Consider this ODE:
$$f_{i}’(r) = f_{i}^2 (r) + O\left(\frac{1}{r^4} \right)$$ As $r \to \infty$. $i=1,2$ and the initial conditions are $f_{i} (1) = C_{i} < 0$
This is equivalent to the ODE: $$w_i’’(r) + O\left(\frac{1}{r^4} \right) w_i(r) = 0$$ Where $f_i = -\frac{w_i’}{w_i}$.
Can we find the behavior at infinity of solutions to this ODE? If I replace $O\left( \frac{1}{r^4} \right) $ with just $\frac{1}{r^4}$, I get that the two linearly independent solutions are $f_{i} = -\frac1r - \frac{1}{r^2} \tan \left(\frac1r \right)$ and $f_{i} = -\frac1r + \frac{1}{r^2} \cot \left(\frac1r \right)$. Does that the mean the solution will always be $-\frac1r + O \left(\frac{1}{r^3} \right)$? (Ignoring initial conditions).
Now consider this system of ODE for the three functions $(\Theta, g_1, g_2)$: $$\Theta’(r) (f_2 - f_1) = O \left(\frac{1}{r^4} \right)$$ $$g_{1}’= -f_1 g_{1} + \Theta’(r) g_{2}$$ $$g_{2}’= -f_2 g_{2} - \Theta’(r) g_{1}$$
I want to find the behaviour of $g_1$ and $g_2$ at infinity (given initial conditions) and not only the leading term. How do I even begin? Any help is appreciated. Also, if you know any reference where I can read more about issues like that, please share it with me.