1550078481-Ordinary_Differential_Equations__Roberts_

The Laplace Trans! arm Method 269

and

1-: 5 ( x - c) dx = l.

Let f(x) be a function which is continuous on some interval about x = c.

We shall define the generalized integral of the product of 5(x - c) and f(x) as

follows:

1-: 5(x - c)f(x) dx = -~~+ 1-: d, (x - c)f(x) dx

l

e+' 1 1

= lim -f(x) dx = lim -f(~)(c + E - c)

,_,a+ c E ,_,a+ E

where c < ~ < c + E. The last equality was obtained by using the mean value

theorem for integrals. Since c < ~ < c+E and f(x) is assumed to be continuous

on some interval about x = c, we have lim,_, 0 + f(~) = f(c). Consequently,

in the context of distribution theory, we have for any function f(x) which is

continuous at x = c,

1-: 5(x - c)f(x) dx = f(c).

Therefore, letting f(x) = e-sx, we have some justification for defining the

Laplace transform of the Dirac delta function 5(x - c), where c > 0, to be

£[5(x - c)] = fo

00

5(x - c)e-sx dx = e-sc.

In what follows we will operate with the Dirac delta function, 5(x - c), as
though it were an ordinary function and we will use the properties discussed
and developed in the preceding paragraphs even though our development of
these properties was not mathematically rigorous. It is perhaps comforting
to know that in the context of distribution theory all these operations and
properties can and have been proven rigorously.

EXAMPLE 1 Solving an Initial Value Problem with an

Impulse Forcing Function

Solve the initial value problem

y" + 2y' + 5y = 35(x - l); y(O) = 0 , y'(O) = 0.

SOLUTION

This initial value problem could represent the damped motion of a mass on

a spring which is initially at rest but which is set in motion at time x = 1