Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 7A / b / Solution

For positive integral $m$, the exponent-zero linear recurrence relation fixes $a_n=(-1{)}^n{a}_0/n!$ for $1\le n<m$, but at $n=m$ its factor $n-m$ vanishes and leaves $a_m$ free. Choose $a_0=1$ and $a_m=0$. All later coefficients then vanish by the same linear recurrence relation.

A polynomial solution of degree $m-1$ isFor $m=1$ this is the constant solution $1$, in agreement with the ordinary point exception. The terminating Frobenius series arises because the coefficient equation at the resonant index becomes $0=0$, permitting truncation rather than forcing a logarithm.