Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 20 / b / Solution

Differentiating $v=x^n{e}^{-x}$ gives $x{v}^{\prime }=(n-x)v$. Applying the Leibniz rule $n+1$ times to this identity givesOnly two terms survive on each side because the multiplying factors are linear. Rearranging yields $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. Since $v^{(n)}=n!e^{-x}{L}_n$, the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substituting and cancelling $n!e^{-x}$ givesThis is deriving the Laguerre equation from Rodrigues' formula, exhibiting $L_n$ as an eigenfunction of the Laguerre differential operator on polynomials.