Page 75 - 4811

Application of Laplace Transform

In general, the Laplace transform method demonstrated above is particularly

applicable to initial-value problems of nth-order linear ordinary differential
equations with constant coefficients, that is,

n
d y d n−1 y
a n + a n−1 + · · · + a y = f(t), (5.1)
0
dt n dt n−1
′
y(0) = y , y (0) = y , . . . , y (n−1) (0) = y n−1 .
0
1
In engineering parlance, f(t) is known as the input, excitation, or forcing function,
and y = y(t) is the output or response. In the event the input f(t) has exponential
order and be continuous, the output y = y(t) to (5.1) can also be shown to have
exponential order and be continuous. This fact helps to justify the application of
the Laplace transform method. More generally, when f ∈ L, the method can still

be applied by assuming that the hypotheses of Theorem (2) are satisfied. While
the solution y = y(t) to (5.1) is given by the Laplace transform method for t ≥ 0,
it is in general valid on the whole real line, −∞ < t < ∞, if f(t) has this domain.
Another important virtue of the Laplace transform method is that the input
.
function f(t) can be discontinuous.

Example 5.3, Solve

′′
y + y = Eu (t), y(0) = 0, y (0) = 1.
′
a
.
. . . . .
Solution. Here the system is receiving an input of zero for 0 ≤ t < a and
E (constant) for t ≥ a. Then

Ee −as
2
′
s L(y) − sy(0) − y (0) + L(y) =
s
and

1 Ee −as 1 ( 1 s )
L(y) = + = + E − e −as .
2
2
2
2
s + 1 s(s + 1) s + 1 s s + 1
Whence
( ) [( ) ]
1 1 s
−1 −1 −as
y = L + EL − e
2
2
s + 1 s s + 1
by the second translation theorem. We can also express y in the form

{
sin t, 0 ≤ t < a,
y =
sin t + E(1 − cos(t − a), t ≥ a.

+
+
′′
−
Note that y(a ) = y(a ) = sin a, y (a ) = y (a ) = cos a, y (a ) = − sin a,
−
−
′
′
2
+
′′
. . . . but y (a ) = − sin a + Ea . Hence y(t) is only piecewise continuous.
74