Let $y(x)$ a $C^1$ function that solves the ODE
\begin{cases} y'(x)= \sin(y(x)+x^2)\ \\ y(0)=0 \end{cases}
i) Show that $y'(0)=y''(0)=0$, while $y'''(0)=2>0$
ii) Show that $y(x)>0$ for every $x \in (0,\sqrt{\pi})$, while $y(x)<0$ for $x \in (-\sqrt{\pi},0)$.
My attempt:
i) I just wrote down the derivatives, I don't wrote the computations as they're easy.
ii) First, I compute the stationary solutions, i.e.
$$\sin(y(x)+x^2) = 0$$ therefore $y(x)=0$ and $y(x)=\pi-x^2$ are the stationary solutions. From $y''(0)=0$ I have that in $x=0$ the solution changes concavity.
Also, from existence and uniqueness, if in a right neigbourhood of $0$ the solution is positive, it will be still positive until $x= \sqrt{\pi}$ since at this point there's the intersetion of the parabola with the $x$-axis.
I don't know how to use the fact that $y'''(0)>0$ in order to prove that in a right neigbourhood of $0$ the solution is positive
How should I move?