Nonlinear differential equation with sine function

Question

If $y\in C^1(\mathbb{R})$ and $y'(x) =\sin(y(x) +x^2)$ for every $x\in\mathbb{R}$ with $y(0)=0$ I proved that $y$ is smooth and that $y'(0)=y''(0)=0$ and that $y'''(0)>0$ but how can I prove that $y>0$ in $(0,\sqrt{\pi}) $ and $y<0$ in $(-\sqrt{\pi}, 0)$?

Answer 1

Hint: The constant function $z(x) = 0$ satisfies $z'(x) < \sin (z(x) + x^2)$ for every $x\in (-\sqrt{\pi}, 0) \cup (0, \sqrt{\pi})$, hence it is a sub-solution on those intervals.