I'm trying to understand singular points at infinity, so looking at the example
$$x^2(1-x^2)y'' - y = 0$$
I am trying to investigate (a) does it have a singular point at infinity, and (b) if it does or doesn't, what can this tell us about solving the DE?
I understand that I want to substitute $x = 1/\xi$ and look for an expansion of the solution about $\xi = 0$. I first note that by the chain rule $\frac{dy}{dx}$ is the same as $\frac{dy}{d\xi}\frac{d\xi}{dx}$ and $\frac{d\xi}{dx}$ can be determined from
$$\frac{d}{dx} x = \frac{d}{dx} \frac{1}{\xi} \qquad \Rightarrow $$ $$1 = -\xi^{-2}\frac{d\xi}{dx}\qquad \Rightarrow $$ $$-\xi^2 = \frac{d\xi}{dx}$$
So we can conclude that
$$\frac{dy}{dx} = -\xi^2\frac{dy}{d\xi}$$
Also we compute $\frac{d^2y}{dx^2}$ by $\frac{d}{dx}\left(-\xi^2\frac{dy}{d\xi}\right)$ which is
$$-\left(2\xi\frac{d\xi}{dx}\frac{dy}{d\xi}+\xi^2\frac{d}{dx}\frac{dy}{d\xi}\right)=$$ $$-\left(-2\xi^3\frac{dy}{d\xi}+\xi^2\frac{d}{dx}\frac{dy}{d\xi}\right)$$
At this point I'm not certain how to compute $\frac{d}{dx}\frac{dy}{d\xi}$ but if I had to guess it'd be
$$\frac{d}{d\xi}\frac{dy}{dx} = \frac{d}{d\xi}\left(-\xi^2\frac{dy}{d\xi}\right)$$ $$=-\left(2\xi\frac{dy}{d\xi}+\xi^2\frac{d^2y}{d\xi^2}\right)$$
Putting all of this together we get that $\frac{d^2y}{dx^2}$ is
$$-\left(-2\xi^3\frac{dy}{d\xi}-\xi^2\left(2\xi\frac{dy}{d\xi}+\xi^2\frac{d^2y}{d\xi^2}\right)\right)=$$ $$4\xi^3\frac{dy}{d\xi}+\xi^4\frac{d^2y}{d\xi^2}$$
That answer disagrees with what I've read elsewhere but I'm not sure how they got their answers elsewhere or what I've done wrong. I'll assume from here that the correct answer is that
$$\frac{d^2y}{dx^2} = 2\xi^3\frac{dy}{d\xi}+\xi^4\frac{d^2y}{d\xi^2}$$
With that in hand the given differential equation at the start becomes
$$(1/\xi^2)(1-1/\xi^2)\left( 2\xi^3\frac{dy}{d\xi}+\xi^4\frac{d^2y}{d\xi^2}\right) - y = 0$$
and the coefficient of the second-order term is then
$$\frac{\xi^4(1-1/\xi^2)}{\xi^2}$$
and we want to know if this has a pole at $\xi=0$. It does not, so presumably that makes infinity not a singular point. Does that mean there is a series expansion of some sort?
What if the point were a regular singular point? Would that then imply some kind of solution? If so, how would you obtain it? Merely by expanding the solution in terms of $\xi$ at $\xi=0$ and then changing out to $\xi = 1/x$ at the end?
An easier way to do the transformation is to first set $y = z(x^{-1})$. Then the chain rule gives \begin{align} \frac{dy}{dx} &=-\frac{z'(x^{-1})}{x^2} \\ \frac{d^2y}{dx^2} &= \frac{z''(x^{-1})}{x^4} + 2\frac{z'(x^{-1})}{x^3}. \end{align} Plug these into the original differential equation, then do the substitution $x\rightarrow \xi^{-1}$. In the case of your differential equation, that gives $$ x^2(1-x^2)\left[\frac{z''(x^{-1})}{x^4} + 2\frac{z'(x^{-1})}{x^3}\right] + z(x^{-1}) = \left(x^{-2}-1\right)z''(x^{-1}) + 2\left(x^{-1} + x\right) z'(x^{-1}) + z(x^{-1}), $$ and applying the substitution gives $$ (\xi^2-1)z'' + 2(\xi + \xi^{-1})z' + z = 0. $$ This has a singular point at $0$, as the coefficient of the first-order term has a singularity there. So we can conclude the original DE has a singular point at infinity.