Prove the Rodrigues formula by using the hypergeometric differential equation

Question

Let us consider the classical orthogonal polynomials (COPs) $p_{n}(x)$, with weight $w(x)$. We have: $$\int_{a}^{b} w(x) p_{n}(x) p_{m}(x) d x= \begin{cases}0 & n \neq m \\ l \neq 0 & n=m\end{cases}$$ As we know, the weights of the COPs satisfy Pearson differential relation: $$ \frac{w^{\prime}(x)}{w(x)}=\frac{D+E x}{A+B x+C x^{2}} $$ whereas the COPs satisfy Rodrigues formula: $$ P_{n}(x)=\frac{1}{k_{n} w(x)} \frac{d^{n}}{d x^{n}}\left[\left(A+B x+C x^{2}\right)^{n} w(x)\right] $$ where $k_{n}$ is a normalization constant. The COPs satisfy also the hypergeometric differential equation $$ \left(A+B x+C x^{2}\right) y^{\prime \prime}+[B+D+(2 C+E) x] y^{\prime}-n[(n+1) C+E] y=0 $$ I would like to prove the Rodrigues formula by using the hypergeometric differential equation. My attempt is based on having written: $$y(x)=\left(A+B x+C x^{2}\right)^{n} w(x)$$ The derivative is $$y'(x)=\left(A+B x+C x^{2}\right)^{n} w'(x)+n(B+2Cx)\left(A+B x+C x^{2}\right)^{n-1} w(x)$$ $$\left(A+B x+C x^{2}\right)y'(x)=w(x)\left[\left(A+B x+C x^{2}\right)^{n}(D+E x)+n(B+2Cx) y(x)\right]$$ But I can not go on. Any suggestions please?