Higher Engineering Mathematics, Sixth Edition

Second order differential equations of the forma

d^2 y

dx^2 +b

dy

dx+cy=f(x)^489

Equating coefficients of cos2xgives:

6 A− 13 B=0(2)

6 ×( 1 )gives :− 78 A− 36 B=36 (3)

13 ×( 2 )gives : 78A− 169 B=0(4)

( 3 )+( 4 )gives : − 205 B= 36

from which, B=

− 36

205

SubstitutingB=

− 36

205

into equation (1) or (2)

givesA=

− 78

205

Hence the P.I.,v=

− 78

205

sin2x−

36

205

cos2x.

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

y=Aex+Be−

5

2 x

−

2

205

(39 sin 2x+18 cos 2x)

Problem 8. Solve

d^2 y

dx^2

+ 16 y=10cos4xgiven

y=3and

dy

dx

=4whenx=0.

Using the procedure of Section 51.2:

(i)

d^2 y

dx^2

+ 16 y=10cos4xin D-operator form is

(D^2 + 16 )y=10cos4x

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

whichm=

√

− 16 =±j4.

(iii) Since the roots are complex the C.F.,

u=e^0 (Acos 4x+Bsin4x)

i.e.u=Acos4x+Bsin4x

(iv) Since sin4x occurs in the C.F. and in the

right hand side of the given differential equa-

tion, let the P.I.,v=x(Csin4x+Dcos 4x)(see

Table 51.1(d), snag case—constantsCandDare

used sinceAandBhave already been used in the

C.F.).

(v) Substituting v=x(Csin4x+Dcos 4x) into

(D^2 + 16 )v=10cos4xgives:

(D^2 + 16 )[x(Csin4x+Dcos4x)]

=10cos4x

D[x(Csin4x+Dcos 4x)]

=x( 4 Ccos4x− 4 Dsin4x)

+(Csin4x+Dcos 4x)( 1 ),

by the product rule

D^2 [x(Csin4x+Dcos4x)]

=x(− 16 Csin4x− 16 Dcos4x)

+( 4 Ccos4x− 4 Dsin 4x)

+( 4 Ccos4x− 4 Dsin 4x)

Hence(D^2 + 16 )[x(Csin4x+Dcos4x)]

=− 16 Cxsin4x− 16 Dxcos4x+ 4 Ccos4x

− 4 Dsin4x+ 4 Ccos4x− 4 Dsin4x

+ 16 Cxsin4x+ 16 Dxcos4x

=10cos4x,

i.e.− 8 Dsin4x+ 8 Ccos4x=10cos4x

Equating coefficients of cos4xgives:

8 C=10, from which,C=

10

8

=

5

4

Equating coefficients of sin4xgives:

− 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+Bsin0+0, i.e.A=3.

dy

dx

=− 4 Asin 4x+ 4 Bcos4x

+^54 x(4cos4x)+^54 sin4x

Whenx=0,

dy

dx

=4, thus

4 =− 4 Asin 0+ 4 Bcos0+ 0 +^54 sin0

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

Hence the particular solution is

y=3cos4x+sin 4x+^54 xsin 4x