1550078481-Ordinary_Differential_Equations__Roberts_

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Ordinary Differential Equations

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11111

-4 -2 0 2 4

x

Figure 2.8 Direction Field for the DE y' = 1 + y^2 and the

Solution of the IVP y' = 1 + y^2 ; y(7r /4) = 1

I EXAMPLE 4 The IVP y' ~ - x/y; y( 1 ) 1

Analyze the initial value problem

(10)

SOLUTION

y' = -x/y ; y(-l) = l.

The functions f(x, y) = -x/y and fy(x, y) = x/y^2 are defined and continu-
ous except for y = 0. Since the point ( - 1, 1) is in the upper ha lf plane (where
y > 0), and since f and fy are defined and continuous for y > 0, by the
continuation theorem the solution of the IVP (10) can be ext ended uniquely

until two of the following four things occur: x-+ - oo, x-+ oo, y(x) -+ 0 , or

y(x) -+ +oo.

Verify that y(x) = )2 - x^2 is the unique solution of the IVP (10) and that
y(x ) is defined, continuous, and differentiable on the interval (-v'2, v'2). Since
y( x) is not defined outside the interval [-v'2,v'2J, (-J2,v'2) is the largest

interval on which the IVP (10) has a solution. Observe tha t as x -+ -v'2 +,

y(x)-+ o+ and as x-+ J2-, y(x)-+ o+. Notice that these results satisfy t he
conclusion of the continuation theorem. In this example, we h ave y( x ) -+ o+
for two different values of x. So in this case the solution approaches the same