1550078481-Ordinary_Differential_Equations__Roberts_

The Initial Value Problem y' = f(x, y); y(c) = d 53

SOLUTION

a. The functions a(x) = 3x^2 and b(x) =ex are both defined and continuous

on ( -oo, oo). So by the previous existence and uniqueness theorem

the given initial value problem has a unique solution on the interval

(-oo, oo ).

b. In this instance, a(x) = 1/((x - l)(x + 2)) is defined and continuous

for x =f=. -2 and x =f=. l. So a( x) is defined and continuous on the

intervals (-oo,-2), (-2,1), and (1,oo). The function b(x) = 1 /x is

defined and continuous for x =f=. 0. Thus, b(x) is defined and continuous

on the intervals (-oo, 0) and (0, oo ). Taking the intersection of these

intervals with the intervals on which a(x) is defined and continuous,

we find a( x) and b( x) are simultaneously defined and continuous on the

intervals (-oo, -2), (-2, 0), (0, 1), and (1, oo). Since -1 is in the interval

(-2, 0), the initial value problem b. has a unique solution on (-2, 0). If

the initial condition y(-1) = 2 were changed to y(l) = 2, then there

would be no solution to the corresponding new initial value problem,

since a( x )- and, therefore, the differential equation- is not defined at

x = l. If the initial condition were changed to y(2) = 2, then the unique

solution to the corresponding initial value problem would exist on the

interval ( 1, oo).

c. In this case, a(x) = cot x is defined and continuous on the inter-

vals (mr, (n + l)n) for n = 0, ±1, ±2, ... and b(x) = tanx is defined

and continuous on the intervals ( (2n - 1 )n /2, (2n + 1 )n /2) for n =

0, ±1, ±2, .... The intersection of these sets of intervals is the set of in-

tervals (nn/2, (n + l)n/2) for n = 0, ±1, ±2, .... Since 2 E (n/2, n), the

initial value problem c. has a unique solution on the interval ( n /2, n).

EXERCISES 2.2

In exercises 1-6 find all points ( c , d) where solutions to the initial

value problem consisting of the given differential equation and the

initial condition y(c) = d may not exist.

l.

3y

y' = +e-x 2. Y I = (x2 XY + y2)

(x-5)(x+3)

3. y I = -^1
   xy
   4. y' = ln(y -1)


4. y' = J(y + 2)(y - 1) 6. Y I = y-x y