

+

(m

1)

+

(m

get

we

,

m

for

Solving

3x

-

x

-

3)

+

4

-

4D

+

1

+

2D

-

(D

sinx)

(e

=

1)

-

D

=

D

(since

3)

+

1)

-

4(D

+

1)

-

((D

sinx)

e

=

3)

+

4D

+

(D

sinx)

(e

=

PI

integral

particular

the

find

to

have

Now we

2

(-x)

2

(-x)

2

(-x)

-1)

=

D

Since

(

5

2cosx))

-

(-sinx

e

=

2D)sinx

-

(-1

)

4D

-

(1

e

=

2D))

-

2D)(-1

+

((-1

2D)sinx

-

)(-1

(e

=

-1)

=

D

Since

(

2D)

+

(-1

sinx)

(e

=

2D)

+

(D

sinx)

(e

=

2

(-x)

2

(-x)

(-x)

2

(-x)

2

(-x)

equation.

al

differenti

the

of

solution

required

the

is

5

2cosx))

-

(-sinx

(e

+

)

e

B

+

e

(A

=

PI

+

CF

=

GS

(-x)

3x

-

x

-

2

)

4.1)

-

(1

±

(-1

2

)

4.1)

-

(1

±

(-1

x/2

-

)e

)x

(3/2

sin(

B

+

)x

(3/2

(Acos(

=

CF

is

function

ary

complement

The

root

real

a

and

conjugates

complex

of

pair

a

are

roots

The

4

cos

2

3

2

sin

cos

4

cos

2

sin

sin

3

2

sin

cos

sin

4

1

3

2

sin

1

sin

1

1

3

1

sin

1

sin

1

1

1

3

1

1

sin

1

1

3

1

sin

2

2

2

2

2

2

2

2

2

2

2

2

2

x)))

(

(

(

-

x))

x-

(x(

=

x))

(-

x-

x+

(-

(

-

x))

x-

(x(

=

x

)

-D)

(

D

(

)

(-

x)

)

-x(D-

=

x

)

-D

(

)

-D)

(

D

(

))

-

((D

x)

)

(-x(D-

=

x

)

(D+

-D)

(

)

-D)

(

D

(

-

))

)(D-

(D+

x)

)

(-x(D-

=

)

(-D-

)

D

(

)-

(-D-

x

x

-

-

equation.

al

differenti

the

of

solution

required

the

is

2

3cosx)

-

sinx)

-

(x(cosx

+

)e

x

3/2

sin

B

+

x

3/2

(Acos(

=

PI

+

CF

=

GS

x/2

-

)e

CF =(A+Bx

nction is

mentary fu

The comple

l.

l and equa

ts are rea

. The roo

,

i.e. m =

=

)

i.e. (m-

=

m +

-

m

on is

ary equati

The auxili

x

)y = x e

D+

-

on is (D

ial equati

different

The given

Solution:

x

)y = x e

D+

Solve (D

x

x

x

-

1

1

0

1

0

1

2

sin

1

2

sin

1

2

2

2

2

2

)

xdx

x

(

D

e

=

x])

([x

D

e

=

)

-

D+

x]

[x

e

=

)

D-

x

xe

PI =

egral

ular

the partic

e to find

Now we hav

x

x

x

x

ò

sin

sin

1

1

(

sin

1

(

sin

int

2

2

2

uation

rential eq

the diffe

olution of

required s

x) is the

x+

(-x

+e

(A+Bx)e

F +PI =

GS = C

x)

x+

(-x

e

=

dx

x)

x+

(

x

[-x

e

=

)dx

x

x+

(-x

e

=

x))

x+

(-x

(e

D

=

x

x

x

x

x

x

cos

2

sin

cos

2

sin

]

sin

sin

sin

sin

cos

sin

cos

1

ò

ò

-

Bx)e

+

(A

=

CF

is

function

ary

complement

The

equal.

and

real

are

roots

The

.

1,1

=

m

i.e.

0

=

1)

-

(m

i.e.

0

=

1

+

2m

-

m

is

equation

auxiliary

The

x

log

e

=

1)y

+

2D

-

(D

is

equation

al

differenti

given

The

:

Solution

x

log

e

=

1)y

+

2D

-

(D

1.

x

2

2

x

2

x

2

dx

x)]

-

[(xlogx

e

=

dx

]

e

x)

-

(xlogx

[e

e

=

1)

-

(D

x))

-

(xlogx

(e

=

1)

-

(D

dx)

e

logx

e

e

=

1

-

D

1

-

D

logx

e

=

1)

-

(D

logx)

(e

=

PI

x

(-x)

x

x

x

(-x)

x

x

x

2

x

ò

ò

ò

3]

-

[2logx

4

e

x

=

]

3x

-

logx

[2x

4

e

=

]

4

3x

-

logx

2

x

[

e

=

]

2

x

-

4

x

-

logx

2

x

[

e

=

]

2

x

-

dx

x

2

1

logx

2

x

[

e

=

x

2

2

2

x

2

2

x

2

2

2

x

2

2

2

x

ò

-

x

ò

dx

)

Xe(

e

=

X

a

-

D

1

ax

-

ax

)

e

B

+

e

(A

=

CF

is

function

ary

complement

The

distinct.

and

real

are

roots

The

2,-1

-

=

m

i.e.

0

=

1)

+

2)(m

+

(m

get

we

,

m

for

Solving

0

=

2

+

3m

+

m

is

equation

auxiliary

The

e

=

2)y

+

3D

+

(D

is

equation

al

differenti

given

The

:

Solution

e

=

2)y

+

3D

+

2.(D

x

-

2x

-

2

)

(e

2

)

(e

2

x

x

2

1

2

2

1

1

1

2

3

int

-PI

=PI

e

D+

e

D+

D+

+

D

e

PI =

egral

ular

the partic

e to find

Now we hav

x

x

x

e

e

e

-

)

(e

(-x)

t

(-x)

x

t

(-x)

x

e

x

-

e

1

x

x

x

e

e

=

e

e

=

e

=

t

where

dt

e

e

=

dx

e

e

e

=

e

1

D

1

=

PI

ò

ò

+

1)

-

.(e

.e

e

=

))

e

-

.(e

(e

e

=

)

e

-

(te

e

=

e

=

t

where

dt

te

e

=

dx

.e

.e

e

e

=

e

2

+

D

1

=

PI

x

)

(e

2x

-

)

(e

x

)

(e

(-2x)

t

t

(-2x)

x

t

(-2x)

x

x

e

2x

-

)

(e

2

x

x

x

x

x

ò

ò

equation.

al

differenti

the

of

solution

required

the

is

tion)

simplifica

(after

.e

e

+

)

e

B

+

e

(A

=

1)

-

.(e

.e

e

-

e

e

+

)

e

B

+

e

(A

=

PI

+

CF

=

GS

)

(e

(-2x)

x

-

2x

-

x

)

(e

(-2x)

)

(e

(-x)

x

-

2x

-

x

x

x

Bsin2x)

+

cos2x

(A

=

CF

is

function

ary

complement

The

conjugates

complex

are

roots

The

2i

±

=

m

i.e.

0

=

2i)

-

2i)(m

+

(m

get

we

,

m

for

Solving

4i

=

4

-

=

m

0,

=

4

+

m

is

equation

auxiliary

The

cos2x

x

=

4)y

+

(D

is

equation

al

differenti

given

The

:

Solution

cos2x

x

=

4)y

+

(D

equation

the

Solve

2

2

2

2

2

2

2

2

(-1)

2ix

2

2

2ix

2

2

2

2ix

2

2

2ix

2

i2x

2

2

2

x

)

4i

D

+

1

1

(

4iD

e

of

part

Real

=

4iD)

+

(D

x

e

of

part

Real

=

4)

+

4i

+

4iD

+

(D

x

e

of

part

Real

=

4)

+

2i)

+

((D

x

e

of

part

Real

=

4)

+

(D

e

x

of

part

Real

=

4)

+

(D

cos2x

x

=

PI

integral

particular

the

find

to

have

Now we

equation.

al

differenti

the

of

solution

required

the

is

cos2x

4

x

-

)

8

x

-

3

x

(

4

sin2x)

(

+

Bsin2x)

+

cos2x

(A

=

PI

+

CF

=

GS

cos2x

4

x

-

)

8

x

-

3

x

(

4

sin2x)

(

=

x/8)

-

4i

x

-

3

x

(

4i

e

of

part

Real

=

)dx

8

1

-

2i

x

-

(x

(

4i

e

of

part

Real

=

)

8

1

-

2i

x

-

(x

4iD

e

of

part

Real

=

)

16i

2

+

4i

2x

-

(x

4iD

e

of

part

Real

=

x

)

16

D

+

4i

D

-

(1

4iD

e

of

part

Real

=

2

3

2

3

2

3

2ix

2

2ix

2

2ix

2

2

2ix

2

(-1)

2

2

2ix

ò

i

)

1

(

2

)

1

(

2

)

1

(

2)

+

2D

-

(D

cos

e

2

2

x

1

+

=

+

+

-

+

=

=

D

D

Since

D

D

e

x

PI

x

)

2

2

2

1

2

(

cos

2

1

+

-

-

+

+

=

D

D

D

x

e

PI

x

2

)

sin

(

x)e

sin

B

x

(Acos

x

x

x

e

PI

isCF

solution

general

The

x

+

+

=

+

2

)

sin

(

x

x

e

PI

x

=

)

1

(

cos

2

1

+

=

D

x

e

PI

x

4)

-

=

D

(since

=

2)

+

3D

-

(-4

3)

+

cos(2x

(2

=

2)

+

3D

-

(D

3)

+

cos(2x

(2

=

2

2

DE.

given

the

of

solution

the

is

10

]

3)

+

sin(2x

3

-

3)

+

[-cos(2x

+

)

e

B

+

e

(A

=

PI

PI

+

CF

=

GS

2x

x

2

1

+

)

(

)

2

1

(

2

)

2

)(

1

(

2

2

case

exception

xe

D

D

e

PI

x

x

-

=

-

-

=

1

)

cos

3

(cos

3

3

8

4

4

4

)

2

sin

sin

2

(

3

3

)

2

(

4

)

2

(

)

2

sin

sin

2

(

3

)

3

4

(

2

sin

sin

6

2

2

2

2

2

2

2

2

-

-

-

=

+

-

+

+

-

-

-

=

+

-

+

-

-

-

=

+

+

=

-

-

-

-

D

x

x

e

D

D

D

x

x

e

D

D

x

x

e

D

D

x

x

e

PI

x

x

x

x

1

1

cos

3

1

9

3

cos

3

1

)

cos

3

(cos

3

2

2

2

2

-

-

+

-

-

-

=

-

-

-

=

-

-

-

x

e

x

e

D

x

x

e

PI

x

x

x

2

cos

3

10

3

cos

3

Be

Ae

2

2

3x

-

x

-

x

e

x

e

PI

CF

GS

x

x

-

-

-

+

+

=

+

=

D

x

e

D

D

x

e

D

D

D

x

e

D

D

x

e

D

D

x

e

PI

x

x

x

x

x

5

4

2

sin

5

2

sin

1

4

5

5

2

2

sin

4

)

1

(

5

)

1

(

2

sin

4

5

2

sin

2

2

2

2

+

-

=

+

=

+

+

-

+

-

=

+

-

+

-

=

+

+

=

-

-

-

-

-

116

)

2

cos

10

2

sin

4

(

)

4

(

25

16

)

2

cos

10

2

sin

4

(

25

16

2

sin

)

5

4

(

)

5

4

)(

5

4

(

2

sin

)

5

4

(

2

x

x

e

x

x

e

D

x

D

e

D

D

x

D

e

PI

x

x

x

x

-

-

=

-

-

-

-

=

-

-

-

=

-

-

+

-

-

-

=

-

-

-

-

116

)

2

cos

10

2

sin

4

(

4

x

x

e

Be

Ae

PI

CF

GS

x

x

x

-

-

+

+

=

+

=

-

-

-

)e

CF =(A+Bx

nction is

mentary fu

The comple

l.

l and equa

ts are rea

. The roo

,

i.e. m =

=

)

i.e. (m-

=

m +

-

m

on is

ary equati

The auxili

x

x e

)y =

D+

-

on is (D

ial equati

different

The given

Solution:

x

x e

)y =

D+

Solve (D

x

x

x

-

1

1

0

1

0

1

2

sin

8

1

2

sin

8

1

2

.

5

2

2

2

2

)

xdx

x

(

D

e

=

x])

([x

D

e

=

)

-

D+

x]

[x

e

=

)

D-

x

xe

PI =

egral

ular

the partic

e to find

Now we hav

x

x

x

x

ò

sin

8

sin

8

1

1

(

sin

8

1

(

sin

8

int

2

2

2

uation

rential eq

the diffe

olution of

required s

x) is the

x+

(-x

e

+

(A+Bx)e

F +PI =

GS = C

x)

x+

(-x

e

=

dx

x)

x+

(

x

[-x

e

=

)dx

x

x+

(-x

e

=

x))

x+

(-x

e

(

D

=

x

x

x

x

x

x

cos

2

sin

8

cos

2

sin

8

]

sin

sin

sin

8

sin

cos

8

sin

cos

8

1

ò

ò

-

8

)

2

sin

2

(cos

32

)

2

sin

4

2

cos

4

(

)

4

(

4

16

)

2

)(cos

2

4

(

)

4

16

(

)

2

)(cos

2

4

(

2

4

)

2

(cos

2

)

2

(cos

2

2

x

x

e

x

x

e

x

D

e

D

x

D

e

D

x

e

D

D

x

e

PI

x

x

x

x

x

x

+

-

=

-

-

=

-

-

+

-

=

-

+

-

=

-

-

=

-

=

D

D

x

e

D

D

D

x

e

D

D

x

e

D

D

x

e

PI

x

x

x

x

2

2

cos

3

4

4

1

2

2

cos

3

)

1

(

4

)

1

(

2

cos

3

4

)

2

(cos

2

2

2

2

-

=

+

-

-

+

+

=

+

+

-

+

=

+

-

=

8

)

2

sin

2

(cos

Be

Ae

3x

x

x

x

e

PI

CF

GS

x

+

-

+

=

+

=

1

9

sin

)

1

3

(

)

1

3

)(

1

3

(

sin

)

1

3

(

1

3

sin

2

3

1

sin

2

3

sin

2

2

1

-

-

=

+

-

-

=

+

=

+

+

-

=

+

+

=

D

x

D

D

D

x

D

D

x

D

x

D

D

x

PI

10

sin

cos

3

1

)

1

(

9

sin

)

1

3

(

1

-

-

=

-

-

-

=

x

x

x

D

PI

]

2

3D)

+

(D

+

[1

2

1

=

)

2

3D)

+

(D

+

(1

2

1

=

3D)

+

D

+

(2

x

=

PI

(-1)

2

1

-

2

2

2

2

]

2

7)

-

6x

-

(x

[

2

1

=

PI

(9)]

2

1

+

3x)

+

(1

-

[x

2

1

=

PI

(9.2)]

4

1

+

6x)

+

(2

2

1

-

[x

2

1

=

PI

)

....]x

)

2

3D)

+

(D

(

+

2

3D)

+

(D

-

([1

2

1

=

2

2

2

2

2

2

2

2

2

2

+

2

7)

-

6x

-

(x

2

1

10

sin

cos

3

+

)

e

B

+

e

(A

=

PI

PI

+

CF

=

GS

2

2x

-

x

-

2

1

+

-

-

+

x

x

7

3

cos

16

9

3

cos

16

3

cos

20

cos

)

15

(

4

cos

3

)

16

1

(

4

cos

3

)

16

(

4

cos

3

)

16

(

4

3

cos

cos

3

)

16

(

cos

2

2

2

1

2

2

3

x

x

D

x

PI

x

x

x

D

x

PI

D

x

x

D

x

PI

=

+

-

=

+

=

=

=

+

-

=

+

=

+

+

=

+

=

7

3

cos

20

cos

4

sin

4

cos

2

1

x

x

x

B

x

A

PI

PI

CF

GS

+

+

+

=

+

+

=

2

/

x

-

3

)

2

3

sin

2

3

(Bcos

Ae

CF

root.

real

a

and

conjugates

complex

are

roots

The

.

2

3

i

±

1

1,

-

=

m

i.e.

-1

m

,

2

4)

-

(1

±

1

=

m

get

we

,

m

for

Solving

0

=

1

m

is

equation

auxiliary

The

x

e

x

C

x

+

+

=

=

+

(

)

12

4

3

1

3

3

2

2

)

2

(

4

2

4

2

3

2

3

2

2

2

2

2

2

2

2

2

2

2

2

x

e

x

e

dx

x

e

x

D

e

dx

x

D

e

D

x

e

D

x

e

D

e

x

PI

x

x

x

x

x

x

x

x

=

=

=

=

=

=

-

+

=

-

=

ò

ò

)

(

4

2

cos

)

4

(

2

sin

2

case

Exception

x

x

D

x

PI

-

=

+

=

(

)

)

1

(sin

1

sin

sin

1

1

sin

)

1

(

sin

2

2

2

2

-

=

-

=

=

-

+

=

-

=

D

ce

x

e

D

x

e

D

x

e

D

e

x

PI

x

x

x

x

(

)

8

4

)

1

(

2

)

1

(

2

1

2

)

1

(

cosh

2

2

2

2

2

x

x

x

x

x

x

e

e

x

D

e

D

e

D

e

e

D

x

PI

-

-

-

+

=

-

+

-

=

-

+

=

-

=

(

)

4

2

sinh

8

8

)

2

)(

2

(

2

)

2

)(

2

(

2

4

2

)

4

(

2

cosh

.

6

2

2

2

2

2

2

2

2

x

x

xe

xe

D

D

e

D

D

e

D

e

e

D

x

PI

x

x

x

x

x

x

=

-

=

+

-

+

+

-

=

-

+

=

-

=

-

-

-

4

1

2

,

2

,

,

0

4

4

1

4

4

)

4

(

1

.

7

2

2

2

2

2

0

2

0

2

-

+

=

+

=

+

=

-

=

=

-

-

=

-

=

-

=

-

=

-

-

x

x

x

x

x

x

Be

Ae

PI

CF

GS

Be

Ae

CF

m

get

we

m

for

solving

m

is

AE

The

e

D

e

D

PI

(

)

)

(

)

2

2

(log

2

)

2

(

2

.

8

log

2

2

log

2

2

log

2

a

x

x

x

x

x

e

a

Since

e

D

e

D

PI

=

-

=

-

=

-

=

)

1

(sin

1

cos

cos

)

1

1

(

cos

)

1

(

cos

.

9

2

2

2

2

-

=

-

=

=

+

-

=

+

=

-

-

-

-

ceD

x

e

D

x

e

D

x

e

D

x

e

PI

x

x

x

x

Documents

ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I... · ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251)-P.VEERAIAH DEPARTMENT OF APPLIED MATHEMATICS