Verifying an eigenvalue of a Sturm-Liouville problem

Question

Consider the Sturm-Liouville problem $$(xy')'+\lambda y=0, 1<x<2$$ with boundary conditions $y(1)=y(2)=0$.

The question of interest is whether $\lambda=0$ is an eigenvalue of the above SL problem.

My attempt: Expand to get $xy''+y'+\lambda y=0$. Seek solutions of the form $y=x^a$. We have $$x[a(a-1)x^{a-2}]+ax^{a-1}+\lambda[x^a]=0.$$ This simplifies to $a(a-1)x^{a-1}+ax^{a-1}+\lambda x^a=0$. I am not sure here for the next step, so I simply substituted $\lambda=0$ to get $$a(a-1)x^{a-1}+ax^{a-1}=0.$$ This will give $a^2-a+a=0$ or $a=0$.

Isn't the trivial solution if $a=0$? Have I done something wrong? I'd appreciate any help or hint.

Answer 1

For $\lambda = 0$, the eigenvalue equation is $(xy')'=0$, which gives $(xy')=C$ where $C$ is a constant. Then $y'=C/x$ gives $y=C\ln(x)+D$. And $y(1)=0$ forces $D=0$. However, $y=C\ln(x)$ satisfies $y(2)=0$ iff $C=0$, which makes $y\equiv 0$. So $\lambda=0$ is not an eigenvalue.