= Solution
{c}
Normalize both leading coefficients to one. For the <indicial root> $a=1/2$, the <recurrence relation> from part (b) gives
$$
c_{2k}=\frac{(-1)^k}{(2k+1)!},\qquad c_{2k+1}=0.
$$
Recognizing the <Taylor series> of the <sine> gives the first <Frobenius solution>
$$
\boxed{y_+(x)=x^{1/2}\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k+1)!}
=\frac{\sin x}{\sqrt x}.}
$$
For the smaller <indicial root> $-1/2$, $n(n-1)d_n=-d_{n-2}$. The equation at $n=1$ is $0=0$; choose the free coefficient $d_1=0$ to avoid adding the first <Frobenius solution>. The remaining coefficients are
$$
d_{2k}=\frac{(-1)^k}{(2k)!},\qquad d_{2k+1}=0.
$$
Thus the <Taylor series> of the <cosine> gives
$$
\boxed{y_-(x)=x^{-1/2}\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}
=\frac{\cos x}{\sqrt x}.}
$$
Their leading terms are $x^{1/2}$ and $x^{-1/2}$, as required for Case IIIb. A direct check uses $y=x^{-1/2}u$: the <differential equation> becomes $x^{3/2}(u''+u)=0$, so these two functions solve it on $x>0$. Their <Wronskian> is $-1/x\ne0$, proving <linear independence>. The <general solution> is $(C_1\sin x+C_2\cos x)/\sqrt x$.
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