Introduction 25EXAMPLE 6 An Initial Value Problem with an Infinite Number
of Piecewise Defined SolutionsOne solution on the interval ( -oo, oo) of the initial value problem(15) y' = 3y2f3; y(O) = 0
is clearly y 1 (x) = 0. A second solution is y 2 (x) = x^3. (Verify this fact.) As
a matter of fact, there are an infinite number of solutions to this initial value
problem. In exercise 6 at the end of this section, you are asked to show that
Yab(x)= 0, a:S:x:S:b!
(x - a)^3 , x <a:::; 0(x-b)^3 , O:S:b<x
is a solution on ( -oo, oo) for every choice of a :::; 0 and b ;:::: 0. This example
illustrates that if an initial value problem has a solution, the solution may not
be unique.
In the following three examples of boundary value problems, we will use
the same differential equation and vary only one boundary condition. First,
we verify that a two-parameter family of solutions of the differential equation
y" + y = 0 is y = Asinx + Bcosx, where A and Bare arbitrary constants.
Differentiating y = A sin x + B cos x twice, we find
y' = A cos x - B sin x and y" = -A sin x - B cos x.
Substitution into the differential equation, yields
y" + y = (-A sin x - B cos x) + (A sin x + B cos x) = 0
for all real x and for all A and B. Since y , y', and y" are all defined on
( -oo, oo), the function y = A sin x + B cos x is a solution of y" + y = 0 on the
interval (-oo, oo ). Suppose in addition to satisfying y" + y = 0, we require
that y(O) = 0. Imposing this condition, results in the equation
y(O) = 0 = AsinO + BcosO = B.
Thus, B = 0 and A is arbitrary. Consequently, the solution of the differential
equation y" +y = O which satisfies the condition y(O) = 0 is the one-parameter
family of functions y =A sin x. A graph of this family on the interval (-5, 5)
is shown in Figure 1.4.