Prove solution of a ODE is bounded I am trying to prove the following statement:
"Let $f:\mathbb{R} \rightarrow \mathbb{R}$ of class $C^1(\mathbb{R})$ verifying that $f(m)=0 \, \, \forall m \in \mathbb{Z}$. Prove that all solutions of the ODE $y'=f(y)$ are bounded and defined in all $\mathbb{R}$."
I know I need to show that $|y|\leq M$ with $M$ a constant but I don't know how to work with the ODE to achieve this. I also looked for inspiration and found this question Proving all solutions of $y'+y=f(x)$ are bounded . Thanks for your attention.

As noticed in the comments, it suffices to observe that if $y$ is not a constant solution (i.e. $f(y(0)) \neq 0$) then its graph cannot intersect any constant solution (by unicity). Therefore $y(x) \not\in \mathbb{Z}$ for all $x$ and so by continuity $\lfloor y(0)\rfloor < y(x) < \lfloor y(0)\rfloor+1$.
Global existence follows by boundedness.
