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, $y_{1}(x)=a_{0}\left(1-\frac{1}{4}x^{2}\right)$?
 A: We have a differential equation that doesn’t have a constant coefficient for the second derivative.
$p\left( x \right) = 4-{x^2} \hspace{0.25in}p\left( 0 \right) = 4 \ne 0$
So, assume that $x_0$ is an ordinary point for this differential equation. We first need the solution and its derivatives,
$y\left( x \right) = \sum\limits_{n = 0}^\infty  {{a_n}{x^n}} \hspace{0.25in}y'\left( x \right) = \sum\limits_{n = 1}^\infty  {n{a_n}{x^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^{n - 2}}}$
Plug these into the differential equation.
$\left( 4-{{x^2}} \right)\sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^{n - 2}}}  +2\sum\limits_{n = 0}^\infty  {{a_n}{x^{n}}}= 0$
Now, break up the first term into two so we can multiply the coefficient into the series and multiply the coefficients of the second and third series in as well.
$4\sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^{n-2}}}  - x^{2}\sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^{n - 2}}}  + 2\sum\limits_{n = 0}^\infty  {{a_n}{x^{n}}}=0$
$4\sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^{n-2}}}  - \sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^n}}  + 2\sum\limits_{n = 1}^\infty  {{a_n}{x^{n}}}=0$
We will only need to shift the first and third series down by two and one respectively to get all the exponents the same in all the series.
$4\sum\limits_{n = 0}^\infty  {(n+2)\left( {n + 1} \right){a_{n+2}}{x^n}}  - \sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){a_n}{x^n}}  + 2\sum\limits_{n = 0}^\infty  {{a_{n+1}}{x^{n}}}=0$
At this point we could strip out some terms to get all the series starting at $n=2$, but that’s actually more work than is needed. Let’s instead note that we could start the second series at $n=0$ if we wanted to because that term is just zero. Likewise, the terms in the second series are zero for both $n=1$ and $n=0$ and so we could start that series at $n=0$. If we do this all the series will now start at $n=0$ and we can add them up without stripping terms out of any series.
$\begin{align*}\sum\limits_{n = 0}^\infty  {\left[ {4{(n+2)}\left( {n + 1} \right){a_{n+2}} - n\left( {n - 1} \right){a_{n}} + 2{a_{n}}} \right]{x^n}}  & = 0\\  
\sum\limits_{n = 0}^\infty  {\left[ {4{(n+2)}\left( {n + 1} \right){a_{n+2}} + \left(n+2 \right)\left(n-1\right){a_{n}}} \right]{x^n}}  & = 0\\ \end{align*}$
Now set coefficients equal to zero.
${4{(n+2)}\left( {n + 1} \right){a_{n+2}} + \left(n+2 \right)\left(n-1\right){a_{n}}}= 0,\hspace{0.25in}n = 0,1,2, \ldots$
Solving this gives,
${a_{n + 2}} =  - \frac{{\left( {n + 2} \right)\left( {n - 1} \right){a_n}}}{4{\left( {n + 2} \right)\left( {n + 1} \right)}},\hspace{0.25in}n = 0,1,2, \ldots$
Now, we plug in values of $n$,
$n = 0:\hspace{0.25in}{a_2} = \frac{a_0}{4}\\
n = 1:\hspace{0.25in}{a_3} =   0{a_1} = 0\\
n = 2:\hspace{0.25in}{a_4} =  - \frac{1}{12}{a_2}\neq0\\
n = 3:\hspace{0.25in}{a_5} =  - \frac{1}{8}{a_3} = 0\\
n = 4:\hspace{0.25in}{a_5} =  - \frac{3}{20}{a_4}\neq0$
Now, from this point we can see that $a_{2n}=0$,for all $n>2$.
Thus,for solution considers, for even $n$, only the terms $a_2$ and $a_0$, that is,
$y(x)=a_{0}\left(1+\frac{1}{4}x^{2}\right)$.
A: 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
$$y_2(x)=-\sum_{n=0}^\infty\frac{x^{2n+1}}{2^{2n-3}(2n+1)(2n-1)}=-16\sum_{n=0}^\infty\frac{1}{4n^2-1}\left(\frac{x}{2}\right)^{2n+1}.$$
