General Existence and Uniqueness of ODE I am trying to make sure that I understand the following question. As well as I am having a bit of trouble understand the partial solutions given.
The question is basically, what can we say about the following IVP ?
$$ y'=y^{1/3}$$ with initial value, $y(0)=0$ for $t \ge 0$
I know that the important theorem to be aware of is the theorem about unique solutions depending on continuity of the partial derivative w.r.t to y and existence depending on the continuity of f(t,y) itself.
So what I am seeing initially is that the partial derivative (w.r.t to y) is continuous everywhere expect at $y=0$. and  $f(t,y)$ itself is continuous for all values of $t$.
So does this Imply that we have a solution for all values of t, but we will only have a unique solution when $y \neq 0 $ ?
Solving, using the fact that we have a separable equation gives me, 
$$y^{-1/3}dy=dt$$
Integrating both sides,
$$\frac{3}{2}y^{2/3}=t+c$$
$$y^{2/3}=\frac{2}{3}(t+c)$$
$$y^2=[(2/3)(t+c)]^{3}$$
$$y= \pm [(2/3)(t+c)]^{3/2}$$
From our initial conditions we can solve easily for c, to get that $c=0$
that is $y=\pm (2/3)t^{3/2}$
Now here is my first point of confusion, the solution says for both these answers, $t \ge 0$  I think I am just lost in it and even though I understood what the theorem stated, I don't understand its implications.
As well, from initial observation we know $y=\psi(t)=0$ is also a solution.
And the final answer is that $y= \lambda (t)= 0$ if $0 \le t \le t_{o}$ and $\pm[(2/3)(t-t_{o})]^{3/2}$ if $t \ge t_{o}$ Is the reason we have that the second part is only for $t \ge t_{0}$ just to avoid taking square root of negatives?
I am still confused about the final answer. I understand how we found that we could have $\pm [(2/3)(t+c)]^{3/2}$ and y=0 but what I am not understand is how from that we can obtain y=0 if $0 \le t \lt t_{o}$ and $\pm [(2/3)(t-t_{o})]^{3/2}$  if $t \ge t_{o}$? How was this done? It just directly stated this. Mainly, how is it that it went directly to saying 
Can someone please help tie this together for me? Thanks a lot 
 A: Your equation is of the form $y' = f(t,y)$ with $f(t,y) = y^{1/3}$.
Theorem (Short time existence): If $f(t,y)$ is continuous at $y_0$, then $y' = f(t,y)$, $y(t_0) = y_0$ has a solution on some interval containing $t_0$.
Your equation satisfies this.
Theorem (Short time uniqueness): If $f(t,y)$ is continuous at $y_0$ and $\frac{\partial f}{\partial y}$ exists and is continuous at $y_0$, then $y' = f(t,y)$, $y(t_0) = y_0$ has a unique solution on some interval containing $t_0$.
Your equation doesn't satisfy this for $y_0 = 0$, because $\frac{\partial}{\partial y} y^{1/3} = \frac{1}{3} y^{-2/3}$ does not exist at $y = 0$. So you don't have uniqueness of solutions starting with $y_0 = 0$.
If you start at some $y_0 \neq 0$, you do have uniqueness of solutions for a short time at least, in other words, on some time interval containing $t_0$. After a while, you might reach $y = 0$ on your solution, and uniqueness would fail. This theorem doesn't say anything in particular about long time uniqueness.
(Note: There are weaker conditions to put on $f$ in the theorems, but these are simple and work.)

Some intuition for what's going on:
If you sketch a slope field for $y' = y^{1/3}$, you'll note that there is a constant solution at $y = 0$. Often in such a case, there will be nearby solutions that asymptote to this solution. But in this case, $\frac{\partial f}{\partial y} = \frac{1}{3} y^{-2/3}$ (the rate at which the slope is changing) gets too large near $y = 0$, and solutions that start negative actually reach this constant solution.
Once that happens, uniqueness is shot: you can switch from the nearby solution $y = \left(\frac{2}{3} t\right)^{3/2}$ that has reached $y = 0$ to the $y = 0$ solution.
In fact, there are uncountably infinitely many solutions: you can stay on $y = 0$ for any distance and then start on one of the $\pm(2/3)^{3/2}(t+c)^{2/3}$ curves.

Note on weaker conditions:
As mentioned in the section above on intuition, the problem was related to the fact that the derivative $\frac{\partial f}{\partial y}$ becomes unbounded (arbitrarily large) near $y = 0$.
A weaker condition in the uniqueness theorem: $f(t,y)$ is Lipschitz in $y$. That is, there's some constant $C$ such that $|f(t,y_1) - f(t,y_2)| \leq C |y_1-y_2|$. (In fact you only need locally Lipschitz, which I'll leave to you to look up.) This really gets at the issue: $f$ should not be allowed to change too quickly, in the sense that secant slopes $\frac{|f(t,y_1) - f(t,y_2)|}{|y_1-y_2|}$ should be locally bounded. The existence and continuity of the derivative guarantees this, so it's a nice simpler-to-state condition to use in the theorem.
A wiki reference to the relevant theorem: Picard-Lindelöf theorem.

Note on why there are uncountably many solutions to $y' = y^{1/3}$, $y(t_0) = 0$:
$y = 0$ is a solution, as you noticed
$y = \pm[(2/3)(t-t_0)]^{3/2}$ is a solution, as you noticed
But so is
$y = 0$ for $t < T$ and $y = \pm[(2/3)(t-T)]^{3/2}$ for $t \geq T$ (for any $T \geq t_0$, so that $y(t_0) = 0$)
and so is
$y = 0$ for $t > T$ and $y = \pm[(2/3)(t-T)]^{3/2}$ for $t \leq T$ (for any $T \leq t_0$, so that $y(t_0) = 0$)
and so is
$y = \pm[(2/3)(t-S)]^{3/2}$ for $t \leq S$ and $y = 0$ for $S < y < T$ and $y = \pm[(2/3)(t-T)]^{3/2}$ for $t \geq T$ (for $S \leq t_0 \leq T$, so that $y(t_0) = 0$)
Why are these last three solutions as well?
The equation $y' = f(t,y)$ is an equation about the derivative of $y$. Both $y = 0$ and $y = \pm[(2/3)(t-T)]^{3/2}$ have the same derivative, namely zero, at $y = T$, so the piecewise function is differentiable with derivative zero there as well, so it must also be a solution to $y' = y^{1/3}$.
In fact, if you think carefully about the fact that the solutions are (on short time scale) unique away from $y = 0$, you'll see these are a complete list of the solutions.
Another comment on this:
If two solutions meet, they must have the same slope, because the equation $y' = f(t,y)$ determines the slope based on $t$ and $y$. So if two solutions meet (which precisely means you don't have uniqueness), you can always switch from one to the other as well.
The equation $y' = y^{1/3}$ is a very special situation where we have a whole line $y = 0$ of cases where uniqueness is violated, and this line is also itself a solution. In this case we get this uncountable family of solutions.
