The given DE is $(x²-3)y"+2xy'=0$
Since there is a singular point ($x=\pm\sqrt{3}$) I used the Frobenius method.
I found two indicial relationships:
$-3r(r+1)=0$ and $-3(r+1)(r+2)=0$ because I have two coefficients outside of the series, $c_{0}$ and $c_{1}$, with respectively $x^{r-2}$ and $x^{r-1}$. So $r$ is either -2,-1 or 0. As the values for $r$ are all separated from each other by 1 integer, I can pick one solution, and I picked $r=0$ as that will makes it easier for me to eliminate the $r$ in the recursive relationship.
The recursive relationship is $c_{n+2}= \frac{(n+r)c_{n}}{3(n+r+2)}$. With $r=0$ it becomes $c_{n+2}= \frac{nc_{n}}{3(n+2)}$.
When I work it out for different values of $n$, all even coefficients (from $c_{2}$ on) end up being $0$. The uneven coefficients reveal the following pattern: $c_{1} \sum_{n=0}^\infty \frac{(2n+1)!!x^{2n+1}}{3^{n+1}(2n+3)!!}$ which I reworked to $2c_{1} \sum_{n=0}^\infty \frac{(2n)!(n+1)x^{2n+1}}{3^{n+1}(2n+2)!}$ using the properties of a double factorial.
My first question is about the $c_{0}$ and $c_{1}$ coefficients. I can't say anything about them because the recursive relationship doesn't say anything about them.
However since $y=\sum_{n=0}^\infty c_{n}x^{n+r}$ and $r=0$ I think, but am not sure, that the total $y=c_{0}+c_{1} (1+2\sum_{n=0}^\infty \frac{(2n)!(n+1)x^{2n+1}}{3^{n+1}(2n+2)!})$. So question 1 is about my assumption here about how to work in $c_{0}$ and $c_{1}$ is correct here.
My second question is a hypothetical one based on this exercise. Let's pretend for a moment that my uneven series doesn't says $(2n+2)!$ but $(2n+1)!$ instead. That would mean I'd have an elementary function hidden in there, namely $sinh(x)$. Am I then allowed to write the uneven series as $2 c_{1} sinh x \sum_{n=0}^\infty \frac{(2n)!(n+1)}{3^{n+1}}$?
Thank you in advance.