Let $p(t)$ be a polynomial with no real roots. I have to prove that the following ODE-IVP: $$\begin{cases} y'=\frac{1}{p(xy)}=f(x,y) \\ y(0)=0 \end{cases} $$
Has a unique solution in $\mathbb{R}$:
$$f_y(x,y)=-\frac{p'(xy)x}{(p(xy))^2}$$
I tried to show that $f_y(x,y)$ is bounded at $J\times(\infty,\infty)$ for all closed intervals $J$, but found out that when $p(t)=t^2+t+1$, $$f_y(x,y)=-\frac{2x^2y+x}{((xy)^2+xy+1)^2},$$
and in that case, $f_y(x,y)$ is not bounded for small values of $y$.
Maybe I should use the Lipschitz condition?
Thanks.
As $p$ is a polynomial without a real root, $f(x,y)$ is locally uniformly Lipschitz continuous in $y$ and continuous in $x$. Therefore, the IVP has a unique maximal solution satisfying $y(0)=0$ according to Picard-Lindelöf theorem.
Remains to prove that the solution is global on $\mathbb R$. For this, just remind that if a solution has a bounded maximal interval, it explodes in this interval. It’s not the case here as for all $x \in \mathbb R$
$$\vert y(x) \vert \le \frac{\vert x \vert}{k}$$ where $k \gt 0$ is such that $\vert p(t) \vert \gt k$ for all $t \in \mathbb R$. The existence of $k$ is a consequence of the fact that $p$ has no real root.