differential equation

484 DIFFERENTIAL EQUATIONS
4. 9 d2 y dy
− 12 + 4y = 3x − 1; when x = 0,
dx2 dx
y = 0 and dy
dx =−4 3
[y =− ( ]
2 +
4 3 x) e 2 3 x + 2 +
4 3 x
5. The charge q in an electric circuit at time t satisfies
the equation L d2 q
dt 2 + Rdq dt + 1 C q = E,
where L, R, C and E are constants. Solve the
equation given L = 2H, C = 200 × 10 −6 F
and E = 250V, when (a) R = 200 and (b) R
is negligible. Assume that when t = 0, q = 0
and dq
dt = 0.
⎡
(a) q = 1 ( 5
⎢ 20 − 2 t + 1 ) ⎤
e −50t
20
⎣
(b) q = 1
⎥
⎦
20 (1 − cos 50t)
6. In a galvanometer the deflection θ satisfies the
d2 θ
differential equation
dt 2 + 4dθ dt + 4 θ = 8.
Solve the equation for θ given that when t = 0,
θ = d θ
dt = 2. [ θ = 2(te−2t + 1)]
(ii) Substituting m for D gives the auxiliary
equation m 2 − 2m + 1 = 0. Factorising gives:
(m − 1)(m − 1) = 0, from which, m = 1 twice.
(iii) Since the roots are real and equal the C.F.,
u = (Ax + B)e x .
(iv) Let the particular integral, v = ke 4x (see
Table 51.1(c)).
(v) Substituting v = ke 4x into
(D 2 − 2D + 1)v = 3e 4x gives:
(D 2 − 2D + 1)ke 4x = 3e 4x
i.e. D 2 (ke 4x ) − 2D(ke 4x ) + 1(ke 4x ) = 3e 4x
i.e. 16ke 4x − 8ke 4x + ke 4x = 3e 4x
Hence 9ke 4x = 3e 4x , from which, k = 1 3
Hence the P.I., v = ke 4x = 1 3 e4x .
(vi) The general solution is given by y = u + v, i.e.
y = (Ax + B)e x + 1 3 e4x .
(vii) When x = 0, y =− 2 3 thus
− 2 3 = (0 + B)e0 + 1 3 e0 , from which, B =−1.
dy
dx = (Ax + B)ex + e x (A) + 4 3 e4x .
When x = 0, dy
dx = 41 13
, thus
3 3 = B + A + 4 3
from which, A = 4, since B =−1.
Hence the particular solution is:
y = (4x − 1)e x + 1 3 e4x
51.4 Worked problems on differential
equations of the form
a d2 y
dx 2 + b dy + cy = f (x) where
dx
f (x) is an exponential function
Problem 4. Solve the equation
d 2 y
dx 2 − 2 dy
dx + y = 3e4x given the boundary
conditions that when x = 0, y =− 2 dy
3
and
dx = 4 1 3
Using the procedure of Section 51.2:
(i) d2 y
dx 2 − 2 dy
dx + y = 3e4x in D-operator form is
(D 2 − 2D + 1)y = 3e 4x .
Problem 5. Solve the differential equation
2 d2 y
dx 2 − dy
dx − 3y = 5e 2 3 x .
Using the procedure of Section 51.2:
(i) 2 d2 y
dx 2 − dy
dx − 3y = 5e 2 3 x in D-operator form is
(2D 2 − D − 3)y = 5e 2 3 x .
(ii) Substituting m for D gives the auxiliary
equation 2m 2 − m − 3 = 0. Factorising gives:
(2m − 3)(m + 1) = 0, from which, m =
2 3 or
m =−1. Since the roots are real and different
then the C.F., u = Ae 2 3 x + Be −x .
(iii) Since e 2 3 x appears in the C.F. and in the
right hand side of the differential equation, let