= Solution
Put $y(x)=w(z)$ with $z=\sin^2x$. The <chain rule> gives
$$
\left(\frac{dz}{dx}\right)^2=4z(1-z),
\qquad
\frac{d^2z}{dx^2}=2(1-2z),
$$
and hence
$$
\frac{d^2y}{dx^2}
=4z(1-z)\frac{d^2w}{dz^2}
+2(1-2z)\frac{dw}{dz}.
$$
After substitution and division by $4z(1-z)$, the <ordinary differential equation> becomes
$$
w''+\left(\frac1{2z}+\frac1{2(z-1)}\right)w'
-\frac{k^2/4}{z(z-1)}w=0.
$$
Comparison with the <Gauss hypergeometric equation> gives, up to interchanging $A$ and $B$,
$$
\boxed{A=\frac{k}{2},\qquad B=-\frac{k}{2},\qquad C=\frac12}.
$$
At $z=0$ and $z=1$, the coefficient of $w'$ has at most a simple pole and that of $w$ has at most a double pole, so both are <regular singular point>[regular singular points]. The <regular singular point at infinity> criterion gives the same conclusion at infinity. The corresponding <Papperitz symbol> is
$$
\boxed{
P\!\left\{
\begin{array}{cccc}
0&1&\infty&z\\
0&0&k/2&\\
1/2&1/2&-k/2&
\end{array}
\right\}.}
$$
Indeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the <Fuchs relation>.
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