Prove that $S(x) = \sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n + 1}$ is solution to $(E): (x^2 - 2)y' + xy + 2 = 0 $ 
Prove that the power series defined as:
$$ S(x) = \sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n + 1} $$
is solution to the differential equation:
$$ (E): (x^2 - 2)y' + xy + 2 = 0 $$
and find its expression.


So far I found:
$ S'(x) = \sum_{n = 0}^{+ \infty} \frac{n! (2n + 1)}{1.3....(2n + 1)} x^{2n} $
$$ \implies S'(x) = \sum_{n = 0}^{+ \infty} \frac{n! (2n)}{1.3....(2n + 1)} x^{2n} + \sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n} $$
$$ \implies x^2 S'(x) = x \sum_{n = 0}^{+ \infty} \frac{n! (2n)}{1.3....(2n + 1)} x^{2n + 1} + x\sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n + 1} $$
$$ \implies x^2 S'(x) = x \sum_{n = 0}^{+ \infty} \frac{n! (2n)}{1.3....(2n + 1)} x^{2n + 1} + xS(x) $$
I do not see how to proceed to the other terms.
 A: $$S(x) = \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)} x^{2n+1} \\
\implies S'(x) = \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n}$$
The left side of the ODE is then
$$\sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n+2} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} + \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)} x^{2n+2} + 2$$
Combine the series with the same-degree terms.
$$2 \sum_{n=0}^\infty \frac{(n+1)!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)} x^{2n+2} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} + 2$$
Shift the index on the first sum.
$$2 \sum_{n=1}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-3)\cdot(2n-1)} x^{2n} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} + 2$$
Join the constant term to the first sum and we're done.
$$2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} = 0$$

To get a closed form for $S$, observe that
$$\frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} = \frac{n!\cdot2\cdot4\cdot\cdots\cdot2n}{1\cdot2\cdot3\cdot4\cdot\cdots\cdot(2n-1)\cdot2n} = \frac{2^n\cdot(n!)^2}{(2n)!} = \frac{2^n}{\binom{2n}n}$$
Compare this to the series expansion of the square of $\arcsin$,
$$\arcsin^2(x) = \sum_{n=0}^\infty \frac{2^{2n-1}}{n^2\binom{2n}n} x^{2n}$$
We can then find $S'$ in terms of derivatives of $\arcsin^2(x)$, and recover $S$ by integrating.

 Differentiate $\arcsin^2$ twice and multiply by $2x$ as needed to obtain
 $$\frac{2x\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=0}^\infty \frac{2^{2n}}{n\binom{2n}n} x^{2n}$$
$$\frac{x^2\sqrt{1-x^2}+x\arcsin(x)}{(1-x^2)^{3/2}} = \sum_{n=0}^\infty \frac{2^{2n}}{\binom{2n}n} x^{2n} = S'\left(\sqrt2\,x\right)$$
 and upon replacing $x\mapsto \frac x{\sqrt2}$,
 $$S'(x) = \frac{x^2\sqrt{2-x^2}+2x\arcsin\left(\frac x{\sqrt2}\right)}{(2-x^2)^{3/2}}$$
 By the fundamental theorem of calculus, noting that $S(0)=0$, we have
 $$S(x) = \int_0^x \frac{y^2\sqrt{2-y^2}+2y\arcsin\left(\frac y{\sqrt2}\right)}{(2-y^2)^{3/2}} \, dy = \frac{2\arcsin\left(\frac x{\sqrt2}\right)}{\sqrt{2-x^2}} - x$$
 as long as $|x|<\sqrt2$.

