How to find the lower and upper bound of a solution of a Cauchy problem?

Question

I am studying some differential equations especially the Cauchy problems, and I encountered this question:
\begin{equation} (E):\bigg\{\begin{split} y'=ysin^{2}(y)\\ y(0)=x_{0} \end{split} \end{equation} They supposed that $0<x_{0}<\pi$, and I have to find the upper and lower bounds of the solution, knowing that it is a maximal solution. I don't know where to start. I need some kind of help.

Answer 1

From $f(y)=y\sin^2(y)$ you know that $f(0)=f(\pi)=0$ with $f(y)>0$ inside that interval, and that this function is locally Lipschitz.

This is all you need to know to solve this task.