Consider the differential equation $y'' + p(x)y'+q(x)y = 0$. If $p(x), q(x) \to \infty$ when $x\to x_0$, then $x_0$ is a singular point of the DE. In particular, all of the textbooks I have checked define $x_0$ to be regular singular if $(x-x_0)p(x)$ and $(x-x_0)^2q(x)$ remain finite when $x\to x_0$.
First we have a DE $a(x)y'' + b(x)y' + c(x)y = 0$. By dividing by $a(x)$ we get the first form of the DE. If $x_0$ is regular singular then $a(x) = (x-x_0)h(x)$, i.e. the singularity can be removed. Of course when we divide by $a(x)$ we are dividing by $(x-x_0)h(x)$. Then when I check the condition $(x-x_0)p(x)=$ constant for $x_0$ to be regular singular, I am actually checking if $p(x_0)/h(x_0)$ remains finite.
One would think that the same analysis is valid with $q(x)$, i.e. checking that $q(x_0)/h(x_0)$ remains finite. But the definition of the criteria is actually stronger, as if in $q(x)$ there would be a denominator with $(x-x_0)$. Why do we need to check that $(x-x_0)^2q(x)$ remains finite instead of $(x-x_0)q(x)$?
Thanks a lot
The condition on $q(x)$ is weaker: it allows for singularities of order $2$, i.e. $q(x)$ is allowed to be of the form \begin{equation} q(x) = \frac{1}{(x-x_0)^2}. \end{equation} We see that in this case, $\lim_{x \to x_0} (x-x_0) q(x)$ does not exist, but $\lim_{x \to x_0} (x-x_0)^2 q(x)$ does. Of course, when $c(x)$ is regular at $x = x_0$, then $q(x) = \frac{c(x)}{(x-x_0) h(x)}$ only has a singularity of order $1$, so it automatically follows that $(x-x_0)^2 q(x)$ remains finite as $x \to x_0$ (if $h(x_0)$ exists). In this formulation, we see that it is allowed that \begin{equation} c(x) \leadsto \frac{1}{x-x_0} \quad \text{as} \quad x \to x_0, \end{equation} and $x_0$ is still a regular singular point.