Second order differential equations of the form a
d 2 y
dx 2 + b
dy
dx + cy = f (x) 489
Equating coefficients of cos 2x gives:
6 A − 13B = 0
(2)
6 × (1)gives : −78 A − 36B = 36
(3)
13 × (2)gives : 78 A − 169B = 0
(4)
(3) + (4)gives :
− 205B = 36
from which,
B =
−36
205
Substituting B =
−36
205
into equation (1) or (2)
gives A =
−78
205
Hence the P.I., v =
−78
205
sin 2x −
36
205
cos 2x.
(vi) The general solution, y = u +v, i.e.
y = Ae
x
+ Be
− 5
2 x
−
2
205
(39 sin 2x + 18 cos 2x)
Problem 8. Solve
d 2 y
dx 2 + 16y = 10 cos4x given
y = 3 and
d y
dx
= 4 when x = 0.
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 + 16y = 10 cos 4x in D-operator form is
(D
2
+ 16)y = 10 cos4x
(ii) The auxiliary equation is m 2 + 16 = 0, from
which m =
√ −16 = ± j 4.
(iii) Since the roots are complex the C.F.,
u =e 0 (A cos 4x + B sin 4x)
i.e. u =Acos 4x + B sin4x
(iv) Since sin 4x occurs in the C.F. and in the
right hand side of the given differential equation, let the P.I., v = x(C sin 4x + D cos 4x) (see
Table 51.1(d), snag case—constants C and D are
used since A and B have already been used in the
C.F.).
(v) Substituting v = x(C sin 4x + D cos 4x) into
(D 2 + 16)v = 10 cos 4x gives:
(D
2
+ 16)[x(C sin 4x + D cos 4x)]
= 10 cos 4x
D[x(C sin 4x + D cos 4x)]
= x(4C cos 4x − 4D sin 4x)
+ (C sin 4x + D cos 4x)(1),
by the product rule
D 2 [x(C sin 4x + D cos 4x)]
= x(−16C sin 4x − 16D cos 4x)
+ (4C cos 4x − 4D sin 4x)
+ (4C cos 4x − 4D sin 4x)
Hence (D 2 + 16)[x(C sin 4x + D cos 4x)]
=−16Cx sin 4x −16Dx cos 4x + 4C cos 4x
− 4D sin 4x + 4C cos 4x − 4D sin 4x
+ 16Cx sin 4x + 16Dx cos 4x
= 10 cos4x,
i.e. −8D sin 4x + 8C cos 4x = 10 cos4x
Equating coefficients of cos 4x gives:
8C = 10, from which, C =
10
8
=
5
4
Equating coefficients of sin 4x gives:
−8D = 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 = A cos 4x + B sin 4x +
5
4 x sin 4x
(vii) When x = 0, y = 3, thus
3 = A cos 0 + B sin 0 + 0, i.e. A = 3.
d y
dx
= −4 A sin 4x + 4B cos 4x
+
5
4 x(4 cos 4x) +
5
4 sin 4x
When x = 0,
d y
dx
= 4, thus
4 = −4 A sin 0 + 4B cos 0 + 0 +
5
4 sin 0
i.e. 4 =4B, from which, B = 1
Hence the particular solution is
y = 3 cos 4x + sin 4x +
5
4 x sin 4x
d 2 y
dx 2 + b
dy
dx + cy = f (x) 489
Equating coefficients of cos 2x gives:
6 A − 13B = 0
(2)
6 × (1)gives : −78 A − 36B = 36
(3)
13 × (2)gives : 78 A − 169B = 0
(4)
(3) + (4)gives :
− 205B = 36
from which,
B =
−36
205
Substituting B =
−36
205
into equation (1) or (2)
gives A =
−78
205
Hence the P.I., v =
−78
205
sin 2x −
36
205
cos 2x.
(vi) The general solution, y = u +v, i.e.
y = Ae
x
+ Be
− 5
2 x
−
2
205
(39 sin 2x + 18 cos 2x)
Problem 8. Solve
d 2 y
dx 2 + 16y = 10 cos4x given
y = 3 and
d y
dx
= 4 when x = 0.
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 + 16y = 10 cos 4x in D-operator form is
(D
2
+ 16)y = 10 cos4x
(ii) The auxiliary equation is m 2 + 16 = 0, from
which m =
√ −16 = ± j 4.
(iii) Since the roots are complex the C.F.,
u =e 0 (A cos 4x + B sin 4x)
i.e. u =Acos 4x + B sin4x
(iv) Since sin 4x occurs in the C.F. and in the
right hand side of the given differential equation, let the P.I., v = x(C sin 4x + D cos 4x) (see
Table 51.1(d), snag case—constants C and D are
used since A and B have already been used in the
C.F.).
(v) Substituting v = x(C sin 4x + D cos 4x) into
(D 2 + 16)v = 10 cos 4x gives:
(D
2
+ 16)[x(C sin 4x + D cos 4x)]
= 10 cos 4x
D[x(C sin 4x + D cos 4x)]
= x(4C cos 4x − 4D sin 4x)
+ (C sin 4x + D cos 4x)(1),
by the product rule
D 2 [x(C sin 4x + D cos 4x)]
= x(−16C sin 4x − 16D cos 4x)
+ (4C cos 4x − 4D sin 4x)
+ (4C cos 4x − 4D sin 4x)
Hence (D 2 + 16)[x(C sin 4x + D cos 4x)]
=−16Cx sin 4x −16Dx cos 4x + 4C cos 4x
− 4D sin 4x + 4C cos 4x − 4D sin 4x
+ 16Cx sin 4x + 16Dx cos 4x
= 10 cos4x,
i.e. −8D sin 4x + 8C cos 4x = 10 cos4x
Equating coefficients of cos 4x gives:
8C = 10, from which, C =
10
8
=
5
4
Equating coefficients of sin 4x gives:
−8D = 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 = A cos 4x + B sin 4x +
5
4 x sin 4x
(vii) When x = 0, y = 3, thus
3 = A cos 0 + B sin 0 + 0, i.e. A = 3.
d y
dx
= −4 A sin 4x + 4B cos 4x
+
5
4 x(4 cos 4x) +
5
4 sin 4x
When x = 0,
d y
dx
= 4, thus
4 = −4 A sin 0 + 4B cos 0 + 0 +
5
4 sin 0
i.e. 4 =4B, from which, B = 1
Hence the particular solution is
y = 3 cos 4x + sin 4x +
5
4 x sin 4x