$ \text { The Hermite polynomials, } H_{n}(x) \text { , satisfy the following: } $ \begin{array}{l}{\text { i. }<H_{N}, H_{M}>=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}} \\ {\text { ii. } \quad H_{n}^{\prime}(x)=2 n H_{n-1}(x)} \\ {\text { iii. } H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)} \\ {\text { iv. } H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right)}\end{array} Using these, show that \begin{array}{l}{\text { a. }\quad H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0 .\ \text { [Use properties ii. and iii.] }}\end{array}
Any help on this would be greatly appreciated!
$H_n^{''}-2xH_n^{'}+2nH_n = (2nH_{n-1})^{'}-4xnH_{n-1}+2nH_n$ (by ii)
= $4n^2 H_{n-2} - 4xnH_{n-1}+2nH_n $ (by ii again)
= $2n(H_n-2xH_{n-1}+2nH_{n-2}) = 0$ (by iii)