Problem related to DE Solve and comment on the solution behaviour at $|x|$ approaches infinity (bounded or unbounded as $x$ approaches $\pm\infty$.
$$\frac{1}{1+x^2} + \sin y + y'\left(x\cos y + \frac{y^2}{2}\right)=0,\quad y(1)=\Pi.$$
Can I have some hints on how to get started on this problem? Thanks
I cannot even separate it.

Deal with the LHS of the ODE and we can obtain the solution
\begin{equation}
\frac{1}{1+x^2}+(\sin(y)+y'x\cos(y))+y'\frac{y^2}{2}=0\\
\frac{1}{1+x^2}+(x\sin(y)+\frac{y^3}{6})'=0\\
\int(x\sin(y)+\frac{y^3}{6})'dx+\int\frac{1}{1+x^2}=0\\
x\sin(y)+\frac{y^3}{6}+\arctan(x)+C=0
\end{equation}
which $C$ is a constant given by the initial condition that $y(1)=\pi$
\begin{equation}
\frac{\pi^3}{6}+\frac{\pi}{4}+C=0\\
C=-(\frac{\pi^3}{6}+\frac{\pi}{4})
\end{equation}
Rewritten the solution by the following form:
\begin{equation}
\frac{y^3}{6}=-x\sin(y)-\arctan(x)-C
\end{equation}
If $|x|\rightarrow+\infty$, we found that the $\arctan(x)$ is bounded, and $x\sin(y)$ is unbounded. So we can claim that $y(x)$ is unbounded when $|x|$ going to infinity.
