Recurrence relations and power series solution I am given the following initial value problem: $$(1-x^2)y''+7xy'-26y=0 \qquad , \qquad  y(0)=0 \qquad  , \qquad y'(0)=4$$
I have solved for the singular points, which are $x= 1, -1$
The question then tells me that I can find a normal power series solution for $y$ about $x=0$ $$y(x)=\sum_{m=0}^{\infty} a_m x^m $$ 
I then went on to find the recurrent relation, to which I determined to be:
$\frac{a(m+1)}{am}=- \frac{(-m^2+8m-26)}{(m+1)(m+2)}$
I plugged in values $m=0$, to $m=3$ into the above derived formula.
So:
$m=0 \implies a(2) = 13 a(0)$,
$m=1   \implies a(3) = \frac{19}{6} a(1)$,
$m=2  \implies  a(4) = \frac{91}{6} a(0)$,
$m=3  \implies  a(5) = \frac{209}{120} a(1)$,
So far, all of my answers are correct. What I am finding difficult to do it using the above results and the given initial conditions to find the first $3$ non-zero terms of the power series solution for $y$.
I appreciate any help, 
thank you for your time.

Considering the differential equation
$$
(1-x^2)y''+\alpha x y' + \beta y = 0
$$
and substituting $y = \sum_{k=0}^n a_k x^k$ we obtain the recurrences
$$
2a_2+\beta a_0 = 0\\
6a_3+(\alpha+\beta)a_1 = 0
$$
and for $k \ge 4$
$$
k(k-1)a_k +((k-2)\alpha +\beta-(k-2)(k-3))a_{k-2} = 0
$$
Attached a plot showing in black the numerical integration for $y$ and in red the series representation with $n=8$

NOTE
This series represent the solution to $y$ for $-1 < x < 1$.
