The IVP, $x'(t)=x^{2/3};x(0)=0$ in an interval arount $t=0$ has a The IVP, $\dot x(t)=x^{2/3};x(0)=0$ in an interval arount $t=0$ has a
(a)No solution
(b)Unique solution
(c)Finitely many solutions
(d)Infinitely many solutions.
Solution:- Applying Picard's -Lindelof Uniqueness and existence theorem. $f(x,t)=3x^{3/2}$, Will it be continuous at $(0,0)$? When $x<0$, $f(x,t)$ no more real valued. So, Discontinuous at $(0,0)$. So, I can not judge from the theorem. I solved using variable separable form.
I got the solution, $3x(t)^{1/3}=t+c$. When I apply initial condition, I get $3x(t)^{1/3}=t$. A unique solution. But in the answer key it is written that equation has infinitely many solutions. How it is possible? Please explain.
 A: I'm reading the given ODE as $x'=|x|^{2/3}$, so that it is defined in the full $(t,x)$-plane. The standard separation of variables technique gives the solutions
$$x_c(t)={1\over27}(t-c)^3\qquad(-\infty<t<\infty),\qquad c\in{\mathbb R}\ .\tag{1}$$
Furthermore it is obvious that $x_*(t)\equiv0$ is a solution as well; but it is at first sight unclear what it has to do with $(1)$. Now the RHS $f(t,x):=|x|^{2/3}$ of the given ODE does along the line $x=0$ not fulfill the  assumption that $f$ should be locally Lipschitz with respect to the variable $x$. As a consequence IVPs starting at points $(t_0,0)$ may have several solutions, and this is indeed the case here. 
The IVP $x'=|x|^{2/3}$, $\>x(0)=0$, so far has the solutions $x_0(t)={1\over27}t^3$ and $x_*(t)\equiv0$. But we may as well consider
$x_+(t)=x_0(t)$ $(t>0)$ and $=0$ otherwise, and similarly $x_-(t)=x_0(t)$ $(t<0)$ and $=0$ otherwise, consider as  solutions. In my view there are just these four solution germs; but in a more liberal counting you can also count the functions $x_{c, +}(t):=x_c(t)$ $(t>c)$ and $=0$ otherwise ($c>0)$, and similarly for $c<0$, as new solutions of the given IVP. In this way you would obtain an infinity of solutions. All solutions described here are $C^1$ (continuously differentiable) but the patched ones are not twice differentiable at the patching point.
A: As you say, you have the solution $x(t)=t^3/27$. You also have the solution $x=0$. As for have $x'(0)=0$, you can patch them at the origin, which gives you four solutions: $x=0$, $x(t)=t^3/27$, and 
$$
x(t)=\begin{cases} t^3/27,&\ x\geq0 \\ 0,&\ x<0\end{cases}
\ \ \ \ \text{ and } \ \ \ x(t)=\begin{cases}0,&\ x\geq0\\  t^3/27,&\ x<0 \end{cases}
$$
Besides these four solutons, all I imagine is that you can patch at other points, with $(t+c_1)^3/27$ on one side and $(t+c_2)^3/27$ on the other, at the cost of your function not being differentiable at the patching point. 
