= Solution
Multiplication by $(1-x^2)^{-1/2}$ puts the equation in <Sturm-Liouville theory> form:
$$
\boxed{
\frac d{dx}\left(\sqrt{1-x^2}\,T_n'\right)
+\frac{n^2}{\sqrt{1-x^2}}T_n=0}.
$$
Multiply the equations for $T_n,T_m$ by $T_m,T_n$, subtract and integrate. The boundary term vanishes because $\sqrt{1-x^2}=0$ at both endpoints and the <polynomial> <derivatives> are bounded. Therefore, for $n^2\ne m^2$,
$$
\boxed{
\int_{-1}^1\frac{T_n(x)T_m(x)}{\sqrt{1-x^2}}\,dx=0}.
$$
This is the usual <Chebyshev polynomial> orthogonality.
Differentiating the original equation and writing $U_n=T_n'$ gives
$$
\boxed{(1-x^2)U_n''-3xU_n'+(n^2-1)U_n=0}.
$$
Its self-adjoint form is
$$
\boxed{
\frac d{dx}\left((1-x^2)^{3/2}U_n'\right)
+(n^2-1)\sqrt{1-x^2}\,U_n=0}.
$$
The same subtraction argument now yields, for $n^2\ne m^2$,
$$
\boxed{
\int_{-1}^1U_n(x)U_m(x)\sqrt{1-x^2}\,dx=0}.
$$
These two relations form the <Chebyshev derivative Sturm-Liouville pair>.
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