= Solution
Normalize the free leading coefficient to $a_0=1$. The <linear recurrence relation> from part (b) gives
$$
4a_2+4a_0=0,\qquad16a_4+2a_0=0,\qquad36a_6+12a_4=0.
$$
Consequently the first four nonzero coefficients are
$$
\boxed{a_0=1,\qquad a_2=-1,\qquad a_4=-\frac18,\qquad a_6=\frac1{24},}
$$
and the single pure <Frobenius solution> is
$$
\boxed{y_1(z)=\frac1z\left(1-z^2-\frac18z^4+\frac1{24}z^6+O(z^8)\right).}
$$
For clarity about the second independent solution, a choice with no added multiple of $y_1$ is
$$
y_2=y_1\log z+\frac52z+\frac38z^3-\frac{11}{144}z^5+\frac{131}{9216}z^7+O(z^9).
$$
These companion coefficients follow by <reduction of order> with the normalized <Wronskian> $W=z^{-3}(1+z^2)^{3/2}\exp[3z^2/(2(1+z^2))]$. The logarithm requires a local branch; it does not supply a second ordinary <power series> solution.
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