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UNIT 2 ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) UNIT 2 -P.VEERAIAH DEPARTMENT OF APPLIED MATHEMATICS 12/23/2014 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 1

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Page 1: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

UNIT 2

ORDINARY DIFFERENTIALEQUATIONS

MATHEMATICS II(MA6251)UNIT 2

-P.VEERAIAHDEPARTMENT OF APPLIED

MATHEMATICS

12/23/2014

SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR

1

Page 2: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

Page 3: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 3

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.

Page 4: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

Second-order linear differentialequations

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 2

0d f df d g dg

a b cf a b cg

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 4

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

Page 5: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

Second-order linear differentialequations

, for and , is a solution to the equation 0mx dyy Ae A m b cy

dx

It is reasonable to consider it as a possible solution for

2

2 0d y dy

a b cydx dx

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 5

2dx dxmxy Ae mxdy

Amedx

2

22

mxd yAm 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.

Page 6: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 6

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

Page 7: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 7

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

Page 8: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

2

2 ( )d y dy

a b cy Q xdx dx

2

2 ( )d g dg

a b cg Q xdx dx

Non-homogeneousSecond-order lineardifferential equations

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 8

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

Page 9: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 9

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

Page 10: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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 2121

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

xx xececy 1121

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 10

Page 11: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

42

2,1

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 11

Page 12: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

Solve the equation

coshx=1)y+2D-(D2

The given differential equation is ( =Coshx

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

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

Page 13: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 13

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 +

Page 14: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 14

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

Page 15: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

)1(12)+2D-(D

e2

x

1 DSincee

PIx

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

e = PISimilarly

x

02

5

22(

52

0

2

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 15

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

Page 16: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 16

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 )

Page 17: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 17

) ) 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

DE.given theofsolution theis 10

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

Page 18: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

Solution: The given differential equation is

y = 2sinx cos3x

The auxiliary equation is

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

22 1)+(D

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 18

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

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

i.e. m2 = -1 = i2 twice

Therefore m = ±i, ±i

The roots are pair of complex conjugates

Page 19: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

The complementary function is

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

Now we have to find the particular integral

2222 1

2sin4sin

1

)3cossin2 =

)+(D

xx- =

)+(D

xx(PI =

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 19

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

Page 20: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

DE.given theof solution theis

9

sin2x+

225

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 20

Page 21: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

42 )4( xyD

42 )4( xyD

xBxAFC 2sin2cos.. 11

24121

1641

4

1

2

4

42

x

xDD

Solve the DE

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 21

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

Page 22: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 22

]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-

Page 23: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

integralparticular thefind tohaveNow we

2

2x

2

2x x

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

.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

Page 24: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

isfunctionary complementThe

distinct.andrealarerootsThe

. ,-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

integralparticular thefind tohaveNow we

2

(-x)

2

(-x)

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 24

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-

Page 25: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

Solve (D3-1)y=x sinx

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

The auxiliary equation is

m3-1= 0

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 25

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

isfunctionary complementThe

rootreala andconjugatescomplex ofpair a arerootsThe

Page 26: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 26

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-

Page 27: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

)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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 27

) 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

Page 28: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

0=1)-(m i.e.

0=1+2m-m

isequationauxiliary The

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 28

Bx)e+(A=CF

isfunctionary complementThe

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

Page 29: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

)e B+e(A=CF

isfunctionary complementThe

distinct.andrealarerootsThe

2,-1-=m i.e.

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

0=2+3m+ m

isequationauxiliary The

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 =

egralular the partice to find Now we hav

xx

x

ee

e

xt(-x)

xex-e1

xx

e= t wheredtee =

dxeee= e1D

1=PI

dx.e.eee=e2+D

1 =PI xxe2x-)(e

2

xx

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 29

)(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

Page 30: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

)

isfunctionary complementThe

conjugatescomplex arerootsThe

2i ±=m i.e.

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

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

isequationauxiliary The

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

integralparticular thefind tohaveNow we

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 30

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

x)

4iD

+1

1(

4iD

e ofpart Real=

Page 31: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 31

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

Page 32: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

Solution: The given differential equation is

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

The auxiliary equation is

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 32

m2-2m + 2= 0

Solving for m , i.e. m = 1 ± i .

The roots are complex conjugates.

The complementary function is

CF = (Acos x + B sin x)ex

Page 33: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

EXTRA PROBLEMS)1(

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

cose22

x

1

DDSinceDD

exPI

x

)22212(

cos21

DDD

xePI

x

2

)sin(x)esin Bx (Acos x xxe

PIisCFsolutiongeneralThex

2

)sin( xxePI

x

)1(

cos21

D

xePI

x

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 33

Page 34: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 34

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 )

Page 35: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 35

)9.(-4)-((4

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

40

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

10

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

DE.given theofsolution theis 10

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

21

)()21(

2

)2)(1(

22 caseexception

xe

DD

ePI

xx

Page 36: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 36

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- xexe

PICFGSxx

Page 37: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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.

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 37

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

The complementary function is

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

Page 38: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

116

)2cos102sin4(

)4(2516

)2cos102sin4(

2516

2sin)54(

)54)(54(

2sin)54(2

xxexxe

D

xDe

DD

xDePI

xx

xx

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 38

116)4(2516

116

)2cos102sin4(4 xxeBeAePICFGS

xxx

Page 39: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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 =

egralular the partice to find Now we hav

x

x

x

x

sin8

sin8

11(

sin8

1(

sin8

int

2

2

2

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 39

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

Page 40: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 40

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 xxe

PICFGSx

The complementary function is CF = Ae +Be

Page 41: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

7. Solve the differential equation

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 41

(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

Page 42: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 42

]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

Page 43: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

8. Solve the differential equation

(D2 +16)y = cos 3 x

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 43

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

Page 44: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 44

7

3cos

20

cos4sin4cos21

xxxBxAPIPICFGS

Page 45: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

isequationauxiliary The

xexCx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 45

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

Page 46: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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(sin1

sinsin

11

sin

)1(

sin 2

222

Dce

xe

D

xe

D

xe

D

exPI

xxxx

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 46

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

Page 47: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

EXTRA PROBLEMS-PARTA

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

7.Find the particular integral of

8. Find the particular integral of

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 47

integral of (D2-4)y =1

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

9.Find the particular integral of

(D+1)2y =e-xcosx

Page 48: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

EXTRA PROBLEMS-PARTA

4

2sinh

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 48

4 BeAePICFGS

)()22(log2)2(

2.8 log

2

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

Page 49: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

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

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 49

Page 50: ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) I YR/UNIT 2(PART I).pdf · 2 2 0 d y dy a b cy dx dx 12/23/2014 SVCE, DEPARTMENT OF APPLIED 5 MATHEMATICS, SRIPERUMBUDUR y Aemx

SOLUTIONS(PART-A & B)

12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 50