Lecture Notes in Differential Equations - Bruce E. Shapiro

340 LESSON 32. THE LAPLACE TRANSFORM
Using Laplace Transforms to Solve Linear Differential
Equations
The idea is this: the Laplace Transform turns derivatives into algebraic
functions. So if we convert an entire linear ODE into its Laplace Transform,
since the transform is linear, all of the derivatives will go away. Then we
can solve for F (s) as a function of s. A solution to the differential equation
is given by any function f(t) whose Laplace transform is given by F (s).
The process is illustrated with an example.
Example 32.10. Solve y ′ + 2y = 3, y(0) = 1, using Laplace transforms.
Let Y (s) = L[y(t)] and apply the Laplace Transform operator to the entire
equation.
From example 32.1, and equation 32.35,
y ′ (t) + 2y(t) = 3t (32.66)
L[y ′ (t) + 2y(t)] = L[3t] (32.67)
L[y ′ (t)] + 2L[y(t)] = 3L[t] (32.68)
sY (s) − y(0) + 2Y (s) = 3 s 2 (32.69)
(s + 2)Y (s) = 3 s 2 + y(0) = 3 s 2 + 1 = 3 + s2
s 2 (32.70)
(32.71)
Solving for Y (s),
Y (s) =
3 + s2
(s + 2)s 2 (32.72)
The question then becomes the following: What function has a Laplace
Transform that is equal to (3 + s 2 )/(s 2 (s + 2))? This is called the inverse
Laplace Transform of Y (s) and gives y(t):
[ ] 3 + s
y(t) = L −1 2
(s + 2)s 2 (32.73)
We typically approach this by trying to simplify the function into smaller
parts until we recognize each part as a Laplace transform. For example, we
can use the method of Partial Fractions to separated the expression:
3 + s 2
(s + 2)s 2 = A + Bs
s 2 + C + Ds
s + 2
=
(A + Bs)(s + 2)
s 2 (s + 2)
+
(C + Ds)s2
s 2 (s + 2)
(32.74)
(32.75)