Given the folllowing ODE, find its eigenvalues and eigenfunctions:
$$ x^2y''-xy'+\lambda y=0,y(1)=y(L)=0,L >1 $$
Doing the analysis for each condition of $\lambda, i.e.,\lambda=0, \lambda>1, \lambda<1, \lambda=1$ (the last three values came using a analysis of euler-cauchy equation).
But I didn't found any solution using this method, always I found only the trivial solution, $y=0$.
Can someone show me a better approach?
Thanks for attention.
The Euler equation $x^2y''(x)-xy'(x)+\lambda y(x)=0$ has a solution $x^{\alpha}$ where $\alpha$ satisfies the indicial equation $$ \alpha(\alpha-1)-\alpha+\lambda=0, \\ \alpha^2-2\alpha+\lambda = 0. $$ This equation has solutions $$ \alpha = 2\pm\frac{\sqrt{4-4\lambda}}{2}=2\pm\sqrt{1-\lambda} $$ You can choose a specific solution $y(x)$ that satisfies $y(1)=0,y'(1)=1$. Such a solution may be written as $$ y(x,\lambda)=\frac{x^{2+\sqrt{1-\lambda}}-x^{2-\sqrt{1-\lambda}}}{2\sqrt{1-\lambda}}. $$ Note that the added condition $y'(1,\lambda)=1$ forces the function $y(x,\lambda)$ to be an entire function of $\lambda$, which means that the eigenvalues are the zeroes of an entire function of $\lambda$. So the eigenvalues have no finite point of accumulation, which can be demonstrated directly by rewriting $y(x,\lambda)$ in the form $$ y(x,\lambda)=x^2\frac{e^{\sqrt{1-\lambda}\ln (x)}-e^{-\sqrt{1-\lambda}\ln(x)}}{2\sqrt{1-\lambda}} = x^2\frac{\sinh(\sqrt{1-\lambda}\ln(x))}{2\sqrt{1-\lambda}} $$ Therefore, the eigenvalue equation $y(L,\lambda)=0$ has the form $$ \sinh(\sqrt{1-\lambda}\ln(L))=0,\;\; \lambda\ne 1, $$ which can be explicitly solved: $$ \sqrt{1-\lambda}\ln(L)=\pm i\pi,\pm 2i\pi,\cdots \\ (1-\lambda)\ln(L)^2=-\pi^2,-4\pi^2,-9\pi^2,\cdots \\ 1-\lambda=-\frac{\pi^2}{\ln(L)^2},-\frac{4\pi^2}{\ln(L)^2},-\frac{9\pi^2}{\ln(L)^2},\cdots \\ \lambda=1+\frac{\pi^2}{\ln(L)^2},1+\frac{4\pi^2}{\ln(L)^2},1+\frac{9^2\pi^2}{\ln(L)^2},\cdots . $$