Solve the equation
$$(1-x^2)y'' -2xy'+2y = 0$$
using power series.
Replacing with $y=\Sigma_{n=0}^\infty a_nx^n$ and the respective derivatives, the procedure returns the recursion formula:
$$a_{n+2}= \frac{n-1}{n+1}a_0$$
Because $a_3=0$, odd coeficients go to $0$ and for even coeficients the recursion formula ends up like this:
$$a_{2n}=\frac{a_0}{2n+1}$$
So the solution would be:
$$y(x)=a_1x+a_0\left(1-\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}\right)$$
However the textbook states that the solution is:
$$y(x)=a_1x+a_0\left(1-x\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\right)$$
I can't see where did I go wrong, neither how can the $2n+1$ end up in the series. Any help/pointing would be appreciated
$\require{cancel}$ You're almost there
\begin{eqnarray} a_2 &=& -a_0 \\ a_4 &=& \frac{1}{3}a_2 = -\frac{1}{3}a_0 \\ a_6 &=& \frac{3}{5}a_4 = -\frac{1}{5}a_0 \\ a_8 &=& \frac{5}{7}a_6 = -\frac{1}{7}a_0 \\ &\vdots&\\ a_{2n} &=& -\frac{1}{2n-1}a_0 \tag{1} \end{eqnarray}
So your solution becomes
\begin{eqnarray} y(x) &=& \sum_{n=0}^{+\infty}a_n x^n = a_0 + a_1 x + \sum_{n=2}^{+\infty}a_n x^n \\ &=& a_0 + a_1 x + \sum_{n=1}^{+\infty}a_{2n} x^{2n} + \cancelto{0}{\sum_{n=1}^{+\infty}a_{2n+1} x^{2n+1}} \\ &=& a_0 + a_1 x + \sum_{n=1}^{+\infty}a_{2n} x^{2n} \\ &=& a_0 + a_1 x + \sum_{n = 0}^{\infty}a_{2n + 2}x^{2n+2} \\ &=& a_0 + a_1 x -\sum_{n = 0}^{+\infty}\frac{a_0}{2n + 1}x^{2n+2} \\ &=& a_1 x + a_0 \left(1 - x\sum_{n = 0}^{+\infty}\frac{1}{2n + 1}x^{2n+1} \right) \tag{2} \end{eqnarray}