13 2nd Order Linear De

8/13/2019 13 2nd Order Linear De

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13 Second Order Linear13 Second Order Linear

Differential EquationsDifferential EquationsObjectives:Objectives:

AppyAppy the second order homogeneousthe second order homogeneousequation for auxiliary equationequation for auxiliary equation

Find the general solution for theFind the general solution for thesecond order nonsecond order non--homogeneoushomogeneous

of the nonof the non--homogeneous equationhomogeneous equation

Solve the coefficients and find theSolve the coefficients and find theparticular solutionsparticular solutions

22ndnd

Order LinearOrder LinearThe general equation for linear differential equationThe general equation for linear differential equation

of nth order isof nth order is

where f(x) and the coefficientswhere f(x) and the coefficients aa ii (x)((x)( ii = 0, 1, 2, , n)= 0, 1, 2, , n)depend on variable x.depend on variable x.

n n 1 2

n n 1 2 1 0n n 1 2

d y d y d y dya a a a a y f ( x )

dxdx dx dx

+ + + + + =


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Types of 2Types of 2ndnd ODEODE

Second order homogeneous DE isSecond order homogeneous DE is

22ndnd OrderOrder

HomogeneousHomogeneous

2

Where a, b and c are constants. Sometimes, it canWhere a, b and c are constants. Sometimes, it canbe denoted bybe denoted by

2

y ya b cy 0

dxdx+ + =

2

=


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Now consider,Now consider,

22ndnd OrderOrderHomogeneousHomogeneous

mxy e= 2d y dy

mxyme

dx=

22 mx

2

d ym e

dx=

2 mx mx mxam e bm e ce 0+ + =

2 dxdx

( )mx 2e am bm c 0+ + =

SinceSince , then is a solution to, then is a solution toequation if m is a root ofequation if m is a root of

mxe 0 y e=

2am bm c 0+ + =

Steps to solve second order homogeneous DESteps to solve second order homogeneous DE

22ndnd OrderOrder


Form a 2ndODE

In form of

Find roots ofauxiliary equation

Compute the solutionaccording to the rootsof auxiliary equations

2

2

d y dya b cy 0

dxdx+ + =

aux ary equat on


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Real anddistinct root

22ndnd OrderOrderHomogeneousHomogeneous

1 2m x m x

Real andrepeated roots

1 2m m

1 2m m= ( ) mxy A Bx e= +

Complex rootsm i= ( )

xy e A cos x B sin x= +

ExamplesExamplesExample 1Example 1Solve the differential equationSolve the differential equation

2

2

d y0

dx=

SolutionSolution2

2

d y0

dx=

m 0=

0 x

=Auxiliar e uation

Real repeated roots

2m 0=

y A Bx= +


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ExamplesExamplesExample 2Example 2

Solve the differential equationSolve the differential equation 22

d y 4 y 0dx

+ =

SolutionSolution2

2

d y4 y 0

dx+ =

m 2i=

Auxiliar e uation

Complex roots

0 x

2m 4 0+ =

=

y A cos 2 x B sin 2 x= +


2d d

SolutionSolution2

2

d y dy6 y 0

dxdx =

( ) ( )m 3 m 2 0 + =

2 y , y , y

dxdx = = =

m 2 , 3=

2m m 6 0 =Auxiliary equation Real distinct roots

2 x 3 xy Ae Be= +

2 x 3 xy 2 Ae 3 Be= +


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SolutionSolution

2 x 3 x ==

2 x 3 xy 2 Ae 3 Be= +

( ) ( )2 0 3 01 Ae Be

= +2 A 3 B 1 + =

B 1 A=

2 A 3 1 A 1 + =1 2 Ae 3 Be

= +

2 A 3 3 A 1 + =5 A 2 = 2A

5=

ExamplesExamplesExample 3Example 3SolutionSolution

B 1 A=

A5

=

2B 1=

2 x 3 x2 3y e e

5 5

= +

2 x 3 x1y 2e 3e

= +

3B

5=


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Solve the differential equationSolve the differential equation2d d

SolutionSolution2

2

d y dy2 y 0

dxdx + =

( )2

m 1 0 =

m 1=

2 y , y , y

dxdx + = = =

2m 2 m 1 0 + =

Auxiliary equation Real repeated roots

( ) x

y A Bx e= +( ) x xy A Bx e Be= + +

ExamplesExamplesExample 3Example 3SolutionSolution

x

( ) x xy A Bx e Be= + +

( )( ) ( )01 A B 0 e= + A 1=

1 B 2+ = 2 1 B 0 e Be = + + 3=

( ) xy 1 3 x e=


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Second order nonSecond order non--homogeneous DE ishomogeneous DE is

2

22ndnd Order NonOrder Non --HomogeneousHomogeneous

WhereWhere a,a, bb andand cc areare constantsconstants.. TheThe relatedrelatedhomogeneoushomogeneousequationequation isis calledca lled thethe complementarycomplementaryequationequation andand isis usedused toto solvesolve thethe nonnon--homogeneoushomogeneous

( ) ( )2

y ya b cy f x , f x 0

dxdx+ + =

equationequation..

2

2d y dya b cy 0

dxdx+ + =

LetLet bebe thethe generalgeneral solutionsolution forfor equationequationandand bebe anyany particularparticular solutionsolution thatthat isisnecessarnecessar toto roduceroduce ex ressionex ression fofo f xf x inin nonnon--

22ndnd Order NonOrder Non --

HomogeneousHomogeneous( )cy y x=

( )py y x=

homogeneoushomogeneous..

Therefore,Therefore, thethe solutionsolution forfor nonnon--homogeneoushomogeneousequationequation isis

c p


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Steps to find complementary solutionSteps to find complementary solution

22ndnd Order NonOrder Non--HomogeneousHomogeneous

Form a second order DE2d d

In form of auxiliary equation

Find the roots of auxiliary equation

2a b cy 0

dxdx+ + =

2am bm c 0+ + =

Compute the solution according to the roots ofauxiliary equations

Steps to find particular solutionSteps to find particular solution

22ndnd Order NonOrder Non--


Form a as same asIf there is any

e uivalent term inf(x) (refer to Table 1).

ompare yp an yc. and yc , multiply ypwith x.

Find yp and yp.

Substitute yp, yp andyp into non-homogeneous

equation.

Collect all thecommon terms.

Equate the coefficientsand determine the

unknown constants(compare thecoefficients).


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F(x) Form of particular solution1

22ndnd Order NonOrder Non--HomogeneousHomogeneous

( ) nf x kx= 2 np 0 1 2 ny C C x C x ... C x= + + + +

2

3 or

4

5or

x ke= py e=

( )f x k co s bx=( )f x k sin bx=

py C c os bx D s in bx= +

( ) n axf x kx e= ( )2 n axp 0 1 2 ny C C x C x ... C x e= + + + +( ) axf x ke cos bx=

( ) axf x ke sin bx= ( )axpy e C cos bx D sin bx= +

6

or

7f(x) is sum of terms in

1-6Sum of the corresponding

( ) nf x kx cos bx=

( )

nf x kx sin bx=( )2 np 0 1 2 ny C C x C x ... C x cos bx= + + + +

( )2 n0 1 2 nC C x C x ... C x sin bx+ + + + +


2

2

d y dy5 6 y 10

dxdx + =

SolutionSolution

2m 5 m 6 0 + =

Complementary solution

) )m 2 m 3 0 =

py C=Particular solution

y 0 =

m 2 , 3=

2 x 3 xcy Ae Be= +

py 0 =


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Solve the differential equationSolve the differential equation2

2d y dy5 6 y 10

dxdx + =

SolutionSolution

2

2

d y dy5 6 y 10

dxdx + = ( ) ( )c py y x y x= +

2 x 3 x 5 =

10 5C6 3

= =

3=

ExamplesExamplesExample 6Example 6SolveSolve

2

2

d y dy3 2 y 6 x

dxdx + =

SolutionSolution

2m 3m 2 0 + =


) )m 1 m 2 0 =py C Dx= +

Particular solution

m 1, 2=

x 2 xcy Ae Be= +

py 0 =p


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SolveSolve 2

2

d y dy3 2 y 6 x

dxdx + =

SolutionSolution

2

2

d y dy3 2 y 6 x

dxdx + =

=

3D 2C 0 + =x0 :( )3 3 9

C2 2

= =

2D 6=

( ) ( )c py y x y x= +

x 2 x 9y Ae Be 3 x2

= + + +Compare coefficientsx : D 3=


25 xd y dy

SolutionSolution

2m 14 m 49 0+ + =


2m 7 0+ =

5 xpy Ce=

Particular solution

5 x

2 y e

dxdx=

m 7=

( ) 7 xcy A Bx e= +

5 xpy 25Ce =p =


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Solve the differential equationSolve the differential equation2

5 xd y dy

SolutionSolution

25 x

2

d y dy14 4 y 4e

dxdx+ + =

5 x 5 x 5 x 5 x=

Compare coefficients

2 y e

dxdx+ + =

=

4 1C144 36

= =( ) 5 x 5 x25C 70C 49C e 4e+ + =5 x 5 x

144Ce 4e=


25 xd y dy

SolutionSolution

( ) ( )c py y x y x= +

2 y e

dxdx=

( ) 7 xcy A Bx e= +

1

( ) 7 x 5 x1y A Bx e e36

= + +py e36=


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Solve the differential equationSolve the differential equation 2 2 x2

d y dy 2 y edxdx

=

SolutionSolution

2m m 2 0 =


) )m 2 m 1 0 + =

2 xpy Ce=

Particular solution

2 x

m 1 , 2=

x 2 xcy Ae Be

= + ( ) 2 x

2C x C e= +

2 x 2 x

py 2Cxe Ce = +


22 x

2

d y dy2 y e

dxdx =

SolutionSolution

2 xpy Cxe=

( ) 2 x2Cx C e= +

2 x 2 xpy 2Cxe Ce = +

( ) 2 x 2 xpy 2 2Cx C e 2Ce = + +

( ) 2 x4Cx 4C e= +


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Solve the differential equationSolve the differential equation 2 2 x2

d y dy 2 y edxdx

=

SolutionSolution

22 x

2

d y dy2 y e

dxdx =

2 x 2 x 2 x 2 x4Cx 4C e 2Cx C e 2Cxe e+ + =

( ) 2 x 2 x

4Cx 4C 2Cx C 2Cx e e+ =2 x 2 x

3Ce e= 3C 1= 1C

3=


22 x

2

d y dy2 y e

dxdx =

SolutionSolution

x x= +

x 2 xcy Ae Be

= + 2 xp1

y xe3

=

x 2 x 2 x1y Ae Be xe3

= + +


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SolveSolve2

2

d yy 12 sin 2 x

dx+ =

SolutionSolution

2m 1 0+ =

Complementary solution2m 1=

m i=

( )

0 x

cy e A cos x B sin x= +

cy A cos x B sin x= +


2

2

d yy 12 sin 2 x

dx+ =

SolutionSolution

py C cos 2 x D sin 2 x= +Particular solution

2C sin 2 x 2 D cos 2 x = +

py 4C cos 2 x 4 D sin 2 x =


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SolveSolve2

2

d yy 12 sin 2 x

dx+ =

SolutionSolution

4C cos 2 x 4 D sin 2 x C cos 2 x D sin 2 x + +12 sin 2 x=

4 D D 12 + =

Sin 2x

3 D 12 =

D 4=

4C C 0 + =Cos 2x

C 0=


2

2

d yy 12 sin 2 x

dx+ =

SolutionSolution

x x= +

py 4 sin 2 x= cy A cos x B sin x= +

y A cos x B sin x 4 sin 2 x= +


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SolveSolve2

2

d yy 12 sin x

dx+ =

SolutionSolution

2m 1 0+ =

Complementary solution2m 1=

m i=

( )

0 x

cy e A cos x B sin x= +

cy A cos x B sin x= +


2

2

d yy 12 sin x

dx+ =

SolutionSolution

py C cos x D sin x= +Particular solution

Cx cos x Dx sin x= +

py Cx sin x C cos x Dx cos x D sin x = + + +

( ) ( )Cx D sin x C Dx cos x= + + +


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SolveSolve2

2

d yy 12 sin x

dx+ =

SolutionSolution

Particular solution

( ) ( )py Cx D cos x C sin x C Dx sin x D cos x = + + +

( ) ( )Cx 2 D cos x 2C Dx sin x + + Cx cos x Dx sin x 12 sin x+ + =


2

2

d yy 12 sin x

dx+ =

SolutionSolution

( ) ( )Cx 2 D cos x 2C Dx sin x + +

Cx cos x Dx sin x 12 sin x+ + =

2C 12 =Sin xC 6=

2 D 0=Cos x0=


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SolveSolve2

2

d yy 12 sin x

dx+ =

SolutionSolution

x x= +

py 6 x cos x= cy A cos x B sin x= +

y A cos x B sin x 6 x cos x= +


3 xy 9 y 72 xe

=

SolutionSolution

2m 9 0 =


) )m 3 m 3 0 + =3 x 3 x

cy Ae Be= +m 3 , 3=


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Solve the differential equationSolve the differential equation 3 xy 9 y 72 xe =

SolutionSolution

( )2 3 x9Cx 9 Dx 6C 12 Dx 2 D e+ +2 3 x 3 x9 Cx Dx e 72 xe

+ =

( ) 3 x 3 x6C 12 Dx 2 D e 72 xe + =


3 xy 9 y 72 xe

=

SolutionSolution

Compare coefficient

( ) 3 x 3 x6C 12 Dx 2 D e 72 xe + =

3 xxe : 12 D 72

=

e : 6C 2 D 0 + = D 3C=

D 6=

6C 2

3= =


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Solve the differential equationSolve the differential equation 3 xy 9 y 72 xe =

SolutionSolution

x x= +

( )2 3 xpy 2 x 6 x e = +3 x 3 xcy Ae Be= +

( )3 x 3 x 2 3 xy Ae Be 2 x 6 x e = + + +


2

2

d x4 x t cos t

dt =

SolutionSolution

2m 4 0 =


m 2 m 2 0 + =

m 2 , 2=

2 t 2 tcx Ae Be

= +


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SolveSolve2

2

d x4 x t cos t

dt =

SolutionSolution

Particular solution

( ) ( )px C Dt cos t E Ft sin t= + + +

p

( ) ( )px D E Ft cos t C Dt F sin t = + + + +


2

2

d x4 x t cos t

dt =

SolutionSolution

Particular solution

( ) ( )px F cos t D E Ft sin t D sin t C Dt F cos t = + + + +

= p

( ) ( )2 F C Dt cos t 2 D E Ft sin t + +

( ) ( )4 C Dt cos t E Ft sin t t cos t + + + =


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SolveSolve2

2

d x4 x t cos t

dt =

SolutionSolution



D 4 D 1 =

tcos t

5 D 1 =5 F 0 =t sin tF 0=

1D

5=


2

2

d x4 x t cos t

dt =

1= =

SolutionSolution



5

2 F C 4C 0 =

cos t

C 0=

2 D E 4 E 0 =sin t2

E25

=


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SolveSolve2

2

d x4 x t cos t

dt =

SolutionSolution

x x t x t= +

p

1 2x t cos t sin t

5 25= +2 t 2 tcx Ae Be

= +

2 t 2 t 1 2x Ae Be t cos t sin t5 25

= + +


2t

2

d xte cos t

dt= +

SolutionSolution

2m 0=


m 0=

( ) 0 tcx A Bt e= +

cx A Bt= +


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SolveSolve2 t

2

d xte cos t

dt= +

SolutionSolution

Particular solution

( ) tpx C Dt e E cos t F sin t= + + +t t = p

( ) t

C D Dt e E sin t F cos t= + + +


2t

2

d xte cos t

dt= +

SolutionSolution

( )t tpx De C D Dt e E cos t F sin t = + + +

( ) tC 2 D Dt e E cos t F sin t= + +

( ) t tC 2 D Dt e E cos t F sin t te cos t+ + = +


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SolveSolve2 t

2

d xte cos t

dt= +

SolutionSolution

Compare coefficient

( ) t tC 2 D Dt e E cos t F sin t te cos t+ + = +

tte : D 1= cos t : E 1 = 1=

te : C 2 D 0+ =

C 2=

sin t : F 0 = F 0=


2t

2

d xte cos t

dt= +

SolutionSolution

( ) tpx 2 t e cos t= +

cx A Bt= +

( ) ( )c px x t x t= +

( ) tx A Bt 2 t e cos t= + + +


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Find the particular solution forFind the particular solution for2

SolutionSolution

2m 1 0+ =


m i=

2 y 10e , y 0 0; y 0 0

dx+ = = =

( )0 x

cy e A cos x B sin x= +A cos x B sin x= +

ExamplesExamplesExample 14Example 14Find the particular solution forFind the particular solution for

2

SolutionSolution

2 y 10e , y 0 0; y 0 0

dx+ = = =

2 xpy Ce=

2 x 2 x 2 x4Ce Ce 10 e+ =

2 xpy 4Ce =

py e= 5Ce 10e=5C 10=

C 2=


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SolutionSolution

2 y 10e , y 0 0; y 0 0

dx+ = = =

2 xpy 2e=cy A cos x B sin x= +

c py y x y x= +

2 xy A cos x B sin x 2e= + +

ExamplesExamplesExample 14Example 14Find the particular solution forFind the particular solution for

2

SolutionSolution

2 y 10e , y 0 0; y 0 0

dx+ = = =

2 xy A cos x B sin x 2e= + +2 xA sin x B cos x 4e= + +

( )2 00 A cos 0 B sin 0 2e= + + A 2=

( )2 00 A sin 0 B cos 0 4e= + + B 4=


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SolutionSolution

2 y 10e , y 0 0; y 0 0

dx+ = = =

2 xy 2 cos x 4 sin x 2e= +

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13 2nd Order Linear De