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Application of Laplace Transform
.

Example 5.1, Consider the initial-value problem

2
d y
′
+ y = 1, y(0) = y (0) = 0.
dt 2
. . . . .
.
Solution. Let us assume for the moment that the solution y = y(t) satisfies
suitable conditions so that we may invoke (2.8). Taking the Laplace
transform of both sides of the differential equation gives

L(y ) + L(y) = L(1).
′′

An application of (2.8) yields

1
2
s L(y) − sy(0) − y (0) + L(y) = ,
′
s
that is,

1
L(y) = .
2
s(s + 1)
Writing
1 A Bs + C
= +
2
2
s(s + 1) s s + 1
as a partial fraction decomposition, we find

1 s
L(y) = − .
2
s s + 1
Applying the inverse transform gives the solution

y = 1 − cos t.

One may readily check that this is indeed the solution to the initial- value
. . . . problem.

Note that the initial conditions of the problem are absorbed into the method,
unlike other approaches to problems of this type (i.e., the methods of variation
of parameters or undetermined coefficients).

General Procedure

The Laplace transform method for solving ordinary differential equations can be

summarized by the following steps.
1. Take the Laplace transform of both sides of the equation. This results in what

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