= Solution
{c}
Write the operator in divergence form:
$$
\mathcal Ly=\big((1-x^2)y'\big)'.
$$
The <Legendre differential equation> therefore gives
$$
\boxed{\mathcal LP_n=-n(n+1)P_n,\qquad n=0,1,2,\ldots}.
$$
These are the <eigenvalues> of $\mathcal L$. With the common positive-sign convention for a <Sturm-Liouville problem>, use $-\mathcal L$ instead:
$$
-\big(pP_n'\big)'=\lambda_n wP_n,\qquad
\boxed{p=1-x^2,\quad q=0,\quad w=1,\quad\lambda_n=n(n+1)}.
$$
This is a singular <Sturm-Liouville problem>, because $p$ vanishes at the endpoints. Bounded <eigenfunctions> select the nonlogarithmic branches and satisfy $pP_n'\to0$ there. The <Lagrange identity for a Sturm-Liouville operator> gives
$$
[m(m+1)-n(n+1)]\int_{-1}^1P_mP_n\,dx
=[(1-x^2)(P_mP_n'-P_nP_m')]_{-1}^1=0.
$$
The <eigenvalues> are distinct, proving the <orthogonality> condition
$$
\boxed{\int_{-1}^1P_m(x)P_n(x)\,dx=0\quad(m\ne n)}.
$$
The weight is one. No choice of normalization of the <Legendre polynomials> is needed for this conclusion.
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