Higher Engineering Mathematics

SECOND ORDER DIFFERENTIAL EQUATIONS (NON-HOMOGENEOUS) 485

I

the P.I., v=kxe

3

2 x (see Table 51.1(c), snag

case (i)).

(iv) Substitutingv=kxe

3

2 xinto (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 xin D-operator form is

(D^2 −4D+4)y=3e^2 x.

(ii) Substituting m for D gives the auxiliary

equationm^2 − 4 m+ 4 =0. Factorising 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 xinto (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)(2kx)

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

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

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

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

Hence (D^2 −4D+4)(kx^2 e^2 x)

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

−4[2ke^2 x(x^2 +x)]+4[kx^2 e^2 x]

=3e^2 x

from which, 2ke^2 x=3e^2 xandk=^32

Hence the P.I.,v=kx^2 e^2 x=^32 x^2 e^2 x.

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

y=(Ax+B)e^2 x+^32 x^2 e^2 x

Now try the following exercise.

Exercise 191 Further problems on differen-

tial equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)wheref(x) is an expo-

nential function

In Problems 1 to 4, find the general solutions of

the given differential equations.

1.

d^2 y

dx^2

−

dy

dx

− 6 y=2ex

[

y=Ae^3 x+Be−^2 x−^13 ex

]

2.

d^2 y

dx^2

− 3

dy

dx

− 4 y=3e−x

[

y=Ae^4 x+Be−x−^35 xe−x

]