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84 CHAPTER 3 The Laplace Transform

To solve for the constants, observe that if we added the fractions on the right the numerator would
have to equal the numerator 2s − 1 of the fraction on the left. Therefore,

A(s + 1)(s + 3) + B(s − 1)(s + 3) + C(s − 1)(s + 1) = 2s − 1.
We can solve for A, B, and C by inserting values of s into this equation. Put s = 1 to get 8A = 1,
so A = 1/8. Put s =−1 to get −4B =−3, so B = 3/4. Put s =−3 to get 8C =−7, so C =−7/8.
Then
1 1 3 1 7 1
Y(s) = + − .
8 s − 1 4 s + 1 8 s + 3
Invert this to obtain the solution
1 3 7
−t
t
y(t) = e + e − e −3t .
8 4 8
Partial fractions decompositions are frequently used with the Laplace transform. The
appendix at the end of this chapter reviews the algebra of this technique.
Notice that the transform method does not first produce the general solution and then solve
for the constants to satisfy the initial conditions. Equations (3.1), (3.2), and (3.3) insert the initial
conditions directly into an algebraic equation for the transform of the unknown function. Still,
we could have solved the problem of Example 3.3 by methods from Chapter 2. The object here
was to illustrate a technique. This technique extends to problems beyond the reach of methods
from Chapter 2, and this is the subject of the next section.

SECTION 3.2 PROBLEMS

2

In each of Problems 1 through 10, use the Laplace trans- 8. y + 9y = t ; y(0) = y (0) = 0

form to solve the initial value problem.
9. y + 16y = 1 + t; y(0) =−2, y (0) = 1

1. y + 4y = 1; y(0) =−3 10. y − 5y + 6y = e ; y(0) = 0, y (0) = 2

−t

2. y − 9y = t; y(0) = 5 11. Prove Theorem 3.1. Hint: Write
3. y + 4y = cos(t); y(0) = 0

∞
−t
4. y + 2y = e ; y(0) = 1 L[ f ](s) = e −st f (t)dt

0
5. y − 2y = 1 − t; y(0) = 4

6. y + y = 1; y(0) = 6, y (0) = 0 and integrate by parts.

7. y − 4y + 4y = cos(t); y(0) = 1, y (0) =−1 12. Derive equation (3.3). Hint: Integrate by parts twice.
3.3 Shifting and the Heaviside Function
The shifting theorems of this section will enable us to solve problems involving pulses and other
discontinuous forcing functions.
3.3.1 The First Shifting Theorem
at
We will show that the Laplace transform of e f (t) is the transform of f (t), shifted a units to the
right. This shift is achieved by replacing s by s − a in F(s) to obtain F(s − a).
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October 14, 2010 14:14 THM/NEIL Page-84 27410_03_ch03_p77-120