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Example 4 - Nonuniqueness

y' = 3y2/3, y(0) = 0

has infinitely many solutions. The graphs split apart, as shown in Figure 14.4.6.


Figure 14.4.6 Example 4

One solution is the constant solution y(t) = 0. A second solution is found by separating variables:

⅓y-2/3 dy = dt,

y1/3 = t + C,       C = 0

y = t3.

This solution can be checked by differentiation:

y' = 3t2 = 3(t3)2/3 = 3y2/3.

The other solutions go along the line y(t) = 0 in some interval and branch off the line y(t) = 0 to the right and left of the interval. The full list of solutions is:


Here a is either a real number or -∞, b is either a real number or +∞, and a ≤ 0 ≤ b. In the case that a = -∞ and b = +∞, the solution is the constant function y(t) = 0.

These solutions all have the same initial value y(0) = 0. The Uniqueness Theorem does not apply in this example because the function f(t, y) = 3y2/3 has no derivative at y = 0, so that f(t, y) is not smooth at y = 0.

Last Update: 2006-11-16