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

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 $\log |1+x|$.