The indirect method utilizes the relationship between the
differential equation and the Laplacetransform, discussed earlier, to find a
solution. The basic idea is to convert the differential
equation into a Laplacetransform, as described above, to get the
resulting output,
Ys
Y
s
.
Then by inverse transforming this and using partialfraction
expansion, we can arrive at the solution.
L
d
d
t
y
(
t
)
=
s
Y
(
s
)

y
(
0
)
L
d
d
t
y
(
t
)
=
s
Y
(
s
)

y
(
0
)
(12)This can be interatively extended to an arbitrary order derivative as in Equation Equation 13.
L
d
n
d
t
n
y
(
t
)
=
s
n
Y
(
s
)

∑
m
=
0
n

1
s
n

m

1
y
(
m
)
(
0
)
L
d
n
d
t
n
y
(
t
)
=
s
n
Y
(
s
)

∑
m
=
0
n

1
s
n

m

1
y
(
m
)
(
0
)
(13)Now, the Laplace transform of each side of the differential equation can be taken
L
∑
k
=
0
n
a
k
d
k
d
t
k
y
(
t
)
=
L
x
(
t
)
L
∑
k
=
0
n
a
k
d
k
d
t
k
y
(
t
)
=
L
x
(
t
)
(14)which by linearity results in
∑
k
=
0
n
a
k
L
d
k
d
t
k
y
(
t
)
=
L
x
(
t
)
∑
k
=
0
n
a
k
L
d
k
d
t
k
y
(
t
)
=
L
x
(
t
)
(15)and by differentiation properties in
∑
k
=
0
n
a
k
s
k
L
y
(
t
)

∑
m
=
0
k

1
s
k

m

1
y
(
m
)
(
0
)
=
L
x
(
t
)
.
∑
k
=
0
n
a
k
s
k
L
y
(
t
)

∑
m
=
0
k

1
s
k

m

1
y
(
m
)
(
0
)
=
L
x
(
t
)
.
(16)Rearranging terms to isolate the Laplace transform of the output,
L
y
(
t
)
=
L
x
(
t
)
+
∑
k
=
0
n
∑
m
=
0
k

1
a
k
s
k

m

1
y
(
m
)
(
0
)
∑
k
=
0
n
a
k
s
k
.
L
y
(
t
)
=
L
x
(
t
)
+
∑
k
=
0
n
∑
m
=
0
k

1
a
k
s
k

m

1
y
(
m
)
(
0
)
∑
k
=
0
n
a
k
s
k
.
(17)Thus, it is found that
Y
(
s
)
=
X
(
s
)
+
∑
k
=
0
n
∑
m
=
0
k

1
a
k
s
k

m

1
y
(
m
)
(
0
)
∑
k
=
0
n
a
k
s
k
.
Y
(
s
)
=
X
(
s
)
+
∑
k
=
0
n
∑
m
=
0
k

1
a
k
s
k

m

1
y
(
m
)
(
0
)
∑
k
=
0
n
a
k
s
k
.
(18)In order to find the output, it only remains to find the Laplace transform X(s)X(s) of the input, substitute the initial conditions, and compute the inverse Laplace transform of the result. Partial fraction expansions are often required for this last step. This may sound daunting while looking at Equation Equation 18, but it is often easy in practice, especially for low order differential equations. Equation Equation 18 can also be used to determine the transfer function and frequency response.
As an example, consider the differential equation
d
2
d
t
2
y
(
t
)
+
4
d
d
t
y
(
t
)
+
3
y
(
t
)
=
cos
(
t
)
d
2
d
t
2
y
(
t
)
+
4
d
d
t
y
(
t
)
+
3
y
(
t
)
=
cos
(
t
)
(19)with the initial conditions y'(0)=1y'(0)=1 and y(0)=0y(0)=0
Using the method described above, the Laplace transform of the solution y(t)y(t) is given by
Y
(
s
)
=
s
(
s
2
+
1
)
(
s
+
1
)
(
s
+
3
)
+
1
(
s
+
1
)
(
s
+
3
)
.
Y
(
s
)
=
s
(
s
2
+
1
)
(
s
+
1
)
(
s
+
3
)
+
1
(
s
+
1
)
(
s
+
3
)
.
(20)Performing a partial fraction decomposition, this also equals
Y
(
s
)
=
.
25
1
s
+
1

.
35
1
s
+
3
+
.
1
s
s
2
+
1
+
.
2
1
s
2
+
1
.
Y
(
s
)
=
.
25
1
s
+
1

.
35
1
s
+
3
+
.
1
s
s
2
+
1
+
.
2
1
s
2
+
1
.
(21)Computing the inverse Laplace transform,
y
(
t
)
=
(
.
25
e

t

.
35
e

3
t
+
.
1
cos
(
t
)
+
.
2
sin
(
t
)
)
u
(
t
)
.
y
(
t
)
=
(
.
25
e

t

.
35
e

3
t
+
.
1
cos
(
t
)
+
.
2
sin
(
t
)
)
u
(
t
)
.
(22)One can check that this satisfies that this satisfies both the differential equation and the initial conditions.
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