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82 CHAPTER 3 The Laplace Transform

THEOREM 3.1 Transform of a Derivative

Let f be continuous for t ≥ 0, and suppose f is piecewise continuous on [0,k] for every k > 0.

Suppose also that lim k→∞ e −sk f (k) = 0if s > 0. Then
L[ f ](s) = sF(s) − f (0). (3.1)

This states that the transform of f (t) is s times the transform of f (t), minus f (0), which
is the original function evaluated at t = 0. This can be proved by integration by parts (see
Problem 11).
If f has a jump discontinuity at 0, as occurs if f is an electromotive force that is switched
on at time zero, then the conclusion of the theorem must be amended to read

L[ f ](s) = sF(s) − f (0+)
where f (0+) = lim t→0+ f (t).
There is an extension of Theorem 3.1 to higher derivatives. If n is a positive integer, let f (n)
denote the nth derivative of f .

THEOREM 3.2 Transform of a Higher Derivative
Let f , f , f (n−1) be continuous for t > 0, and suppose f (n) is piecewise continuous on [0,k] for

every k > 0. Suppose also that
lim e −sk f ( j) (k) = 0
k→∞
for s > 0 and j = 1,2,··· ,n − 1. Then

n

L[ f (n) ](s) = s F(s) − s n−1 f (0) − s n−2 f (0) − ··· − sf (n−2) (0) − f (n−1) (0). (3.2)

The second derivative case n = 2 occurs sufficiently often that we will record the formula
separately for this case:
2
L[ f ](s) = s F(s) − sf (0) − f (0). (3.3)

We are now prepared to use the Laplace transform to solve some initial value problems.

EXAMPLE 3.2

We will solve y − 4y = 1; y(0) = 1.

We already know how to solve this problem, but we will apply the Laplace transform to
illustrate the idea. Take the transform of the differential equation using the linearity of L and
equation (3.1) to write

L[y − 4y](s) = L[y ](s) − 4L[y](s)
= (sY(s) − y(0)) − 4Y(s) = L[1](s).
Insert the initial data y(0) = 1, and use the table to find that L[1](s) = 1/s. Then
1
(s − 4)Y(s) − 1 = .
s

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October 14, 2010 14:14 THM/NEIL Page-82 27410_03_ch03_p77-120