Second order differential equations of the form a
d 2 y
dx 2 + b
dy
dx + cy = 0 481
(c) Since the roots are real and different, the general
solution is
V = Ae
ωt
+ Be
−ωt
(d) When t = 0, V = 7 hence 7 = A + B
(1)
dV
dt
= Aωe ωt − Bωe −ωt
When
t = 0,
dV
dt
= 3ω,
thus
3ω = Aω − Bω,
i.e.
3 = A − B
(2)
From equations (1) and (2), A = 5 and B = 2
Hence the particular solution is
V = 5e
ωt
+ 2e
−ωt
Since
sinh ωt =
1
2 (e ωt − e −ωt )
and
cosh ωt =
1
2 (e ωt + e −ωt )
then sinh ωt + cosh ωt = e ωt
and cosh ωt − sinh ωt = e
−ωt from Chapter 5.
Hence the particular solution may also be
written as
V = 5(sinh ωt + cosh ωt )
+ 2(cosh ωt − sinh ωt )
i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt
i.e. V = 7 coshωt + 3 sinh ωt
Problem 6. The equation
d 2 i
dt 2 +
R
L
di
dt
+
1
LC
i = 0
represents a current i flowing in an electrical circuit
containing resistance R, inductance L and
capacitance C connected in series. If R = 200 ohms,
L =0.20 henry and C = 20 ×10 −6 farads, solve the
equation for i given the boundary conditions that
when t = 0, i = 0 and
di
dt
= 100.
Using the procedure of Section 50.2:
(a)
d 2 i
dt 2 +
R
L
di
dt
+
1
LC
i = 0 in D-operator form is
D
2
+
R
L
D +
1
LC
i = 0 where D ≡
d
dt
(b) The auxiliary equation is m 2 +
R
L
m +
1
LC
= 0
Hence m =
−
R
L
±
R
L
2
− 4(1)
1
LC
2
When R = 200, L =0.20 and C = 20 ×10 −6 , then
m =
−
200
0.20
±
200
0.20
2
−
4
(0.20)(20 × 10 −6 )
2
=
−1000 ±
√
0
2
= −500
(c) Since the two roots are real and equal (i.e. −500
twice, since for a second order differential equation there must be two solutions), the general
solution is i = (At +B)e −500t .
(d) When t = 0, i = 0, hence B = 0
di
dt
= (At + B)(−500e −500t ) + (e −500t )(A),
by the product rule
When t = 0,
di
dt
= 100, thus 100 =−500B + A
i.e. A = 100, since B = 0
Hence the particular solution is
i = 100te
−500t
Problem 7. The oscillations of a heavily damped
pendulum satisfy the differential equation
d 2 x
dt 2 + 6
dx
dt
+ 8x = 0, where x cm is the
displacement of the bob at time t seconds.
The initial displacement is equal to +4 cm and the
initial velocity
i.e.
dx
dt
is 8 cm/s. Solve the
equation for x.
d 2 y
dx 2 + b
dy
dx + cy = 0 481
(c) Since the roots are real and different, the general
solution is
V = Ae
ωt
+ Be
−ωt
(d) When t = 0, V = 7 hence 7 = A + B
(1)
dV
dt
= Aωe ωt − Bωe −ωt
When
t = 0,
dV
dt
= 3ω,
thus
3ω = Aω − Bω,
i.e.
3 = A − B
(2)
From equations (1) and (2), A = 5 and B = 2
Hence the particular solution is
V = 5e
ωt
+ 2e
−ωt
Since
sinh ωt =
1
2 (e ωt − e −ωt )
and
cosh ωt =
1
2 (e ωt + e −ωt )
then sinh ωt + cosh ωt = e ωt
and cosh ωt − sinh ωt = e
−ωt from Chapter 5.
Hence the particular solution may also be
written as
V = 5(sinh ωt + cosh ωt )
+ 2(cosh ωt − sinh ωt )
i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt
i.e. V = 7 coshωt + 3 sinh ωt
Problem 6. The equation
d 2 i
dt 2 +
R
L
di
dt
+
1
LC
i = 0
represents a current i flowing in an electrical circuit
containing resistance R, inductance L and
capacitance C connected in series. If R = 200 ohms,
L =0.20 henry and C = 20 ×10 −6 farads, solve the
equation for i given the boundary conditions that
when t = 0, i = 0 and
di
dt
= 100.
Using the procedure of Section 50.2:
(a)
d 2 i
dt 2 +
R
L
di
dt
+
1
LC
i = 0 in D-operator form is
D
2
+
R
L
D +
1
LC
i = 0 where D ≡
d
dt
(b) The auxiliary equation is m 2 +
R
L
m +
1
LC
= 0
Hence m =
−
R
L
±
R
L
2
− 4(1)
1
LC
2
When R = 200, L =0.20 and C = 20 ×10 −6 , then
m =
−
200
0.20
±
200
0.20
2
−
4
(0.20)(20 × 10 −6 )
2
=
−1000 ±
√
0
2
= −500
(c) Since the two roots are real and equal (i.e. −500
twice, since for a second order differential equation there must be two solutions), the general
solution is i = (At +B)e −500t .
(d) When t = 0, i = 0, hence B = 0
di
dt
= (At + B)(−500e −500t ) + (e −500t )(A),
by the product rule
When t = 0,
di
dt
= 100, thus 100 =−500B + A
i.e. A = 100, since B = 0
Hence the particular solution is
i = 100te
−500t
Problem 7. The oscillations of a heavily damped
pendulum satisfy the differential equation
d 2 x
dt 2 + 6
dx
dt
+ 8x = 0, where x cm is the
displacement of the bob at time t seconds.
The initial displacement is equal to +4 cm and the
initial velocity
i.e.
dx
dt
is 8 cm/s. Solve the
equation for x.