Consider $$y'=(y-1) (y-3)^{\frac{1}{3}} $$ with $y(0)=y_0$
Determine for which values of $y_0 \in \mathbb{R}$ the maximal solutions are unique and global.
My attempt:
First, I notice that the stationary solutions are $y(t)=1$ and $y(t)=3$. I can therefore have three possibilities:
$y_0 \in (1,3)$. Then $y'(t) <0$ and hence, by existence and uniqueness, I must have $$lim_{t \rightarrow \infty} y(t)= 1$$
$y_0 < 1$: In that case $y'(t)<0$ and hence I'd say that $lim_{t \rightarrow \infty} y(t)=-\infty$
$y_0 >3$: With a similar argument as above, I find that $lim_{t \rightarrow \infty} y(t)=+\infty$
I'm tempted to say that I always have global solutions... but that r.h.s does not seem to Lipschitz to me, so there must be some problems, but I don't know what to do honestly