The operator $$Au = -\frac{d}{dx}((1-x^2)\frac{du}{dx}), \quad -1<x<1$$ supposedly has eigenfunctions $$P_0 = 1, \quad P_1 = x, \quad P_2 = \frac{3}{2}x^2-\frac{1}{2}x$$ with eigenvalues $$\lambda_0 = 0, \quad \lambda_1 = 2, \quad \lambda_2 = 6.$$
Controlling this, the first two are ok, but the third one seems wrong,
$$AP_2 = -((1-x^2)(3x-\frac{1}{2}))' = -(3x-3x^3-\frac{1}{2}+\frac{x^2}{2})' = (-3+9x^2 -x) \neq \lambda_2\frac{3}{2}x^2 - \lambda_2\frac{1}{2}x \implies AP_2 \neq \lambda_2P_2.$$
What am I missing?
A sanity check: The Legendre polynomials have the same parity as their degree. In particular, the second Legendre polynomial should be an even function (i.e., only contain even powers of $x$). But your $P_2(x)$ is not even, and so cannot be correct.