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.