Can all differential equations be solved using power series? To elaborate, does the differential equation have to be of some form to be solvable by power series.  More specifically, if I wanted to solve this equation by power series would I be able to?
$$
(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)=0
$$
Also, $n$ is a constant in the stated equation.
 A: If the differential equation is analytical, then one can try to find local solutions as power series.
However, in points of singularity, here in $x=\pm1$, one may have to multiply the power series by a fractional power, giving Puisseux-series as solutions.
A: $(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)
=0
$
I'll try
$y(x)
=\sum_{k=0}^{\infty} a_k x^k
=a_0+a_1x+\sum_{k=2}^{\infty} a_k x^k
$.
$y'(x)
=\sum_{k=1}^{\infty} ka_k x^{k-1}
=\sum_{k=0}^{\infty} (k+1)a_{k+1} x^{k}
$
so
$xy'(x)
=\sum_{k=0}^{\infty} (k+1)a_{k+1} x^{k+1}
=a_1x+\sum_{k=2}^{\infty} ka_{k} x^{k}
$.
$y''(x)
=\sum_{k=2}^{\infty} k(k-1)a_k x^{k-2}
=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
$
so
$\begin{array}\\
(1-x^2)y''(x)
&=(1-x^2)\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}\\
&=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
-x^2\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}\\
&=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
-\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k+2}\\
&=\sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2} x^{k}
-\sum_{k=2}^{\infty} (k-1)ka_{k} x^{k}\\
&=2a_2+6a_3x
+\sum_{k=2}^{\infty} ((k+1)(k+2)a_{k+2}-(k-1)ka_{k}) x^{k}\\
\end{array}
$.
Adding these all up
$\begin{array}\\
0
&=(1-x^2)y''(x)-3xy'(x)+n(n+2)y(x)\\
&=n(n+1)(a_0+a_1x)-3(a_1x)+2a_2+6a_3x
+\sum_{k=2}^{\infty} x^k\left(   n(n+2)a_k-3ka_{k}+((k+1)(k+2)a_{k+2}-(k-1)ka_{k})\right)\\
&=n(n+1)a_0+2a_2+x(n(n+1)a_1-3a_1+6a_3)
+\sum_{k=2}^{\infty} x^k\left(   (n(n+2)-3k-k(k-1))a_k+(k+1)(k+2)a_{k+2})\right)\\
&=n(n+1)a_0+2a_2+x(n(n+1)a_1-3a_1+6a_3)
+\sum_{k=2}^{\infty} x^k\left(   (n(n+2)-k(k+2))a_k+(k+1)(k+2)a_{k+2}\right)\\
\end{array}
$
Therefore,
assuming no errors
($P(\text{no errors}) < 1-1/e)$),
$0
=n(n+1)a_0+2a_2
$,
$0
=(n(n+1)-3)a_1+6a_3
$,
and
$0
= (n(n+2)-k(k+2))a_k+(k+1)(k+2)a_{k+2}
$.
These recurrences
show how to compute
the $a_k$
in terms of
$a_0$ and $a_1$.
Note that
$y(x)$ is the sum of
two independent series,
one with even exponents
and one with odd.
The last recurrence has
an obvious problem
at $k=n$,
so LutzL's suggestion
of multiplying the series
by $x^{\alpha}$
might have to be done.
I'll leave it at this.
