Eigen Value of differential equation

Question

How can I find eigenvalue and eigenvector of the following:

$$Sp(x)= p''(x)-2xp'(x)$$

In class we only covered matrices for eigenvector. I have no clue how to approach differential equation. Someone please help.

Answer 1

$$ S 1 = 0\\ S x = -2 x\\ S x^2 = 2-4x^2\\ = 2*1 + 0*x + (-4)*x^2 $$

If you let $p(x)=1$, you get $Sp(x) = 0 = 0*p(x)$ so $\lambda=0$ for the eigenvector $p(x)=1$.

Another one is also apparent already. Let $p(x)=x$, you get $Sp(x)=-2x$, so $\lambda=-2$ for the eigenvector $p(x)=x$.

Put the matrix together in this basis $\{ 1,x,x^2\}$

$$ S = \begin{pmatrix} 0 & 0 & 2\\ 0 & -2 & 0\\ 0 & 0 & -4\\ \end{pmatrix} $$

This is the matrix you seek to find the eigenvalues and eigenvectors for. There are the two that we've already found. The only one left will turn out to be $\lambda=-4$ with $p(x)=-1+0*x+2*x^2$. In that case $Sp(x)=4-8x^2=(-4)*p(x)$.