Lecture Notes in Differential Equations - Bruce E. Shapiro

342 LESSON 32. THE LAPLACE TRANSFORM
2. Evaluate all of the Laplace transforms, including any initial conditions,
so that the resulting equation has all y(t) removed and replaced
with Y (s).
3. Solve the resulting algebraic equation for Y (s).
4. Find a function y(t) whose Laplace transform is Y (s). This is the
solution to the initial value problem.
The reason why this works is because of the following theorem, which we
will accept without proof.
Theorem 32.10 (Lerch’s Theorem 1 ). For any function F (s), there is at
most one continuous function function f(t) defined for t ≥ 0 for which
L[f(t)] = F (s).
This means that there is no ambiguity in finding the inverse Laplace transform.
The method also works for higher order equations, as in the following example.
Example 32.11. Solve y ′′ − 7y ′ + 12y = 0, y(0) = 1, y ′ (0) = 1 using
Laplace transforms.
Following the procedure outlined above:
0 = L[y ′′ − 7y ′ + 12y] (32.88)
= L[y ′′ ] − 7L[y ′ ] + 12L[y] (32.89)
= s 2 Y (s) − sy(0) − y ′ (0) − 7(sY (s) − y(0)) + 12Y (s) (32.90)
= s 2 Y (s) − s − 1 − 7sY (s) + 7 + 12Y (s) (32.91)
= (s 2 − 7s + 12)Y (s) + 6 − s (32.92)
s − 6
Y (s) =
s 2 − 7s + 12 = s − 6
(s − 3)(s − 4) = 3
s − 3 − 2
s − 4
(32.93)
where the last step is obtained by the method of partial fractions. Hence
[ 3
y(t) = L −1 s − 3 − 2 ]
= 3e 3t − 2e 4t . (32.94)
s − 4
1 Named for Mathias Lerch (1860-1922), an eminent Czech Mathematician who published
more than 250 papers.