Solving $(4-x^2)y''+2y=0$ by series I know that there are other ways to solve this, but I want to solve the ODE $(4-x^2)y''+2y=0$ by series close to $0$, that is, I'm looking for solutions of the form $y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}.$
Substituing in the ODE, I got:
$$(8a_{2}+2a_{0})+(24a_{3}+2a_{1})x+\sum_{n=2}^{\infty}[4(n+2)(n+1)a_{n+2}+2a_{n}-n(n-1)a_{n}]x^{n}=0\implies \\
a_2=-\frac{a_0}{4};\quad a_3=-\frac{a_{1}}{12}; \quad a_{n+2}=\frac{(n-2)a_{n}}{4(n+2)}. $$
But this recorrence relation give me a problem: $a_4=0\implies a_{2n}=0$ for all $n>2 $.
My question is: Did I made a mistake? Or this solution considers, for even $n$, only the terms $a_2$ and $a_0$, that is, $y_{1}(x)=a_{0}\left(1-\frac{1}{4}x^{2}\right)$?

Since you correctly obtained a solution $y_1(x)=x^2-4$ (I set $a_0=-4$ in your notation), here is a way to find all solutions.  Suppose that a general solution $y(x)$ takes the form $y_1(x)z(x)=(x^2-4)z(x)$.  We get
$$y'(x)=y_1(x)z'(x)+y_1'(x)z(x)=(x^2-4)z'(x)+2xz(x)$$
and
$$y''(x)=y_1(x)z''(x)+2y_1'(x)z'(x)+y''_1(x)z(x)=(x^2-4)z''(x)+4xz'(x)+2z(x).$$
From the original ODE $-(x^2-4)y''(x)+2y(x)=0$, so
$$-(x^2-4)\big((x^2-4)z''(x)+4xz'(x)+2z(x)\big)+2\big((x^2-4)z(x)\big)=0.$$
That is
$$-(x^2-4)z''(x)-4xz'(x)=-\big((x^2-4)z''(x)+4xz'(x)+2z(x)\big)+2z(x)=0.$$
Hence if $u(x)=z'(x)$, then
$$(x^2-4)u'(x)+4xu(x)=0$$
or
$$\frac{u'(x)}{u(x)}=-\frac{4x}{x^2-4}=-\frac{2}{x-2}-\frac{2}{x+2}.$$
This implies
$$\ln u(x)=-2\ln|x-2|-2\ln|x+2|+c'=-\ln\big((x-2)^2(x+2)^2\big)+c'$$
for some constant $c'$.  That is, if $c=e^{c'}$, we have
$$z'(x)=u(x)=\frac{c}{(x-2)^2(x+2)^2}=\frac{c}{32}\left(\frac{1}{x+2}+\frac{2}{(x+2)^2}-\frac{1}{x-2}+\frac{2}{(x-2)^2}\right).$$
That is, if $b=-c/32$ and $a$ is a constant, then
$$z(x)=a-b\left(\ln |x+2|-\frac{2}{x+2}-\ln|x-2|-\frac{2}{x-2}\right)=a+b\left(\frac{4x}{x^2-4}-\ln\left|\frac{x+2}{x-2}\right|\right).$$
So
$$y(x)=y_1(x)z(x)=ay_1(x)+by_2(x),$$
where
$$y_2(x)=4x-(x^2-4)\ln\left|\frac{x+2}{x-2}\right|.$$
It is possible to find $y_2(x)$ through the series method.  Note that
$$y_2(x)=\sum_{n=0}^\infty b_nx^n$$
where $b_0=0$, $b_1=8$, and $b_n=\frac{n-4}{4n}b_{n-2}$ for $n\ge3$.  So $b_{2n}=0$ for all integers $n\ge0$.  And for $n\ge 1$,
$$b_{2n+1}=8\left(\frac{2n-3}{4(2n+1)}\right)\left(\frac{2n-5}{4(2n-1)}\right)\cdots \left(\frac{1}{4\cdot 5}\right)\left(\frac{-1}{4\cdot 3}\right)=-\frac{1}{2^{2n-3}(2n+1)(2n-1)}.$$  That is
$$y_2(x)=-\sum_{n=0}^\infty\frac{x^{2n+1}}{2^{2n-3}(2n+1)(2n-1)}=-16\sum_{n=0}^\infty\frac{1}{4n^2-1}\left(\frac{x}{2}\right)^{2n+1}.$$
