Hermite's equation of order $\alpha$ Show that the general solution of Hermite's equation of order $\alpha$:
$${y}''-2x{y}'+2\alpha y=0$$
$$is$$
$$y(x)=c_{0}y_{1}(x)+c_{1}y_{2}(x)$$
where $y_{1}(x)$ and $y_{2}(x)$ are power series solutions centered at the ordinary point 0.
Find the two power series.
I am stuck at this question. I don't really know what to do.
 A: Assume that $y(x)=\sum_{n=0}^{\infty}{a_nx^n}$. Substituting this into the equation, we find
$$\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}-2\sum_{n=0}^{\infty}na_nx^{n}+2\alpha\sum_{n=0}^{\infty}a_nx^{n}=0$$
or, equivalently,
$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}na_nx^{n}+2\alpha\sum_{n=0}^{\infty}a_nx^{n}=0.$$
This means that whenever 
$$a_{n+2}=\frac{2(n-\alpha)}{(n+2)(n+1)}a_n,\tag{1}$$
the series will solve the differential equation. Obviously, $a_0$ and $a_1$ fix the rest of the series. In particular, we can set $a_1=0$ - then the series will contain only even powers of $x$. Similarly, for $a_0=0$ it will contain only odd powers of $x$. The general solution is an arbitrary linear combination of this odd and even part, so you can interpret $a_0$ and $a_1$ as two arbitrary integration constants.

Added: You can also explicitly solve the recursion (1): for example,
$$a_{2k}=\frac{\Gamma(\frac12)\Gamma(k-\frac{\alpha}{2})}{k!\Gamma(-\frac{\alpha}{2})\Gamma(k+\frac12)}a_0.$$
[Exercise: show this - for example, by induction]. Hence the series
$$\sum_{k=0}^{\infty}\frac{\Gamma(\frac12)\Gamma(k-\frac{\alpha}{2})}{k!\Gamma(-\frac{\alpha}{2})\Gamma(k+\frac12)}x^{2k}$$
gives one solution of your differential equation.
One can similarly determine the coefficients with odd indices and write the second solution.
