# Problem with Nonunique Solutions

Hello all,

I am just having a problem understanding a problem in my textbook on nonunique solutions. Let me get to the problem:

So, consider the initial value problem:

$$y' = y^{1/3}\mbox{, } y(0) = y_0 = 0$$
$$\mbox{for t}\geq 0$$

So, solving for the differential equation, I get:

$$y = [\frac{2}{3}(t + C)]^{3/2}$$

So, satisfying initial condition, $$0 = [\frac{2}{3}(0 + C)]^{3/2}$$

So, $$C = 0$$

$$y = [\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

So, that's all understandable to me.
But the answer in the book goes on to say that:

$$y = -[\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

is also a solution. And:

$$y = 0$$
, for $$t\geq 0$$

is also a solution. Finally, the answer says you can generalize the solultion to:

$$y = \chi (t) =\left\{\begin{array}{cc}0,&\mbox{ if } 0\leq t< t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, & \mbox{ if } t\geq 0\end{array}\right$$

This last part is very confusing for me. If someone could explain it, it would be very helpful. For example, if the value of $$t_0$$ was given, and it followed the generalization above, wouldn't values for $$0 \leq t < t_0$$ not be 0, but instead be undefined? Since, you can't do $$(negative number)^{3/2}$$
Right?

Thanks in advance for all the help.
-mk

HallsofIvy
Homework Helper
mkkrnfoo85 said:
Hello all,

I am just having a problem understanding a problem in my textbook on nonunique solutions. Let me get to the problem:

So, consider the initial value problem:

$$y' = y^{1/3}\mbox{, } y(0) = y_0 = 0$$
$$\mbox{for t}\geq 0$$

So, solving for the differential equation, I get:

$$y = [\frac{2}{3}(t + C)]^{3/2}$$

So, satisfying initial condition, $$0 = [\frac{2}{3}(0 + C)]^{3/2}$$

So, $$C = 0$$

$$y = [\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

So, that's all understandable to me.
But the answer in the book goes on to say that:

$$y = -[\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

is also a solution.
Yes, it is.
$$(-[\frac{2}{3}t]^{3/2})= -\frac{2}{3}[\frac{3}{2}t^{\frac{1}{2}$$
while $$(-[\frac{2}{3}t]^\frac{3}{2})^\frac{1}{3}]$$ is exactly the same thing. Also, of course, y(0)= 0.

And:

$$y = 0$$
, for $$t\geq 0$$

is also a solution.
Absolutely: (0)'= 0 and (0)1/3= 0.

Finally, the answer says you can generalize the solultion to:

$$y = \chi (t) =\left\{\begin{array}{cc}0,&\mbox{ if } 0\leq t< t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, & \mbox{ if } t\geq 0\end{array}\right$$

This last part is very confusing for me. If someone could explain it, it would be very helpful. For example, if the value of $$t_0$$ was given, and it followed the generalization above, wouldn't values for $$0 \leq t < t_0$$ not be 0, but instead be undefined? Since, you can't do $$(negative number)^{3/2}$$
Right?

Well, yes. Perhaps that is why they specifically said $$t\geq 0$$?

Thanks in advance for all the help.
-mk

You'er welcome.

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