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

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 $\boxed{x=\infty \text{ is an irregular singular point}}$.

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 $\log (x-m)$.

Define the Wronskian by $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. Differentiating and using the normal-form differential equation givesIntegrating this first-order linear differential equation proves Abel's identity:Thus, on any interval avoiding zero and $m$,Absolute-value signs from real logarithms can be absorbed into the constant on each such interval. For linearly independent solutions, $C\ne 0$.

At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.