Differential Equation $dy/dx = y^{1/3}$ What is the difference between the number of solutions of the system $$\frac{dy}{dx} = y^{1/3}$$ with initial value $y(0)=0$ in one case and $y(0)=1$ in another case?
And why does this difference occur? 
I know there's a theorem about uniqueness of solutions of first order ODE (although my understanding of that theorem isn't too good). 
 A: The Cauchy problem $$\begin{cases} y'=\sqrt[3]y\\ y(0)=c\end{cases}$$
Has unique solution $y_c(t)=\frac{c}{\vert c\rvert}\sqrt{\left(\frac23t+\sqrt[3]{c^2}\right)^3}$ for $c\ne 0$ in the interval $(-\frac32\lvert c\rvert^{2/3},\infty)$. This may be found by observing that, by Cauchy-Lipschitz, a unique solution exists in some interval $I_c\ni 0$ which depends on $c$. Said solution $y$ must satisfy $y(t)\ne 0$ for all $t$ in some subinterval of $I$ (and therefore either $y'(t)>0$ or $y'(t)<0$ for all $t$ in that interval), which allows you to separate variables and calculate by substitution $$\int_0^t\frac{y'(x)}{\sqrt[3]{y(x)}}\,dx=\int_0^t 1\,dx\\ \frac23\sqrt[3]{y^2(t)}-\frac23\sqrt[3]{y(0)^2}=t\\ \frac23\sqrt[3]{y^2(t)}-\frac23\sqrt[3]{c^2}=t$$
The small details are just algebra/routine.
On the other hand, if $c=0$, there are infinitely many solutions in all intervals $(-\varepsilon,\varepsilon)$, and they are exactly the functions in the form $$y_0^{\delta,\pm}(t)=\begin{cases} 0&\text{if }t\le \delta\\ \pm\sqrt{(\frac23t-\frac23\delta)^3}&\text{if } t>\delta\end{cases}$$
where $\delta\in [0,\varepsilon]$ and and $\pm$ is just a choice of sign.
It is clear that these functions satisfy the ODE for $c=0$. On the other hand, by Peano's theorem at least one solution to the ODE must exist in any interval containing $0$. Now, let $y$ be any such solution. And let there be some $t_0>0$ such that $y(t_0)\ne 0$. Also, by continuity, we may choose $t_0\in(0,\delta)$ such that $y(t_0)$ is arbitrarily close to $0$. By the previous remark, $y$ must satisfy $y(t)=y_{y(t_0)}(t-t_0)$ for all $t>t_0-\frac32\lvert y(t_0)\rvert^{2/3}$ (which must necessarily be a positive number). Taking the limit as $t\to \delta=t_0-\frac32\lvert y(t_0)\rvert^{2/3}$, we see that $y(\delta)=0$. Also, since $y_c(t-\beta)\ne 0$ for all $t>\beta$, it is clear that $y(t)$ must be zero for all $0\le x<\delta$. For the same reason, $y(x)$ cannot be $\ne 0$ for any $x<0$ (otherwise $y(0)\ne 0$). Therefore indeed $y=y_0^{\alpha,\operatorname{sgn}(y(t_0))}$.

Added: Actually, the argument may be simplified as such: let $A(c)$ be our Cauchy problem $$\begin{cases}y'=\sqrt[3]y\\ y(0)=c\end{cases}.$$
Then:

*

*for $c\ne0$, the problem $A(c)$ has the unique solution on $\Bbb R$ (and all neighbourhoods of $0$) $$y_c(t)=\begin{cases}\frac c{\lvert c\rvert}\sqrt{\left(\frac23t+\sqrt[3]{c^2}\right)^3}&\text{if }t>-\frac32\lvert c\rvert^{2/3}\\ 0&\text{if }t\le-\frac32\lvert c\rvert^{2/3}\end{cases}$$


*for $c=0$, the set of solutions is the family of function which are either $0$ or in the form $$y_0^{\delta,\pm}(t)=\begin{cases}\pm\sqrt{\frac23 t-\frac23\delta}&\text{if }t>\delta\\ 0&\text{if }t\le \delta\end{cases}$$ for $\pm$ a choice of sign and $\delta\ge 0$.
The argument for (1) is the same as above for $t>\alpha=-\frac32\lvert c\rvert^{2/3}$. Then, you use the fact that, if by any chance $y_c(\beta)\ne 0$ for some $\beta<\alpha$, then $y_c(t)=y_{y_c(\beta)}(t-\beta)$ for all $t>\beta$, and in particular $y_c(\alpha)=y_{y_c(\beta)}(\alpha)\ne0$, against the hypothesis $\lim_{t\to \alpha} y_c(t)=0$.
This has the advantage that for (2) one immediately has that, if $y$ is a non-zero solution of $A(0)$ then necessarily $y(t)=y_{y(t_0)}(t-t_0)$ for all $t\in \Bbb R$ and for any $t_0$ such that $y(t_0)\ne 0$.

