Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 8 / d

If $N=|n|$ is an integer, the recurrence has zero numerator at $j=N$. Choosing the parity containing $a_N$ therefore makes all later coefficients vanish. Conversely, for a nonzero polynomial of degree $N$, the coefficient of $x^N$ in the differential equation is $(n^2-N^2)a_N$, so $n^2=N^2$ is necessary.

Normalizing the terminating series by $y(1)=1$ gives the first Chebyshev polynomials:Substitution verifies each differential equation and endpoint value. More generally, $T_N(x)=\cos (N\arccos x)$; its recurrence $T_{N+1}=2x{T}_N-T_{N-1}$ also establishes polynomiality and the normalization.