differential equation

488 DIFFERENTIAL EQUATIONS
5. A differential equation representing the
motion of a body is d2 y
dt 2 + n2 y = k sin pt,
where k, n and p are constants. Solve the
equation (given n ̸= 0 and p 2 ̸= n 2 ) given that
when t = 0, y = dy
dt = 0.
[
y =
k (
n 2 − p 2 sin pt − p ) ]
n sin nt
6. The motion of a vibrating mass is given by
d 2 y
dt 2 + 8dy + 20y = 300 sin 4t. Show that the
dt
general solution of the differential equation is
given by:
y = e −4t (A cos 2t + B sin 2t)
+ 15 ( sin 4t − 8 cos 4t)
13
7. L d2 q
dt 2 + Rdq dt + 1 C q = V 0 sin ωt represents
the variation of capacitor charge in an
electric circuit. Determine an expression
for q at time t seconds given that R = 40 ,
L = 0.02 H, C = 50 × 10 −6 F, V 0 = 540.8V
and ω = 200 rad/s and given the boundary
conditions that when t = 0, q = 0 and
dq
dt = 4.8
[ q = (10t + 0.01)e
−1000t
]
+ 0.024 sin 200t − 0.010 cos 200t
51.6 Worked problems on differential
equations of the form
a d2 y
dx 2 + b dy + cy = f (x) where
dx
f (x) is a sum or a product
(D 2 + D − 6)y = 12x − 50 sin x
(ii) The auxiliary equation is (m 2 + m − 6) = 0,
from which,
(m − 2)(m + 3) = 0,
i.e. m = 2orm =−3
(iii) Since the roots are real and different, the C.F.,
u = Ae 2x + Be −3x .
(iv) Since the right hand side of the given differential
equation is the sum of a polynomial and a
sine function let the P.I. v = ax + b + c sin x +
d cos x (see Table 51.1(e)).
(v) Substituting v into
(D 2 + D − 6)v = 12x − 50 sin x gives:
(D 2 + D − 6)(ax + b + c sin x + d cos x)
= 12x − 50 sin x
D(ax + b + c sin x + d cos x)
= a + c cos x − d sin x
D 2 (ax + b + c sin x + d cos x)
=−c sin x − d cos x
Hence (D 2 + D − 6)(v)
= (−c sin x − d cos x) + (a + c cos x
−d sin x) − 6(ax + b + c sin x + d cos x)
= 12x − 50 sin x
Equating constant terms gives:
a − 6b = 0 (1)
Equating coefficients of x gives: −6a = 12,
from which, a =−2.
Hence, from (1), b =− 1 3
Equating the coefficients of cos x gives:
−d + c − 6d = 0
(2)
i.e. c − 7d = 0
Problem 9. Solve
d 2 y
dx 2 + dy − 6y = 12x − 50 sin x.
dx
Equating the coefficients of sin x gives:
−c − d − 6c =−50
i.e. −7c − d =−50
(3)
Using the procedure of Section 51.2:
(i) d2 y
dx 2 + dy − 6y = 12x − 50 sin x in D-operator
dx
form is
Solving equations (2) and (3) gives: c = 7 and
d = 1.
Hence the P.I.,
υ =−2x − 1 3
+ 7 sin x + cos x