1550078481-Ordinary_Differential_Equations__Roberts_

Introduction

I EXAMPLE 3 A Solution D e fined Piecewise

Verify that the piecewise defined function

y( x ) = - 2 '

{

x 2

x,

x < 0

0:::; x

13

is differentiable on the interval ( -oo, oo) and is a solution of the differential

equation xy' - 2y = 0 on (-oo, oo).

SOLUTION

On the interval (-oo, 0) , y(x) = -x^2 ; therefore, on (-oo, 0) its derivative is

y' (x) = -2x. On the interval (0, oo), y(x ) = x^2 ; hence, on (0 , oo) its derivative

is y' (x) = 2x. To determine if y(x) is differentiable at x = 0, we compute the

left-hand derivat ive at 0 and the right-hand derivative at 0. Doing so, we find

y' (0) = lim -(O + h)

2

+

02

= lim _-h-

2

= lim - h = 0

• fi,Q- h fi ,Q - h h-->O-

and

'(O) 1. (O+h)2-02

Y+ = h-->Oi m + I i

h2

li m - lim h = 0.

h-->O+ h h-->O+

Since y'_ ( 0) = 0 = y~ ( 0), the function y( x) is differentiable at x = 0. Since

y( x ) is differentiable on (-oo,O), on (O,oo), and at x = 0, the function y( x )

is differentiable on the interval (-oo, oo ), and its derivative is

Y '( x ) = {-22xx, , x<O

0:::; x.

A graph of y( x) and y' ( x ) is shown in Figure 1.1.

-4 - 2

y

4

2

0

• 2

-4

x

2 4 - 4 - 2

4

2

0

• 2

-4

F igure 1.1 Graph of y( x) and y' ( x)

y'

x

2 4