Unique solution to an Initial Value Problem

All solutions are given by $$y(x) = begin{cases} 0 & x le L, \ (x-L)^3, & x gt L, end{cases} $$ where $L in [0, infty]$. (For $L=infty$, $y equiv 0$; thanks to John Hughes for catching this case missing in an earlier version of this answer.)

One way to see that (rough sketch): Suppose $y(x_0) > 0$ for some $x_0 > 0$ and solve (using your fancy theorems, if you want to – but separation of variables will do just fine) backwards in time. This works until you hit 0, from where you can only continue with $0$ since $y$ is increasing and nonnegative.

This is one where just guessing an answer helps. From the first equation, it sure looks as if $y = x^n$ might be a solution for some $n$, and doing out the algebra, it turns out that $y = x^3$ is actually a solution. But so is $y = 0$. So ... not a lot of uniqueness there.

Sometimes brute force instead of fancy theorems is the way to go.