= Solution
The origin is an <ordinary point>. Substitute $y=\sum_{n\geq0}a_nx^n$ into the equation and collect the coefficient of $x^n$:
$$
(n+2)(n+1)a_{n+2}+(k^2-n^2)a_n=0.
$$
Hence the <recurrence relation> is
$$
\boxed{a_{n+2}=\frac{n^2-k^2}{(n+2)(n+1)}a_n,\qquad n\geq0.}
$$
There are two free initial coefficients. The independent even and odd <power series> begin
$$
\begin{aligned}
y_{\rm e}(x)&=1-\frac{k^2}{2!}x^2+\frac{k^2(k^2-4)}{4!}x^4-\frac{k^2(k^2-4)(k^2-16)}{6!}x^6+\cdots,\\
y_{\rm o}(x)&=x+\frac{1-k^2}{3!}x^3+\frac{(1-k^2)(9-k^2)}{5!}x^5+\cdots.
\end{aligned}
$$
The general solution is $\boxed{y=a_0y_{\rm e}+a_1y_{\rm o}}$. As a useful independent check, the substitution $x=\sin t$ reduces the equation to $d^2y/dt^2+k^2y=0$. Thus
$$
y_{\rm e}=\cos(k\arcsin x),\qquad y_{\rm o}=\begin{cases}\sin(k\arcsin x)/k,&k\neq0,\\\arcsin x,&k=0.\end{cases}
$$
For a nonterminating parity series, $|a_{n+2}/a_n|\to1$, so the <ratio test> in $x^2$ gives $\boxed{R=1}$. The <regular singular points> at $\pm1$ explain the natural obstruction.
There is an important exception to an unconditional radius-one answer. If $k^2=m^2$ for an integer $m\geq0$, the series of parity $m$ terminates at degree $m$ and is a multiple of a <Chebyshev polynomial>. Its <radius of convergence> is $\boxed{R=\infty}$; the other independent parity solution still has radius one. For $m=0$ the even solution is constant. Any nonzero combination containing the nonterminating solution has radius one. These are the full <power series of the Chebyshev differential equation>, including the terminating cases.
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