= Solution
At the smaller <indicial root> $\sigma=1/2$, the $k=1$ equation is $0=0$: it leaves $a_1$ free. For $k\geq2$, the valid recurrence is
$$
a_k=-\frac{k-1}{k}a_{k-1}.
$$
Choosing $a_1=0$ makes every later coefficient zero. The resulting <terminating Frobenius series> is therefore
$$
\boxed{y(x)=a_0\sqrt{x}.}
$$
Substitution gives $y'=a_0/(2\sqrt{x})$ and $y''=-a_0/(4x^{3/2})$; 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 $\log|1+x|$.
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