I'm studying the following ODE (for those interested, it is a Klein-Gordon eq with $\partial_t =0$ for a test field around a black hole). $$ \partial_x (x(x+2)\partial_x \phi) = x(x+2)\ddot{\phi} + 2(x+1)\dot{\phi} = \tilde{m}^2 (x+2)^2 \phi .$$ For $\tilde{m}=0$ I found the general solution $$ \phi = A \ln \left(\frac{x}{x+2} \right) + B,\quad \quad \quad \tilde{m}=0,\quad A,B \in R$$ which diverges at $x=0$, unless one takes the constant solution.
Back to the full first equation, I'm interested in finding non-diverging solutions. I try study what happens near $x=0$ now by guessing $\phi \sim x^{\alpha}$ in this neighborhood
$$ x(x+2) \alpha (\alpha -1) x^{\alpha-2} + 2(x+1) \alpha x^{\alpha-1} = (x+2)^2 \tilde{m}^2 x^{\alpha}$$
The lowest power of $x$ dominates near $x=0$. Notice the term $\propto \tilde{m}^2$ does not contribute.
$$\implies 2\alpha (\alpha -1) + 2\alpha = 0 \implies \alpha =0$$
As the $\tilde{m}^2$ term drops out this method has recaptured the $\phi = B$ solution above (there is no sign of the log solution as this does not have a Laurent series). This gives me hope that $\exists$ a solution which is near constant near the horizon but receives corrections from the $\tilde{m}$ term for $x>0$. (Is this not justified?)
With this hope I explore a perturbative solution of the form $$\phi = B + \tilde{m}^2 \phi_1 \quad \quad \quad \tilde{m}\ll1 $$ plugging in $$\implies \partial_x (x(x+2)\partial_x \phi_1) = (x+2)^2 B $$ $$\implies \partial_x \phi_1 = \frac{B}{3} \frac{(x+2)^2}{x} + \frac{C}{x(x+2)}$$ $$\implies \phi_1 = \frac{B}{3} \left( \frac{1}{2}x(8+x) + 4 \ln x \right) + C\ln\left( \frac{x}{x+2}\right) +D$$ So $\phi_1$ diverges at $x=0$, breaking the perturbative solution. I'm confused, From further above it seemed like the term $\propto \tilde{m}$ should not drastically affect the solution at $x=0$, yet when I look for an effect of a small $\tilde{m}$ I get something that diverges at $x=0$.
What part of my reasoning is wrong? Can I still hope for a solution well-behaved at $x=0$?