Second order differential equations of the forma
d^2 y
dx^2 +b
dy
dx+cy=^0481
(c) Since the roots are real and different,the general
solution is
V=Aeωt+Be−ωt
(d) Whent=0,V=7 hence 7=A+B (1)
dV
dt
=Aωeωt−Bωe−ωt
When t= 0 ,
dV
dt
= 3 ω,
thus 3ω=Aω−Bω,
i.e. 3 =A−B (2)
From equations (1) and (2),A=5andB= 2
Hencethe particular solution is
V=5eωt+2e−ωt
Since sinhωt=^12 (eωt−e−ωt)
and coshωt=^12 (eωt+e−ωt)
then sinhωt+coshωt=eωt
and coshωt−sinhωt=e−ωt from Chapter 5.
Hence the particular solution may also be
written as
V= 5 (sinhωt+coshωt)
+ 2 (coshωt−sinhωt)
i.e.V=( 5 + 2 )coshωt+( 5 − 2 )sinhωt
i.e.V=7coshωt+3sinhωt
Problem 6. The equation
d^2 i
dt^2
+
R
L
di
dt
+
1
LC
i= 0
represents a currentiflowing in an electrical circuit
containing resistanceR, inductanceLand
capacitanceCconnected in series. IfR=200ohms,
L= 0 .20 henry andC= 20 × 10 −^6 farads, solve the
equation forigiven the boundary conditions that
whent=0,i=0and
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=0whereD≡
d
dt
(b) The auxiliary equation ism^2 +
R
L
m+
1
LC
= 0
Hencem=
−
R
L
±
√√
√
√
[(
R
L
) 2
− 4 ( 1 )
(
1
LC
)]
2
WhenR=200,L= 0 .20 andC= 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 equa-
tion there must be two solutions),the general
solution isi=(At+B)e−^500 t.
(d) Whent=0,i=0, henceB= 0
di
dt
=(At+B)(−500e−^500 t)+(e−^500 t)(A),
by the product rule
Whent=0,
di
dt
=100, thus 100=− 500 B+A
i.e.A=100, sinceB= 0
Hence the particular solution is
i= 100 te−^500 t
Problem 7. The oscillations of a heavily damped
pendulum satisfy the differential equation
d^2 x
dt^2
+ 6
dx
dt
+ 8 x=0, wherexcm is the
displacement of the bob at timetseconds.
The initial displacement is equal to+4cmandthe
initial velocity
(
i.e.
dx
dt
)
is 8cm/s. Solve the
equation forx.