Higher Engineering Mathematics, Sixth Edition

Second order differential equations of the forma

d^2 y

dx^2 +b

dy

dx+cy=f(x)^487

−^23 =( 0 +B)e^0 +^13 e^0 , from which,B=−1.

dy

dx

=(Ax+B)ex+ex(A)+^43 e^4 x.

Whenx=0,

dy

dx

= 4

1

3

, thus

13

3

=B+A+

4

3

from which,A=4, sinceB=−1.

Hence the particular solution is:

y=( 4 x− 1 )ex+^13 e^4 x

Problem 5. Solve the differential equation

2

d^2 y

dx^2

−

dy

dx

− 3 y=5e

3

2 x.

Using the procedure of Section 51.2:

(i) 2

d^2 y

dx^2

−

dy

dx

− 3 y=5e

3

2 x in D-operator form is

(2D^2 −D− 3 )y=5e

3

2 x.

(ii) Substituting m for D gives the auxiliary

equation 2m^2 −m− 3 =0. Factorizing gives:

( 2 m− 3 )(m+ 1 )=0, from which, m=^32 or

m=−1. Since the roots are real and different then

the C.F.,u=Ae

3

2 x+Be−x.

(iii) Since e

3

2 x appears in the C.F. and in the

right hand side of the differential equation, let the

P.I.,v=kxe

3

2 x(see Table 51.1(c), snag case (i)).

(iv) Substitutingv=kxe

3

2 x into(2D^2 −D− 3 )v=

5e

3

2 xgives:(2D^2 −D− 3 )kxe

3

2 x=5e

3

2 x.

D

(

kxe

3

2 x

)

=(kx)

(

3

2 e

3

2 x

)

+

(

e

3

2 x

)

(k),

by the product rule,

=ke

3

2 x

( 3

2 x+^1

)

D^2

(

kxe

3

2 x

)

=D

[

ke

3

2 x

( 3

2 x+^1

)

]

=

(

ke

3

2 x

)

( 3

2

)

+

( 3

2 x+^1

)

(

3

2 ke

3

2 x

)

=ke

3

2 x

( 9

4 x+^3

)

Hence(2D^2 −D− 3 )

(

kxe

3

2 x

)

= 2

[

ke

3

2 x

( 9

4 x+^3

)

]

−

[

ke

3

2 x

( 3

2 x+^1

)

]

− 3

[

kxe

3

2 x

]

=5e

3

2 x

i.e.^92 kxe

3

2 x+ 6 ke

3

2 x−^32 xke

3

2 x−ke

3

2 x

− 3 kxe

3

2 x=5e

3

2 x

Equating coefficients of e

3

2 xgives: 5k=5, from

which,k=1.

Hence the P.I.,v=kxe

3

2 x=xe

3

2 x.

(v) The general solution is y=u+v,i.e.

y=Ae

3

2 x+Be−x+xe

3

2 x.

Problem 6. Solve

d^2 y

dx^2

− 4

dy

dx

+ 4 y=3e^2 x.

Using the procedure of Section 51.2:

(i)

d^2 y

dx^2

− 4

dy

dx

+ 4 y=3e^2 x in D-operator form is

(D^2 −4D+ 4 )y=3e^2 x.

(ii) Substituting m for D gives the auxiliary

equation m^2 − 4 m+ 4 =0. Factorizing gives:

(m− 2 )(m− 2 )=0, from which,m=2 twice.

(iii) Since the roots are real and equal, the C.F.,

u=(Ax+B)e^2 x.

(iv) Since e^2 xandxe^2 xboth appear in the C.F. let the

P.I.,v=kx^2 e^2 x(see Table 51.1(c), snag case (ii)).

(v) Substitutingv=kx^2 e^2 x into(D^2 −4D+ 4 )v=

3e^2 xgives:(D^2 −4D+ 4 )(kx^2 e^2 x)=3e^2 x

D(kx^2 e^2 x)=(kx^2 )(2e^2 x)+(e^2 x)( 2 kx)

= 2 ke^2 x(x^2 +x)

D^2 (kx^2 e^2 x)=D[2ke^2 x(x^2 +x)]

=( 2 ke^2 x)( 2 x+ 1 )+(x^2 +x)( 4 ke^2 x)

= 2 ke^2 x( 4 x+ 1 + 2 x^2 )

Hence(D^2 −4D+ 4 )(kx^2 e^2 x)

=[2ke^2 x( 4 x+ 1 + 2 x^2 )]

−4[2ke^2 x(x^2 +x)]+4[kx^2 e^2 x]

=3e^2 x