1550078481-Ordinary_Differential_Equations__Roberts_

262 Ordinary Differential Equations

SOLUTION

Using the trigonometric identity cos(x - 7r) = - cosx, we see that

f(x) = cosx - u(x - 7r) cosx = cosx + u(x - 7r) cos(x - 7r).

Then taking the Laplace transform, using the linearity property, and the
second translation property, we find

.C[f(x)] = .C[cosx + u(x - 7r) cos(x - 7r)] = .C[cosx] + .C[u(x - 7r) cos(x - 7r)]

(1 + e-?rs )s

= .C[cosx] + e-7rs .C[cosx] = (1 + e-7rs).C[cosx] = 2.

s + 1

EXAMPLE 2 Using the Second Translation Property

to Find an Inverse Laplace

Find a function f(x) whose Laplace transform is ~-?rs.

s + 4

SOLUTION

2

Since ~

4

is the Laplace transform of sin 2x and since the factor e-?rs in-
s +
dicates the function sin 2x should be delayed 7f units, we calculate the Laplace

1. transform of
   2

u(x - 7r) sm 2(x - 7r) and find

1 1 e-7rS

.C[

2

u(x-7r)sin2(x-7r)] =

2

e-7r^8 .C[sin2s] = s 2 +

4

.

1

Consequently, f(x) =

2

u(x-1f) sin 2(x-7r). Noting that sin(2x-27r) =sin 2x,

we can write f(x) more conventionally a~

{

0,

f(x) =

1

2

sin2x, 7f:::; x

Now let us consider finding a solution of the initial value problem

(2) y" + y = 2u(x - 1) - u(x - 2) = h(x); y(O) = 0, y'(O) = 1.

In this instance, the forcing function is the step function h(x) of equation (1).
This function is discontinuous at x = 1 and x = 2 and, therefore, is not the
solution of any linear homogeneous differential equation with constant coeffi-
cients. So here we have our first example of an initial value problem which we