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.
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.