Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 20 / a / Solution

Direct differentiation of $v=(x^2-1{)}^l$ gives $(x^2-1)v^{\prime }=2lxv$; the identity also holds for $l=0$. Differentiate it $l+1$ times using the Leibniz rule. The quadratic factor has only its first two derivatives nonzero, soWriting $y=v^{(l)}$ and collecting terms gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$. Multiplication by $-1$ gives the Legendre differential equation: