44 Ordinary Differential Equations
EXAMPLE 1 An Initial Value Problem for which the Interval of
Existence Depends on the Initial Condition
Consider the initial value problem(2) y' = y^2 ; y(O) = d.
Let R b e any finite rectangle containing the point (0 , d), which li es on the
y -axis. The function f(x, y) = y^2 is continuous on R. So by the fundamental
existence theorem there exists a solution to the IVP (2). In this case, we can
easily verify that
(3)1y(x) = K - x'
where K is an arbitrary constant, is a one-parameter family of solutions of
the differential equation y' = y^2. Differentiating (3), we find
1 2
y'(x) = (K - x)2 = Y (x)for any K So, indeed, y(x) = l/(K - x) is the solution of y' = y^2. Notice
that y( x) is defined, continuous, and differentiable for all real x ~ K. That
is, y(x) = l/(K - x) is defined, continuous, differentiable, and the solution
of y' = y^2 on the intervals (-oo,K) and (K,oo). To solve the IVP (2), J{
must be chosen to satisfy the initial condition y(O) = d = 1/ K. Consequently,
K = 1/ d; and, therefore, the interval of existence depends solely on the initial
valued and not on the function f(x, y) = y^2.
Notice that when d = 2, the solution of the IVP y' = y^2 ; y(O) = 2 on the
interval (-oo, 1 /2) is
1
y(x) = -1 - ·
2-x
Since the interval of existence specified in the fundamental existence theorem
is symmetric with respect to the initial point <;, which in this example is the
point 0, the value of h specified in the theorem must of necessity be less than
or equal to 1 /2 when d = 2. Likewise, for arbitrary d > 0 the solution of
the IVP (2) exists on the interval (-oo, l/d); and , therefore, the value of h
specified by the existence theorem must be less than or equal to 1 / d. The
direction field for y' = y^2 and the solution of the IVP y' = y^2 ; y(O) = 2 are
displayed in Figure 2.6.