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