We are given the differential equation $\large\frac{\textrm dy}{\textrm dx}(x)=\frac{1}{3y(x)+x^2}+\frac{2}{4+x^2}$ and its directional field (see image below).
If this equation is solvable in closed form, I suspect it is not an easy solution and Wolfram Alpha confirms this idea by not giving such a solution.
We are asked:
$\lim\limits_{x\to \infty}y(x)$ for the solution with $y(0)=-1$ and the solution with $y(0)=1$. For both of these I simply look at the directional field and follow the lines. So for $y(0)=-1$, I guess $\lim\limits_{x\to \infty}y(x)=-\infty$ and for $y(0)=1$, I guess $\lim\limits_{x\to \infty}y(x)=3$.
for what initial conditions the solution $y(x)$ is bounded for all $x\in\mathbb{R}$. Once again I look at the directional field and guess that all solutions with initial conditions under or on the parabola that we see aren't bounded. So I say that all solutions with $y(0)>0$ are bounded.
The way I have "solved" these problems does not feel very nice and I am not sure it is right. I would like to hear if there are ways I could find the things I am asked using the differential equation, without relying as much on the directional field. I would also appreciate suggestions on how to find the solution which corresponds to the parabola (I'm not sure of course it's a parabola, but I think so) that we see.