Reading the Courant-Hilbert Methods of Mathematical Physics (p.326) we encounter:
"...If we differentiate equation $$ \left[(1-x^2)u')\right]'+\lambda u=0 $$ with respect to $x$, we obtain a differential equation for the function $u'(x)$. ....It follows that a solution which is regular at both end points of the interval exist only for $\lambda=n(n+1)$and is given in this case by $P'_n(x)$. "
Here $P'_n(x)$ is the derivative of the Legendre polynomial, solution of the preceding equation.
I understand that this way. The equation is $(L+\lambda)u=0$ with $L$ the linear operator $L u=\left[(1-x^2)u' \right]'$. If $D$ is the linear operator $Du=u'$ than we have $D(Lu+\lambda u)=DL u+\lambda Du=0$.
But I don't see how to prove the statement from here because, obviously $DL\ne LD$.
Explicitly we have $$ DL u=\left[ (1-x^2)u''-2xu' \right]' $$ so the equation becomes: $$ \left[ (1-x^2)u'' \right]'-2u'-2xu''+\lambda u'=0 $$ or, with $u'=v$ $$ \left[ (1-x^2)v' \right]'-2xv'+(\lambda-2) v=0 $$ and also here i don't see how to proceed.