Showing a power series is a solution to a differential equation. Show that the following power series is a solution to $y''-2xy'+2y = 0$. $$y = \sum_{n=0}^{\infty} \frac{-x^{2n}}{n!(2n-1)}$$.
Okay so differentiating and substituting in gives $$-2\sum_{n=0}^{\infty} \frac{nx^{2n-2}}{n!}+4\sum_{n=0}^{\infty} \frac{nx^{2n}}{n!(2n-1)}-2\sum_{n=0}^{\infty} \frac{x^{2n}}{n!(2n-1)}$$ but I can't seem to simplify this further.
Any help would be great thanks!
 A: Recall that within the radius of convergence, a power series $\sum_{n=0}^\infty a_nx^n$ can be differentiated term-by-term, that is,
$$
\frac{\mathsf d}{\mathsf dx}\left[\sum_{k=0}^\infty a_nx^n\right] = \sum_{n=0}^\infty \frac{\mathsf d}{\mathsf dx} [a_nx^n].
$$
Differentiating $y(x)$ we have
$$
y'(x) = \sum_{n=0}^\infty \frac{-2nx^{2n-1}}{n!(2n-1)} = \sum_{n=0}^\infty \frac{-2x^{2n}}{(n-1)!(2n-1)},
$$
and differentiating again we have
$$
y''(x) = \sum_{n=0}^\infty \frac{-2n(2n-1)x^{2n-2}}{n!(2n-1)}= \sum_{n=0}^\infty \frac{-2nx^{2(n-1)}}{n!} = \sum_{n=0}^\infty \frac{-2(n+1)x^{2n}}{(n+1)!} = \sum_{n=0}^\infty \frac{-2x^{2n}}{n!}.
$$
Adding $y''(x) - 2xy'(x) +2y(x)$, we have
$$
\sum_{n=0}^\infty \frac{-2x^{2n}}{n!} + \sum_{n=0}^\infty \frac{4x^{2n}}{(n-1)!(2n-1)} - \sum_{n=0}^\infty \frac{2x^{2n}}{n!(2n-1)},
$$
which is equal to
$$
2\sum_{n=0}^\infty\left(-\frac1{n!}+\frac 2{(n-1)!(2n-1)}-\frac1{n!(2n-1)} \right)x^{2n}
$$
It remains to show that the coefficients of $x^{2n}$ in the above expression are identically zero. We have
$$
-\frac1{n!}+\frac 2{(n-1)!(2n-1)}-\frac1{n!(2n-1)} = \frac{-(n-1)!(2n-1)+2n!-(n-1)!}{n!(n-1)!(2n-1)},
$$
and so it suffices to show that the numerator of the above expression is zero. Proceeding,
$$
-(n-1)!(2n-1)+2n!-(n-1)! = -2n! +(n-1)! +2n! - (n-1)! = 0,
$$
and so it follows that the given power series is a solution to the differential equation $y''(x) -2xy'(x)+2y(x)=0$.
