Lecture Notes in Differential Equations - Bruce E. Shapiro

336 LESSON 32. THE LAPLACE TRANSFORM
Example 32.9. Find the Laplace Transform of the step function
f(t) =
{ 0, 0 ≤ t ≤ 3
2, t ≥ 3
(32.28)
Using the definition of L[f(t)], we compute
L[f(t)] =
∫ ∞
2
3e −ts dt = − 3 s e−ts ∣ ∣∣∣
∞
2
= 3e−2s
s
(32.29)
Laplace Transforms of Derivatives
The Laplace Transform of a derivative is what makes it useful in our study
of differential equations. Let f(t) be any function with Laplace Transform
F (s). Then
[ ] df(t)
L =
dt
∫ ∞
0
df(t)
e −st dt (32.30)
dt
We can use integration by parts (which means we use equation A.3) to solve
this problem; it is already written in the form ∫ udv with
Since this gives v = f and du = −se −st dt, we obtain
which we will write as
L[f ′ (t)] = e −st f(t) ∣ ∣ ∞ 0
u = e −st (32.31)
dvf ′ (t)dt (32.32)
∫ ∞
+ s f(t)e st dt (32.33)
0
= −f(0) + sF (s) (32.34)
L[f ′ (t)] = sF (s) − f(0) (32.35)
Thus the Laplace Transform of a Derivative is an algebraic function
of the Laplace Transform. That means that Differential equations
can be converted into algebraic equations in their Laplace representation,
as we will see shortly.
Since the second derivative is the derivative of the first derivative, we can