LRC Circuit: Roots of
Characteristic Equation
LCL
R
L
R
r1
22
2
−
−=
Have finished Case 1: two real roots
LRC Circuit: Repeated Roots
Recall the solution to the LRC ODE in Case 1 (two real roots to the
characteristic equation) is
Consider this solution as approaches zero; i.e., as the two roots
+
+
−
−= +−−− tt
ceeVtV )()(
02
)(
2
)(
)(
Equation 7
LRC Circuit: Repeated Roots (2)
Divide Equation 7 by to obtain
Write and in series expansions:
t
eV
−
0
ttt
ceee
V
tV
2
)(
2
)()(
0
+
+
−
−= −
t
e
t
e
−
Equation 8
LRC Circuit: Repeated Roots (3)
Which becomes
...)2/)(1(
2
)(
...)2/)(1(
2
)()(
2
2
0
=+++
+
+−+−
−
−=
tt
tte
V
tV t
c
LRC Circuit: Repeated Roots (4)
In the limit as approaches zero this becomes
t
c
te
V
tV
+=
1
)(
LRC Circuit: Repeated Roots (5)
Note that since
and
L
R
2
=
LRC Circuit: Repeated Roots (6)
Now
so
)(
)1()(
2
0
0
t
c
t
c
teV
dt
dV
etVtV
−
−
−=
+=
)(
32
0
++−
−−
tt
teeV
LRC Circuit: Repeated Roots (7)
Checking initial conditions:
Hence
is a solution of the LRC ODE when
)0(
)1()(
0
0
=
+=
−
VV
etVtV
c
t
c
t
cetVtV
−
+= )1()( 0
L
R
2
=
where
LRC Circuit: Repeated Roots (8)
The results of this example extend to any ODE of the form
02 2
2
2
=++ y
dt
dy
dt
yd
In-Class Example Problems
Solve:
1)0(,2)0(
065
2
2
=
=
=++
yy
y
dt
dy
dt
yd
Homework Assignment 7
Prepare for Test #1 in next class
Read: In text, Chapter 5; beginning of chapter through Section 5.1.2
Work :