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

First simplify the coefficient of $y$ usingAfter dividing by $x(x-m)$, the second-order linear differential equation isAt zero, $xP\to 1$ and $x^{2}Q\to 0$, and both products are analytic there. At $m$, $(x-m)P\to 1$ and $(x-m{)}^{2}Q\to m^2{\omega }^2$, again with analytic extensions. Thus $\boxed{x=0\text{ and }x=m\text{ are regular singular points}}$, and there are no other finite singular points.

Infinity must also be considered because the question asks for all singular points. Put $t=1/x$ and $Y(t)=y(1/t)$. The coefficient of $Y$ in its normal-form equation is $Q(1/t)/t^4\sim {\omega }^2/t^4$. Multiplication by $t^2$ does not make it analytic at zero. Hence $\boxed{x=\infty \text{ is an irregular singular point}}$.