= Solution
\b[Boundary form.] For real <functions> and $\mathcal Ly=-(py')'+qy$, <integration by parts> gives
$$
\begin{aligned}
\int_a^b(y_1\mathcal Ly_2-y_2\mathcal Ly_1)\,dx&=\int_a^b[-y_1(py_2')'+y_2(py_1')']\,dx\\
&=[p(y_2y_1'-y_1y_2')]_a^b.
\end{aligned}
$$
The integral terms containing $py_1'y_2'$ cancel, as do the $q$ terms. This is the <Lagrange identity for a Sturm-Liouville operator>. The <boundary conditions> must make this boundary form zero for every admissible pair. Several suitable choices are homogeneous <Dirichlet boundary conditions> at both endpoints; homogeneous <Neumann boundary conditions> at both endpoints; one of each; or separated homogeneous <Robin boundary conditions>
$$
\alpha_a y(a)+\beta_a p(a)y'(a)=0,\qquad\alpha_b y(b)+\beta_b p(b)y'(b)=0,
$$
where each fixed real pair $(\alpha,\beta)$ is nonzero. At each endpoint the two boundary data vectors are then proportional, making their antisymmetric product zero. Another choice is periodic data $y(a)=y(b)$ and $p(a)y'(a)=p(b)y'(b)$, which cancel the two endpoint contributions. These domains give the required <self-adjoint differential operator> in the regular setting.
\b[Eigenfunctions.] Set $x=\cos\theta$, $0\leq\theta\leq\pi$, and $Y(\theta)=y(\cos\theta)$. The <chain rule> gives
$$
y'=-\frac{Y_\theta}{\sin\theta},\qquad(1-x^2)y''-xy'=Y_{\theta\theta}.
$$
The <Chebyshev differential equation> therefore becomes $Y''+n^2Y=0$, with $Y(0)=Y(\pi)=0$. For a positive integer $n$, the first condition excludes the cosine term and the second is satisfied by $\sin(n\theta)$. Thus
$$
\boxed{y_n(x)=N_n\sin(n\arccos x),\qquad\lambda_n=n^2,\qquad n=1,2,\ldots.}
$$
There is no nonzero $n=0$ <eigenfunction>: the transformed solution is linear in $\theta$ and vanishes at both endpoints only if identically zero. Negative $n$ merely repeats a positive mode with the opposite sign, so it is excluded from the indexing of an independent <eigenbasis>.
\b[Weight and normalization.] Multiplying the differential equation by $-1/\sqrt{1-x^2}$ puts it into <Sturm-Liouville form>:
$$
-\frac d{dx}\left(\sqrt{1-x^2}\,y'\right)=\lambda\frac{y}{\sqrt{1-x^2}}.
$$
Hence $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2)^{-1/2}$. Under $x=\cos\theta$, the measure $w(x)\,dx$ becomes $-d\theta$, so
$$
\int_{-1}^1w(x)y_n(x)y_m(x)\,dx=N_nN_m\int_0^\pi\sin(n\theta)\sin(m\theta)\,d\theta=N_nN_m\frac\pi2\delta_{nm}.
$$
Therefore
$$
\boxed{w(x)=\frac1{\sqrt{1-x^2}},\qquad N_n=\sqrt{\frac2\pi}\quad(n\geq1).}
$$
These endpoints are singular because $p$ vanishes and $w$ diverges there, but the weight is integrable. Moreover $py_n'=-N_n n\cos(n\theta)$ remains finite, so the vanishing values of $y_n$ make the boundary form zero. These are the <Dirichlet eigenfunctions of the Chebyshev differential operator>, not the first-kind <Chebyshev polynomials> $\cos(n\arccos x)$, which have nonzero endpoint values.
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