114 Ordinary Differential Equations
we see that a solution exists as long as (x, y(x)) remains in the first or third
quadrant or y(x) = 0. Furthermore, the solution is unique as long as (x, y(x))
remains in the first or third quadrant. Once the solution, y(x), reaches the
x-axis- that is , once y(x) = 0- the solution may no longer be unique.
a. The point ( -1, 0), which corresponds to the initial condition y( -1) = 0,
is on the x-axis so the solution to the IVP (29a) exists, but the solution
may not be unique. Notice that y(x) = 0 is a solution of (29a). Verify
that for all r:::; -1 and alls::'.:'. 0
-[(-x)^312 - r]2 /9, r:::; -1
Yrs(x) = 0 -1 < x < 0
(x^312 -s)^2 /9 o:::;s
is a solution of the IVP (29a). Hence (29a) has an infinite number of
so lutions on ( -oo, oo).
b. Since (-1, -1 /9) is in the third quadrant, the IVP (29b) has a unique
solution until x , -oo, x , o-, y , -oo, or y , o-. On the interval
(-oo,O) the unique solution to the IVP (29b) is y(x) = x^3 /9. (Verify
this fact.) At x = 0, y(x) = 0 and the so lution may no longer be unique.
Show that for all u ::'.:'. 0
l
x3/9
Yu(x) = 0 :
(x3/2 - u)2 /9 '
O<x<u
u:::;x
is a solution of the IVP (29b). Hence, (29b) has an infinite number of
solutions on (-oo, oo ).
Numerical Solution
We generated numerical approximations to the IVPs (29a) and (29b) using
both MAPLE and SOLVEIVP.
a. The initial point (-1, 0) is on the x-axis and as we saw in the math-
ematical analysis section, there are an infinite number of solutions to
the IVP (29a). SOLVEIVP generated the so lution y(x) = 0. The soft-
ware gave us no indication that this is not a unique solution. MAPLE
indicated floating point overflow no matter how small the interval of
integration was chosen. This result would often lead one to assume the
IVP (29a) has no solution.