Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 8 / b / Solution

For the Frobenius solution, differentiate term by term:The unshifted terms in the equation have coefficient $[4(n+\sigma )(n+\sigma -1)-4(n+\sigma )+3]a_n$. The extra factors of $x$ give the shifted contribution $[4(n+\sigma )(n+\sigma -1)+1]a_n{x}^{n+\sigma +1}$. Equating the coefficient of $x^{k+\sigma }$ for $k\ge 1$ gives the linear recurrence relationFactoring both coefficients is especially useful:The separate lowest-power equation is $(4{\sigma }^2-8\sigma +3)a_0=0$. Do not cancel $2k+2\sigma -3$ when it vanishes: that is the possible undetermined coefficient at Frobenius resonance.