1550078481-Ordinary_Differential_Equations__Roberts_

224 Ordinary Differential Equations

The Laplace transform is an operator which assigns to one function, f(x),

another function, F(s). It can be shown that if the improper integral appear-

ing in the definition converges for some fixed value so, then it will converge

for all s > s 0. For some functions the Laplace transform exists for all real s.

While for other functions the Laplace transform does not exist for any real s.

Recall from calculus that the improper integral

(oo g(x) dx is defined to be the limit lim ( B g(x) dx,

Jo B-++^00 Jo

provided the limit exists. When the limit exists, we say the integral exists

and its value is the limit. When the limit does not exist, we say the integral

does not exist. Consider the improper integral f 000 e-sxdx. For s = 0, the

function e-sx = 1 and the improper integral diverges, since

f

00

1 dx = lim ( B 1 dx = lim B = +oo.

Jo B-++oo } 0 B-+oo

For s =f. 0,

1

(^00) e-sx dx = lim 1B e-sx dx = lim ( e -sx IB)
0 B-++oo 0 B-++oo -S 0
= lim -l(e-^88 -1)={ ~ '
B-++oo S
diverges ,
ifs> 0
ifs< 0
Instead of writing the limit limB-++oo( )I~ over and over again, we will de-
note this limit simply by ( )ID"· Writing the integrand of the integral under
consideration as le-sx, we see that the integral is the Laplace transform of

the function f(x) = l. Thus, we have shown that

1

00

.C[l] = le-sx dx = -(-1 e-sx - 1) loo^1 ) provided s > 0.

0 s 0 s

EXAMPLE 1 Laplace Transform of eax, .C[eax]

Find the Laplace transform of f ( x) = eax.

SOLUTION

.C[eax] = 100 eaxe-sx dx = 100 e-(s-a)x dx = s -=-la e-(s-a)x[

provided s >a= so. Ifs'."::: a, then .C[eax] does not exist.

1

s-a