= Solution
For the larger <indicial root> $\sigma=3/2$, the recurrence in part (b) becomes
$$
2k[(2k+2)a_k+2ka_{k-1}]=0,\qquad k\geq1.
$$
No factor being cancelled is zero in this range, so
$$
\boxed{a_k=-\frac{k}{k+1}a_{k-1}=\frac{(-1)^k}{k+1}a_0.}
$$
Thus the resulting <Frobenius solution> is
$$
\boxed{y(x)=a_0x^{3/2}\sum_{k=0}^\infty\frac{(-x)^k}{k+1}=a_0x^{3/2}\left(1-\frac x2+\frac{x^2}3-\cdots\right).}
$$
There is exactly \b[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.
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