Consider the differential operator given by $L_{\epsilon}u := -u'' + \epsilon xu$ with $u(0) = u(\pi) = 0$. For $\epsilon = 0$, then the smallest eigenvalue of $L_0$ is $1$ with eigenfunction $\sin x$.
Now suppose $\epsilon > 0$. Let $\lambda$ be an eigenvalue of $L = L_{\epsilon}$ which one can expand as $\lambda = 1 + \epsilon\lambda_{1} + O(\epsilon^{2})$ with eigenfunction $\phi = \sin x + \epsilon\phi_{1} + O(\epsilon^{2})$.
Is there a way to find $\lambda_{1}$ and $\phi_{1}$? Comparing the $\epsilon$ terms, I get that $$-\phi_{1}'' + x\sin x = \phi_{1} + \lambda_{1}\sin x,$$ which I can then solve for $\phi_{1}$ but I don't know a way to find $\lambda_{1}$.