I'm interested in examining the asymptotic behavior of Riccati equations of the form
$$ y'(x) = f(x) + g(x) y^2(x) $$
for $x \to \infty$. I've done some digging but I can't seem to find a simple explanation of how to go about computing something like this. If that's too general, then take this example
$$ y'(x) = 10x e^{-2x} - \frac{10}{x^2} y^2 $$ It seems "obvious" just by looking at this that as $x \to \infty$ we have $y'(x) \to 0$ so $y \to \text{constant}$. Some numerical simulations have shown that $y \sim 0.17$ for large $x$ but is it possible to show this (most importantly, the actual value of the constant) directly from the equation?
A quick answer would be that this highly depends on $f$ and $g$, but that's rather obvious. In this case, there is a clear difference between the rates of convergence with which $f$ and $g$ tend to zero. While $f$ converges exponentially fast to zero, $g$ approaches zero in an algebraic fashion, i.e. much slower. Therefore, if you take $x$ large enough, the ODE can be approximated by \begin{equation} y'(x) = -\frac{10}{x^2} y(x)^2 + \mathcal{O}(e^{-\alpha x}) \end{equation} for any $0<\alpha<2$. The solution of this ODE is \begin{equation} y_\text{far}(x) = \frac{x}{\frac{1}{y_\infty}x - 10}, \end{equation} where $\lim_{x \to \infty} y_\text{far}(x) = y_\infty$. So, from the behaviour of the ODE for large $x$, we cannot readily infer its limiting value.
For small $x$, $f$ can be approximated as \begin{equation} f(x) = 10 x-20 x^2 + \mathcal{O}(x^3) \end{equation} Obviously, $g(x)$ is (again) the dominant term in the ODE, since $g$ diverges as $x \to 0$. However, neglecting the contribution of $f$ would give again the solution $y_\text{far}$, which converges to zero as $x \to 0$. Moreover, the divergence of $g$ implies that any solution $y$ which has bounded derivative should converge to zero at least as quickly as $x^2$.
If you're only interested in the asymptotic behaviour of solutions as $x \to \infty$, I can imagine that you don't care so much if $y$ converges at all if $x \to 0$. In that case, if you take an initial condition which is large enough (in this case taking $x_0 > 3$ would suffice), you can get a good approximation for $y_\inf$ by calculating \begin{equation} y_0 \approx y_\text{far}(x_0) = \frac{x_0}{\frac{1}{y_\inf}x_0 - 10}, \end{equation} so \begin{equation} y_\inf \approx y_0 \frac{1}{1+10 \frac{y_0}{x_0}}. \end{equation}