Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 8 / b / ii / Solution

Put $t=1-x$ and $Y(t)=y(x)$. The differential equation becomesSubstitute the Frobenius solution $Y=t^{\sigma }{\sum }_{n\ge 0}{a}_n{t}^n$. The lowest power has coefficient ${\sigma }^2{a}_0$, soEquating successive coefficients gives, with $a_{-1}=0$,Thus $a_0\ne 0$ is the only free coefficient: $a_1=-\lambda a_0$, $a_2=({\lambda }^2-c^2)a_0/4$, and so on. Since every denominator is nonzero, there is exactly one independent pure Frobenius solution of this form. The repeated root supplies no second free coefficient; the other solution is logarithmic, as parts (iii) and (iv) demonstrate.