differential equation

480 DIFFERENTIAL EQUATIONS
Thus 4 = A + B (1)
Velocity,
dx
dt =−2Ae−2t − 4Be −4t
dx
dt
= 8 cm/s when t = 0,
thus 8 =−2A − 4B (2)
From equations (1) and (2),
A = 12 and B =−8
Hence the particular solution is
x = 12e −2t − 8e −4t
i.e. displacement, x = 4(3e −2t − 2e −4t )cm
Now try the following exercise.
Exercise 189 Further problems on second
order differential equations of the form
a d2 y
dx 2 + b dy
dx + cy = 0
1. The charge, q, on a capacitor in a certain electrical
circuit satisfies the differential equation
d2 q
dt 2 + 4dq + 5q = 0. Initially (i.e. when
dt
t = 0), q = Q and dq = 0. Show that the
dt
charge in the circuit can be expressed as:
q = √ 5 Qe −2t sin (t + 0.464)
2. A body moves in a straight line so that its
distance s metres from the origin after time
t seconds is given by d2 s
dt 2 + a2 s = 0, where a
is a constant. Solve the equation for s given
that s = c and ds
dt = 0 when t = 2π a
[s = c cos at]
3. The motion of the pointer of a galvanometer
about its position of equilibrium is represented
by the equation
I d2 θ
dt 2 + K dθ + Fθ = 0.
dt
If I, the moment of inertia of the pointer about
its pivot, is 5 × 10 −3 , K, the resistance due to
friction at unit angular velocity, is 2 × 10 −2
and F, the force on the spring necessary to
produce unit displacement, is 0.20, solve the
equation for θ in terms of t given that when
t = 0, θ = 0.3 and dθ
dt = 0.
[θ = e −2t (0.3 cos 6t + 0.1 sin 6t)]
4. Determine an expression for x for a differential
equation d2 x
dt 2 + 2ndx dt + n2 x = 0 which
represents a critically damped oscillator,
given that at time t = 0, x = s and dx
dt = u.
[x ={s + (u + ns)t}e −nt ]
5. L d2 i
dt 2 + R di
dt + 1 i = 0 is an equation representing
current i in an electric circuit. If
C
inductance L is 0.25 henry, capacitance C
is 29.76 × 10 −6 farads and R is 250 ohms,
solve the equation for i given the boundary
conditions that when t = 0, i = 0 and di
[
dt = 34.
i = 1 (
e −160t − e −840t)]
20
6. The displacement s of a body in a damped
mechanical system, with no external forces,
satisfies the following differential equation:
2 d2 s
dt 2 + 6ds dt + 4.5s = 0
where t represents time. If initially, when
t = 0, s = 0 and ds = 4, solve the differential
dt
equation for s in terms of t. [s = 4te − 2 3 t ]