How do I write a power series from a recurrence relation? I have ODE
$$y''- 2xy'+ 2ky = 0$$
From this I have found a recurrence relation
$$C_{n+2} = \dfrac {(2n+2k)}{(n+1)(n+2)}C_n$$
How would I write this out in series summation form? I am trying to relate all terms bacl to $C_0$, but am struggling to write out in summation form.
 A: Starting with $$C_{n+2} = \dfrac {(2n+2k)}{(n+1)(n+2)} \, C_n,$$ apply $n = 0,1,2, \cdots$ to determine the pattern. As seen by:
\begin{align}
C_{n+2} &= \dfrac {(2n+2k)}{(n+1)(n+2)} \, C_n \\
C_{2} &= \dfrac {(2k)}{(1)(2)} \, C_0 \\
C_{3} &= \dfrac {2k+2}{(2)(3)} \, C_1 \\
C_{4} &= \dfrac {2k+4}{(3)(4)} \, C_2 = \frac{(2 k)(2 k + 4)}{4!} \, C_{0} = \frac{2^2 \, k (k + 2)}{4!} \, C_{0} \\
C_{5} &= \dfrac {2k+6}{(4)(5)} \, C_3 = \dfrac {2^2 \, (k+1)(k+3)}{5!} \, C_1.
\end{align}
This leads to
\begin{align}
y(x) &= C_{0} + C_{1} \, x + C_{2} \, x^2 + \cdots \\
&= C_{0} + C_{1} \, x + \dfrac {(2k)}{(1)(2)} \, C_0 \, x^2 + \dfrac {2k+2}{(2)(3)} \, C_1 \, x^3 + \cdots \\
&= C_{0} \, \sum_{n=0}^{\infty} \frac{a_{n} \, 2^{n} \,  x^{2 n}}{(2 n)!} + C_{1} \, \sum_{n=0}^{\infty} \frac{b_{n} \, 2^{n} \, x^{2 k + 1}}{(2 k + 1)!}, 
\end{align}
where $a_{n}$ and $b_{n}$ are coefficients involving $k$.
Further work shows that
\begin{align}
C_{2n} &= \frac{4^{n}}{(2 n)!} \, \left(\frac{k}{2}\right)_{n} \, C_{0} \\
C_{2 n +1} &= \frac{4^{n}}{(2 n +1)!} \, \left(\frac{k+1}{2}\right)_{n} \, C_{1}.
\end{align}
This leads to
$$ y(x) = C_{0} \, \sum_{n=0}^{\infty} \frac{\left(\frac{k}{2}\right)_{n} \, (2 x)^{2n}}{(2 n)!} + \frac{C_{1}}{2} \, \sum_{n=0}^{\infty} \frac{\left(\frac{k+1}{2}\right)_{n} \, (2 x)^{2 n+1}}{(2 n+1)!}, $$
where $(a)_{n}$ is the Pochhammer symbol.
Note:
There seems to be an error, or missing components, with the coefficient recurrence relation. This is based upon the solution of
$$ y'' - 2 \, x \, y' + 2 \, k \, y = 0$$
is $$ y(x) = c_{0} \, H_{k}(x) + c_{1} \, {}_{1}F_{1}\left(- \frac{k}{2}, \frac{1}{2}; x \right),$$
where $H_{n}(x)$ are the Hermite polynomials.
