ECE 382
REVIEW OF SOLUTIONS OF LINEAR ORDINARY DIFFERENTIAL
EQUATIONS WITH CONSTANT COEFFICIENTS
We shall consider ordinary differential equations of the form
a
n
d
n
x
dt
n
+
a
n

1
d
n

1
x
dt
n

1
+
···
+
a
1
dx
dt
+
a
0
x
=
q
(
t
)
(1)
where the coefficients
a
0
,
a
1
,...,
a
n
are constant, and discuss some physical problems giving rise
to such equations. If we employ the symbols
D
,
D
2
,
D
3
,...,
etc. to denote
d
dt
,
d
2
dt
2
,
d
3
dt
3
,...,
respec
tively, then Eq. (1) may be written as
(
a
n
D
n
+
a
n

1
D
n

1
+
···
+
a
1
D
+
a
0
)
x
=
q
(
t
)
.
(2)
1
The Homogeneous Case
We shall suppose that Eq. (2) is the mathematical model for some physical plant which may be
represented pictorially as in Figure 1.
+
_
q(t)
t=0
PLANT
y(t)
+
_
S
Figure 1: Mathematical model of a physical
plant.
Here we suppose that it is desired to find the
output voltage
y
(
t
)
. The plant has been operating
for some time prior to
t
=
0. At
t
=
0, the switch
S is opened and the input to the plant, the voltage
q
(
t
)
, is zero for
t
≥
0
.
For simplicity we shall work
with secondorder examples. Figure 2 represents
a concrete example of such a secondorder plant.
+
_
t=0
y(t)
E
x(t)
C
R
L
S
r
Figure 2: A secondorder plant example.
The situation after
t
=
0 can be described by
using Kirchhoff’s voltage law around the closed
circuit. If
x
(
t
)
represents the inductor current, then
L
dx
dt
+
Rx
+
1
C
Z
t

∞
xdt
=
0
.
(3)
If we differentiate this equation with respect to
t
,
we have a secondorder linear homogeneous ordi
nary differential equation with constant coefficients:
L
d
2
x
dt
2
+
R
dx
dt
+
1
C
x
=
0
.
(4)
If we assign values to
L
,
R
and
C
of 1 henry, 3 ohms and 1/2 farad respectively (not very realistic
numbers, but convenient for illustrative purposes), we may write Eq. (4) using operator notation as
(
D
2
+
3
D
+
2
)
x
= (
D
+
2
)(
D
+
1
)
x
=
0
(5)
1
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Setting
(
D
+
1
)
x
=
z
,
(6)
we obtain
(
D
+
2
)
z
=
0
.
(7)
From Eqs. (6) and (7), it appears that the building blocks of all linear differential equations with
constant coefficients are the firstorder differential equations of the form
˙
x
+
ax
=
p
(
t
)
.
(8)
By inspection, one can see that the function
e
at
is an
integrating
factor of Eq. (8). This means that
if one multiplies Eq. (8) by
e
at
, then the lefthand side becomes derivative. We obtain
(
˙
x
+
ax
)
e
at
=
p
(
t
)
e
at
(9)
or
d
dt
(
xe
at
) =
e
at
p
(
t
)
.
(10)
Integrating this last equation with respect to
t
from
t
=
0 to some generic time
t
, we obtain
x
(
t
)
e
at

x
(
0
) =
Z
t
0
e
a
τ
p
(
τ
)
d
τ
(11)
where
τ
is the dummy variable of integration. Thus,
x
(
t
) =
e

at
x
(
0
)+
e

at
Z
t
0
e
a
τ
p
(
τ
)
d
τ
(12)
where
x
(
0
)
is an arbitrary constant. It is to be determined by the supposedly known initial value of
the variable
x
(
t
)
. As an example of the application of this formula, consider the circuit in Figure 3.
+
_
R
L
x(t)
p(t)
Figure 3: An example circuit.
If the voltage
p
(
t
)
is given by
p
(
t
) =
E
sin
ω
t
,
then Kirchhoff’s law applied to this loop gives us
L
dx
dt
+
Rx
=
E
sin
ω
t
,
(13)
or
dx
dt
+
R
L
x
=
E
L
sin
ω
t
.
(14)
Equation (14) is now in the form of Eq. (8). An
integrating factor is
e
R
L
t
.
Applying the formula in
Eq. (12) we obtain
x
(
t
) =
x
(
0
)
e

R
L
t
+
e

R
L
t
Z
t
0
(
E
L
sin
ω
s
)
e
R
L
s
ds
.
(15)
From a table of integrals, we obtain
x
(
t
) =
x
(
0
)
e

R
L
t
+
E
L
e

R
L
t
"
e
R
L
t
(
R
L
sin
ω
t

ω
cos
ω
t
)
+
ω
(
R
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 Spring '08
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