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

After division by the leading coefficient, write $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, whereThe only finite singular points of a differential equation are $\boxed{x=0\text{ and }x=-1}$. At zero, $xP=-1/(1+x)$ and $x^{2}Q=(x+3)/[4(1+x)]$ are analytic. At $-1$, $(x+1)P=-1/x$ and $(x+1{)}^{2}Q=(x+1)(x+3)/(4x^2)$ are analytic. The regular singular point criterion for a second-order equation therefore makes both points regular.

If infinity is included, it is a regular singular point at infinity as well: $P(x)=O(x^{-2})$ and $Q(x)=1/(4x^2)+O(x^{-3})$. Equivalently, with $t=1/x$ the transformed coefficients are $\tilde{P}=2/t+1/(1+t)$ and $\tilde{Q}=(1+3t)/[4t^2(1+t)]$, which satisfy the same regular-singular criterion at $t=0$.

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.

Because $a_0\ne 0$, the lowest-power equation from part (b) requires the indicial equationThe two candidate indicial roots are thereforeTheir difference is the integer one, so the smaller-root coefficient equation at $k=1$ needs examination rather than automatic division.

For the larger indicial root $\sigma =3/2$, the recurrence in part (b) becomesNo factor being cancelled is zero in this range, soThus the resulting Frobenius solution isThere is exactly one free parameter, $a_0$, setting the overall scale. The analytic series factor has radius of convergence one. For a real-valued expression take $x>0$; a complex version requires a local square-root branch.

The Taylor series of the logarithm givesTherefore the closed form of the larger-root Frobenius solution isThe series derivation holds initially for $0<x<1$; the closed form extends as a real solution to all $x>0$. To check it without series, put $y=\sqrt{x}v(x)$ in the original equation. All undifferentiated $v$ terms cancel and the remaining expression is $4x^{5/2}[(1+x)v^{\prime \prime }+v^{\prime }]$. Hence $[(1+x)v^{\prime }{]}^{\prime }=0$, and $v=\log (1+x)$ indeed solves it. Its logarithm is analytic near zero, so this is not a forced $\log x$ companion term at the Frobenius point.

At the smaller indicial root $\sigma =1/2$, the $k=1$ equation is $0=0$: it leaves $a_1$ free. For $k\ge 2$, the valid recurrence isChoosing $a_1=0$ makes every later coefficient zero. The resulting terminating Frobenius series is thereforeSubstitution gives $y^{\prime }=a_0/(2\sqrt{x})$ and $y^{\prime \prime }=-a_0/(4x^{3/2})$; the three terms in the equation sum to $a_0\sqrt{x}[-(1+x)-2+(x+3)]=0$.

Termination is a permitted choice, not a condition forced on every smaller-root solution. Retaining the free $a_1$ adds the solution in part (e), so the general real solution on $x>0$ is $\boxed{y=\sqrt{x}[C_1+C_2\log (1+x)]}$. This is exactly the freedom explained by undetermined coefficient at Frobenius resonance. On either negative interval avoiding $-1$ and $0$, analogous real formulas use $\sqrt{|x|}$ and $\log |1+x|$.