1550078481-Ordinary_Differential_Equations__Roberts_

The Laplace Trans! orm Method 241

The third statement converts the representation of the function g to the

proper format for use with the laplace command. The output is

1 1 2e(-s}

G·=---+--. s s (^2) x 2

8. Show that if J(x) and g(x) are both of exponential order a as x---> +oo,

then f(x) - g(x) is also of exponential order a as x---> +oo.

9. For each of the following functions F(s) find a function f(x) such that
   .C[f(x)] = F(s). That is, for each given function F(s) find an inverse
   Laplace transform .c-^1 [F(s)]. If you have a CAS available which cal-
   culates the inverse Laplace transform, a lso use the CAS to calculate
   the inverse Laplace transforms of each of the given functions F(s) and
   compare those answers to the ones you obtained by hand.
   3
   b.
   4 -2s
   a.
   s3 (s + 2)^2
   c.
   s^2 +3

1 s-l

f.

2

d.

s^2 (s + 1)

e.

s^2 - 2s + 5 s^2 - 2s + 5

-4 2s + 5 1 2

g.

s(s^2 + 1)

h.

s^2 + 2s + 2

i. -+--

s^2 s^2 - 1

j.

3s

s^2 - 4s + 3

!comments on Computer Software! The following three MAPLE

statements compute the inverse Laplace transform of

2

F(s) = -s(-s +-1)

in two different ways.

with(inttrans):

f:=invlaplace(2/(s * (s + 1)), s, x);

f:=invlaplace(2/s - 2/(s + 1), s, x);

The output of the second statement is

. 1

f-=. 4e(-^1 /^2 x) smh(- 2 x)

2 2

s s + 1

while the output of the third statement is

f:= 2 - 2e(-x}

Notice the two results are equ al but are expressed differently.