= Solution
Seek a positive weight such that the <differential operator> becomes $-w^{-1}(py')'$. Matching the second <derivative> gives $p=w(1-x^2)$, while matching the first <derivative> requires $p'=-xw$. Consequently
$$
(1-x^2)w'-2xw=-xw,\qquad \frac{w'}w=\frac{x}{1-x^2}.
$$
Choosing the multiplicative constant to be one gives
$$
\boxed{w(x)=\frac1{\sqrt{1-x^2}},\qquad p(x)=\sqrt{1-x^2},\qquad -(py_n')'=n^2wy_n}.
$$
This is <Sturm-Liouville form> with zero potential term. Multiply the equations for $y_n,y_m$ by $y_m,y_n$, subtract, and integrate. The <Lagrange identity for a Sturm-Liouville operator> gives
$$
(n^2-m^2)\int_{-1}^1wy_ny_m\,dx
=\left[p(y_ny_m'-y_my_n')\right]_{-1}^{1}=0.
$$
The boundary expression vanishes because $p\to0$ and the functions and their <derivatives> are finite. Since $n,m\geq0$ are distinct, their <eigenvalues> $n^2,m^2$ are distinct, proving
$$
\boxed{\int_{-1}^{1}\frac{y_n(x)y_m(x)}{\sqrt{1-x^2}}\,dx=0\quad(n\ne m)}.
$$
The endpoint singularities of the weight are integrable, so these <weighted inner products> exist.
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