Consider the following ODE:
$$f'(r) = f^2(r) + O \left( \frac{1}{r^4} \right)$$
as $r$ goes to infinity. The initial conditions are $f(1) = C <0$. 


What is the behaviour of a solution $f$ at infinity? (not only the leading term). 


This ODE is equivalent to: 
$$w'' + O \left( \frac{1}{r^4} \right) w = 0$$


where $f = -\frac{w'}{w}$. 


If I replace $O \left( \frac{1}{r^4} \right)$ with just $\frac{1}{r^4}$, then I know that the 2 linearly independent solutions are $f(r) = -\frac1r -\frac{1}{r^2} \tan{\frac1r}$ and $f(r) = -\frac1r + \frac{1}{r^2} \cot{\frac1r}$


Does that mean that the solution will always be $f(r) = -\frac1r + O \left( \frac{1}{r^3} \right)$ maybe for some family of initial conditions?


Any help is appreciated. If there is a reference that does things like that in detail, please share it with me. 



Asymptotic Behaviour of Solutions to a Riccati-type ODE with Small Forcing Term
In your example you obtained two linearly independent solutions with different behavior: $\cot(1/r)\sim r,\; r\to\infty$, so your second solution is $O(r^{-2})$.


This is the general pattern if you assume that your perturbation is analytic at $\infty$. Write your equation as 
$$f'=f^2+q(r),\quad q(r)=r^{-4}\sum\_{0}^\infty a\_kr^{-k}.$$
When this is so, make the change of the variable $r=1/\zeta$, $y(\zeta)=w(1/\zeta)$ in your linear equation for $w$ ($f=-w'/w$)
and obtain
$$\zeta^2y''+2\zeta y'+\zeta^2(q\_0+\ldots)y=0.$$
The indicial equation $\rho(\rho-1)+2\rho=0$ has roots $0,-1$,
so solutions $y$ can be bounded or behave as $c/\zeta$ near $\zeta=0$.
Returning to your original equation, this means that some solutions $f$ are like 
$-1/r$, while others are like $O(r^{-2})$.


Which is the case for a particular initial condition is impossible to decide because your restriction $O(r^{-4})$ tells nothing about $q$ on a long finite interval. 


My argument shows that $O(r^{-4})$ can be relaxed to $O(r^{-3})$ with the same conclusion.


My assumption that $q$ is analytic at $\infty$ is of course too strong, and can be relaxed, depending on your needs, but I don't think it can be completely dropped.


