I was recently posing myself this question. Given the Lagrange DE $$[(1-x^2)u']'+\lambda u=0,$$ where $\lambda$ is a real parameter and $x\in[-1,1]$, it is well known that, if $\lambda=n(n+1)$ for some integer $n$, then we get the Legendre polynomials as solutions of the DE.
However, if we consider a general parameter $\lambda$ and we consider the solution $u=u_\lambda$ which solves the DE with that particular parameter, then it is true that $u\in L^2([-1,1])$? Moreover, do we still have some boundary conditions like $$\lim_{x\to\pm 1}(1-x^2)u(x)=0?$$
Thanks for your attention. Best regards,
-Guido-
I am not sure about the behavior at $\pm 1$, but the solutions to these differential equations are called Legendre functions of the first and second kind. There is a lot of literature about them, and they are implemented in Mathematica and Maple.