Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 8 / b / Solution

Set $t=x-m$ and use the Frobenius solution $y=t^{\nu }(a_0+a_{1}t+\cdots )$ with $a_0\ne 0$. From part (a), the leading coefficients are $P\sim 1/t$ and $Q\sim m^2{\omega }^2/t^2$. The coefficient of the lowest power $t^{\nu -2}$ is thereforeThe indicial equation reduces to ${\nu }^2+m^2{\omega }^2=0$, givingSince $m\omega >0$, the two indicial roots differ by a nonzero imaginary number, not an integer. Thus the two local Frobenius solutions exist independently on a chosen branch of $\log (x-m)$.