Higher Engineering Mathematics

SECOND ORDER DIFFERENTIAL EQUATIONS (NON-HOMOGENEOUS) 487

I

Using the procedure of Section 51.2:

(i)

d^2 y

dx^2

+ 16 y=10 cos 4xin D-operator form is

(D^2 +16)y=10 cos 4x

(ii) The auxiliary equation ism^2 + 16 =0, from

whichm=

√

− 16 =±j4.

(iii) Since the roots are complex the C.F.,
u=e^0 (Acos 4x+Bsin 4x)

i.e.u=Acos 4x+Bsin 4x

(iv) Since sin 4xoccurs in the C.F. andin the

right hand side of the given differential equa-

tion, let the P.I.,v=x(Csin 4x+Dcos 4x) (see

Table 51.1(d), snag case—constantsCandD

are used sinceAandBhave already been used

in the C.F.).

(v) Substituting v=x(Csin 4x+Dcos 4x) into

(D^2 +16)v=10 cos 4xgives:

(D^2 +16)[x(Csin 4x+Dcos 4x)]

=10 cos 4x

D[x(Csin 4x+Dcos 4x)]

=x(4Ccos 4x− 4 Dsin 4x)

+(Csin 4x+Dcos 4x)(1),

by the product rule

D^2 [x(Csin 4x+Dcos 4x)]

=x(− 16 Csin 4x− 16 Dcos 4x)

+(4Ccos 4x− 4 Dsin 4x)

+(4Ccos 4x− 4 Dsin 4x)

Hence (D^2 +16)[x(Csin 4x+Dcos 4x)]

=− 16 Cxsin 4x− 16 Dxcos 4x+ 4 Ccos 4x

− 4 Dsin 4x+ 4 Ccos 4x− 4 Dsin 4x

+ 16 Cxsin 4x+ 16 Dxcos 4x

=10 cos 4x,

i.e.− 8 Dsin 4x+ 8 Ccos 4x=10 cos 4x

Equating coefficients of cos 4xgives:

8 C=10, from which,C=

10

8

=

5

4

Equating coefficients of sin 4xgives:

− 8 D=0, from which,D=0.

Hence the P.I.,v=x

(

5

4 sin 4x

)

.

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

y=Acos 4x+Bsin 4x+^54 xsin 4x

(vii) Whenx=0,y=3, thus

3 =Acos 0+Bsin 0+0, i.e.A=3.

dy

dx

=− 4 Asin 4x+ 4 Bcos 4x

+^54 x(4 cos 4x)+^54 sin 4x

Whenx=0,

dy

dx

=4, thus

4 =− 4 Asin 0+ 4 Bcos 0+ 0 +^54 sin 0

i.e. 4= 4 B, from which,B= 1

Hence the particular solution is

y=3 cos 4x+sin 4x+^54 xsin 4x

Now try the following exercise.

Exercise 192 Further problems on differen-

tial equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)wheref(x) is a sine

or cosine function

In Problems 1 to 3, find the general solutions of

the given differential equations.

1. 2

d^2 y

dx^2

−

dy

dx

− 3 y=25 sin 2x

[

y=Ae

3

2 x+Be−x

−^15 (11 sin 2x−2 cos 2x)

]

2.

d^2 y

dx^2

− 4

dy

dx

+ 4 y=5 cosx

[

y=(Ax+B)e^2 x−^45 sinx+^35 cosx

]

3.

d^2 y

dx^2

+y=4 cosx

[y=Acosx+Bsinx+ 2 xsinx]

4. Find the particular solution of the differential

equation

d^2 y

dx^2

− 3

dy

dx

− 4 y=3 sinx; when

x=0,y=0 and

dy

dx

=0.

⎡

⎢

⎣

y=

1

170

(6e^4 x−51e−x)

−

1

34

(15 sinx−9 cosx)

⎤

⎥

⎦