I need to find a third-degree approximation of $x^3y'' + x^2y' - (e^x-1)y=0$ at some singular point. I've found that the singular points of $x^3y'' + x^2y' - (e^x-1)y=0$ are 0 and $\infty$. However, at infinity, the equation after substituting $x=1/\xi$ and $y(x)=z(1/\xi)$ is
$$\xi z''+(2/\xi-1)z'-\left(\sum_{i=1}^\infty\frac{1}{i!}\frac{1}{\xi^i}\right)z=0$$
The point $\xi=0$ is a singular point, but since
$$\lim_{\xi\rightarrow 0}\frac{\xi(2/\xi -1)}{\xi}=\infty$$
so that $\infty$ is an irregular singular point, which I assume means that we cannot find a series solution expanded there.
On the other hand $x=0$ is a regular singular point of the original DE and therefore I should be able to find a series solution of the form $y=x^r\sum_{i=0}^\infty a_ix^i$. First defining
$$p_0=\lim_{x\rightarrow 0}\frac{x\cdot x^2}{x^3} = 1$$
$$q_0=\lim_{x\rightarrow 0}\frac{x^2\cdot (-e^x-1)}{x^3} = \lim_{x\rightarrow 0}\frac{-e^x}{1} = -1$$
then we can solve the indicial equation
$$r(r-1)+r-1 = (r+1)(r-1)\Rightarrow r=\pm 1$$
It makes sense to me that $r=-1$ would be difficult to investigate because in the Taylor expansion we would need derivatives of a non-polynomial, so I attempt to find a solution for $r=1$, and I would think this shouldn't be very hard (well, except for the appearance of $e^x$).
$$y = \sum_{i=0}^\infty a_ix^{i+1}$$ $$y' = \sum_{i=0}^\infty (i+1)a_ix^i$$ $$y'' = \sum_{i=1}^\infty i(i+1)a_ix^{i-1}$$
So
$$x^3\sum_{i=1}^\infty i(i+1)a_ix^{i-1}+x^2\sum_{i=0}^\infty (i+1)a_ix^i-\left(\sum_{i=1}^\infty \frac{x^i}{i!}\right)\sum_{i=0}^\infty a_ix^{i+1}$$
The lowest-degree term I'm going to get out of this is a first-degree term, coming only from $\left(\sum_{i=1}^\infty \frac{x^i}{i!}\right)\sum_{i=0}^\infty a_ix^{i+1}$. This will imply that $a_0=0$.
That might not be so bad, but after looking for the $x^2$ terms, I find that $a_1=0$ and then from there I find $a_2=0$ and $a_3=0$ and so on... it feels as though I must have made a mistake somewhere but I can't see where.