The Initial Value Problem y' = f(x, y); y(c) = d 47
I EXAMPLE 2 The IVP y' = x.jY; y(c) = d
Analyze initial value problems of the form(5) y' = x.JY; y(c) = d.SOLUTION
In this example, the function j(x, y) = x-JY is defined and a continuous
function of x and yon any finite rectangle in the xy-plane where y ;::: 0. Thus,
by the fundamental existence theorem, for every point (c, d) where d ;::: 0,
there is some interval I with center c in which the initial value problem (5)
has a solution. Calculating the partial derivative off with respect to y, we
find fy(x, y) = x/(2-JY). The function fy(x, y) is defined and continuous for
y > 0. Thus, by the fundamental existence and uniqueness theorem, when
d > 0 the solution of the IVP (5) is guaranteed to be unique. Hence, the
solution of the initial value problem
(6) y' = x.JY; y(l) = 2
is unique on some interval with center 1. However, the solution to the initial
value problem
(7) y' = x.JY; y(l) = 0
may or may not b e unique, since d = 0. In this case, we easily find the solution
of the IVP (7) is not unique, since both y 1 (x) = 0 and y 2 (x ) = (x^2 - 1)^2 /16
are solutions of (7).
The fundamental existence and uniqueness theorem is called a local the-
orem, because the solution is guaranteed to exist and be unique only on a
"small" interval. The following theorem, which we again state without proof,
is call ed a continuation theorem.
CONTINUATION THEOREMIf f(x, y) and fy(x, y) are both continuous functions of x and yin a finite
rectangle Rand if (c, d) ER, then the solution of the initial value problem
(8) y' = f(x, y); y(c) = d
can be ext ended uniquely until the boundary of R is reached.