Non-uniqueness of solutions of an ordinary differential equation

Question

For instance, consider the following initial value problem $$x'=3x^{2/3}, \ \ \ \ x(0)=0$$

This initial value problem has infinitely many solutions given by $$ x(t) = \begin{cases} 0 & t<c \\ (t- c)^3 & t \geq c \end{cases} \ \ \ \ \ (*) $$ for any non-negative $c$. Picard's existence theorem says that the initial value problem $x'=f(x,t), x(t_0)=x_0$ has a unique solution if $f(x,t)$ is continuous at $(x_0,t_0)$ and Lipschitz continuous in $x$. So the non-uniqueness comes from the fact that $3x^{2/3}$ is not Lipschitz at $x=0$. But I was wondering why a student taking a basic undergrad course in differential equations is not able to get the totality of solutions (*) ? i.e. one proceeds like this $$ \frac{1}{3}x^{-2/3}dx=dt \ \ \ \Rightarrow \ \ \ \ x^{1/3}=t+C $$ applying the initial condition we get $C=0$. So $$x(t)=t^3$$
Thanks

Answer 1

In the derivation that the undergraduate student has found (which is sometime called separation of variables) they are assuming that $x\neq 0$. In fact the solution they find: $$ x(t) = (t+C)^3 $$ is valid and unique in every point where $x(t)\neq 0$. The same undergraduate student, before dividing the equation by $x$, would check that they are not losing solutions. And they find, that $$ x(t) = 0 $$ is in fact a solution.

Now they notice that solutions of the first kind can touch the solution $x=0$ and when this happens the derivatives of the two solutions agree. Hence they can merge the solutions together to get the general solution you wrote at the beginning.