482 Higher Engineering Mathematics
Using the procedure of Section 50.2:
(a)
d 2 x
dt 2 + 6
dx
dt
+ 8x = 0 in D-operator form is
(D 2 + 6D + 8)x = 0, where D ≡
d
dt
.
(b) The auxiliary equation is m 2 + 6m + 8 =0.
Factorising gives: (m + 2)(m + 4) = 0, from
which, m =−2 or m = −4.
(c) Since the roots are real and different, the general
solution is x =Ae −2t + Be −4t .
(d) Initial displacement means that time t = 0. At this
instant, x = 4.
Thus 4 = A + B
(1)
Velocity,
dx
dt
= −2 Ae
−2t
− 4Be
−4t
dx
dt
= 8 cm/s when t = 0,
thus 8 = −2 A − 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 188 Further problems on second
order differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
1. The charge, q, on a capacitor in a certain
electrical circuit satisfies the differential equation
d 2 q
dt 2 + 4
dq
dt
+ 5q = 0. Initially (i.e. when
t = 0), q = Q and
dq
dt
= 0. Show that the
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
d 2 s
dt 2 + a 2 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
d 2 θ
dt 2 + K
dθ
dt
+ Fθ = 0.
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
d 2 x
dt 2 + 2n
dx
dt
+ n 2 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
d 2 i
dt 2 + R
di
dt
+
1
C
i = 0 is an equation representing current i in an electric circuit. If
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
20
e −160t − e −840t
6. The displacement s of a body in a damped
mechanical system, with no external forces,
satisfies the following differential equation:
2
d 2 s
dt 2 + 6
ds
dt
+ 4.5s = 0
where t represents time. If initially, when
t = 0, s = 0 and
ds
dt
= 4, solve the differential
equation for s in terms of t .
[s = 4t e
−
3
2 t ]
Using the procedure of Section 50.2:
(a)
d 2 x
dt 2 + 6
dx
dt
+ 8x = 0 in D-operator form is
(D 2 + 6D + 8)x = 0, where D ≡
d
dt
.
(b) The auxiliary equation is m 2 + 6m + 8 =0.
Factorising gives: (m + 2)(m + 4) = 0, from
which, m =−2 or m = −4.
(c) Since the roots are real and different, the general
solution is x =Ae −2t + Be −4t .
(d) Initial displacement means that time t = 0. At this
instant, x = 4.
Thus 4 = A + B
(1)
Velocity,
dx
dt
= −2 Ae
−2t
− 4Be
−4t
dx
dt
= 8 cm/s when t = 0,
thus 8 = −2 A − 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 188 Further problems on second
order differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
1. The charge, q, on a capacitor in a certain
electrical circuit satisfies the differential equation
d 2 q
dt 2 + 4
dq
dt
+ 5q = 0. Initially (i.e. when
t = 0), q = Q and
dq
dt
= 0. Show that the
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
d 2 s
dt 2 + a 2 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
d 2 θ
dt 2 + K
dθ
dt
+ Fθ = 0.
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
d 2 x
dt 2 + 2n
dx
dt
+ n 2 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
d 2 i
dt 2 + R
di
dt
+
1
C
i = 0 is an equation representing current i in an electric circuit. If
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
20
e −160t − e −840t
6. The displacement s of a body in a damped
mechanical system, with no external forces,
satisfies the following differential equation:
2
d 2 s
dt 2 + 6
ds
dt
+ 4.5s = 0
where t represents time. If initially, when
t = 0, s = 0 and
ds
dt
= 4, solve the differential
equation for s in terms of t .
[s = 4t e
−
3
2 t ]