TITLE: Determining whether x=1 is a regular singular point in the differential equation
QUESTION [2 upvotes]: Consider differential equation: $x^3(x^2-1)^2(x^2+1)y" + (x-1)xy' + y = 0$. Determine
whether $x=0$ is a regular singular point. Determine whether $x=1$ is a regular singular
point. Are there any regular singular points that are complex numbers? Justify
conclusions.
I guess I just don't know where to start with this problem. I can't find anything in my textbook about determining whether certain values of $x$ are regular singular points.
Any help is appreciated.

REPLY [1 votes]: First, rewrite the equation in standard form:
$$ y'' + P(x) y' + Q(x) y = 0$$
A point $x_0$ is a singular point if $P(x)$ and $Q(x)$ diverge as $x \rightarrow x_0$.
A singular point is regular if $(x - x_0) P(x)$ and $(x - x_0)^2 Q(x)$ do not diverge as $x \rightarrow x_0$.
So in your case,
$$ P(x) = \frac{ (x - 1) x}{x^3 (x^2 - 1)^2 (x^2 + 1)}, \quad Q(x) = \frac{1}{x^3 (x^2 - 1)^2 (x^2 + 1)}.$$
Note that these expressions can be simplified a little bit. Now you can check whether or not these expressions diverge or converge using the usual limit computations, which are made much nicer since these expressions are rational functions.
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