whose coefficients are polynomials in
s.
Moreover, if the coefficients of the
given equation are polynomials of degree 1 in
t
, then (regardless of the order of the given equa-
tion) the differential equation for
is just a linear
first-order
equation. Since we know how
to solve this first-order equation, the only serious obstacle we may encounter is obtaining the
inverse Laplace transform of
[This problem may be insurmountable, since the solution
may
not
have a Laplace transform.]
In illustrating the technique, we make use of the following fact.
If
is piecewise
continuous on
and of exponential order, then
(12)
(You may have already guessed this from the entries in Table 7.1, page 359.) An outline of the
proof of (12) is given in Exercises 7.3, Problem 26.
Solve the initial value problem
(13)
Let
and take the Laplace transform of both sides of the equation in (13):
(14)
Using the initial conditions, we find
E
y
–
F
A
s
B
s
2
Y
A
s
B
sy
A
0
B
y
¿
A
0
B
s
2
Y
A
s
B
E
y
–
F
A
s
B
2
E
ty
¿
A
t
B
F
A
s
B
4
Y
A
s
B
1
s
.
Y
A
s
B
E
y
F
A
s
B
y
–
2
ty
¿
4
y
1
,
y
A
0
B
y
¿
A
0
B
0
.
lim
s
S
E
f
F
A
s
B
0
.
3
0,
q
B
f
A
t
B
y
A
t
B
Y
A
s
B
.
Y
A
s
B
Y
A
s
B
y
A
t
B
s
2
Y
¿
A
s
B
2
sY
A
s
B
y
A
0
B
.
d
ds
3
s
2
Y
A
s
B
sy
A
0
B
y
¿
A
0
B
4
E
ty
–
A
t
B
F
A
s
B
d
ds
E
y
–
F
A
s
B
f
A
t
B
y
–
A
t
B
,
n
1
d
ds
3
sY
A
s
B
y
A
0
B
4
sY
¿
A
s
B
Y
A
s
B
.
E
ty
¿
A
t
B
F
A
s
B
d
ds
E
y
¿
F
A
s
B
f
A
t
B
y
¿
A
t
B
,
n
1
E
t
n
f
A
t
B
F
A
s
B
A
1
B
n
d
n
F
ds
n
A
s
B
.
3
e
t
p
cos 2
t
4
e
t
p
sin 2
t
e
p
t
.
w
A
t
B
y
A
t
p
B
3
e
t
p
cos
3
2
A
t
p
B
4
4
e
t
p
sin
3
2
A
t
p
B
4
e
A
t
p
B
t
p
w
A
t
B
y
A
t
p
B
.
w
A
t
p
B
y
A
t
B
,
Section 7.5
Solving Initial Value Problems
379
Example 4
Solution

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and
Substituting these expressions into (14) gives
(15)
Equation (15) is a linear first-order equation and has the integrating factor
(see Section 2.3). Multiplying (15) by
we obtain
Integrating and solving for
yields
(16)
Now if
is the Laplace transform of a piecewise continuous function of exponential order,
then it follows from equation (12) that
For this to occur, the constant
C
in equation (16) must be zero. Hence,
and taking
the inverse transform gives
We can easily verify that
is the solution to
the given initial value problem by substituting it into (13).
◆
We end this section with an application from
control theory.
Let’s consider a servomech-
anism that models an automatic pilot. Such a mechanism applies a torque to the steering
control shaft so that a plane or boat will follow a prescribed course. If we let
be the true
direction (angle) of the craft at time
t
and
be the desired direction at time
t
, then
denotes the
error
or
deviation
between the desired direction and the true direction.
Let’s assume that the servomechanism can measure the error
and feed back to the
steering shaft a component of torque that is proportional to
but opposite in sign (see
Figure 7.6 on page 381). Newton’s second law, expressed in terms of torques, states that
A
moment of inertia
B
A
angular acceleration
B
total torque.