= Deriving the Laguerre equation from Rodrigues' formula
{title2=$xL_n''+(1-x)L_n'+nL_n=0$}
Set $v=x^ne^{-x}$. Direct <differentiation> gives $xv'=(n-x)v$. Apply the <Leibniz rule> to differentiate $n+1$ times:
$$
xv^{(n+2)}+(n+1)v^{(n+1)}
=(n-x)v^{(n+1)}-(n+1)v^{(n)}.
$$
Hence $xv^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. By <Rodrigues' formula>, $v^{(n)}=n!e^{-x}L_n$, so the next two <derivatives> are $n!e^{-x}(L_n'-L_n)$ and $n!e^{-x}(L_n''-2L_n'+L_n)$. Substitution gives the displayed <Laguerre differential operator on polynomials> eigenvalue equation. This derivation also covers $n=0$.
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