= Solution
Divide the equation by $\sqrt{1-x^2}$ to obtain its <Sturm-Liouville form>:
$$
\boxed{\frac d{dx}\left(\sqrt{1-x^2}\,y'\right)+\frac{n^2}{\sqrt{1-x^2}}y=0.}
$$
Thus $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2)^{-1/2}$. For $L=-w^{-1}(p\,d/dx)'$, the <Lagrange identity for a Sturm-Liouville operator> gives the endpoint form $[p(\bar f'g-\bar f g')]_{-1}^1$. A natural self-adjoint choice is
$$
\boxed{\lim_{x\to-1^+}\sqrt{1-x^2}\,y'(x)=\lim_{x\to1^-}\sqrt{1-x^2}\,y'(x)=0.}
$$
Under $x=\cos\theta$, these become the ordinary zero-derivative conditions at $\theta=0,\pi$ for $-d^2/d\theta^2$. They therefore define the realization whose <Neumann eigenfunctions of the Chebyshev differential operator> are $T_n(x)=\cos(n\arccos x)$, with $n=0,1,\ldots$.
Boundedness alone does not choose these endpoint conditions: $\sin(n\arccos x)$ is also bounded but generally has a nonzero endpoint value of $py'$. Requiring bounded $y'$ is a sufficient stronger regularity condition for the displayed zero-flux limits. Other self-adjoint endpoint choices are possible; the boundary form and the operator domain must specify which is intended.
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