The Green's function of following differential operator
$$(\mathcal{L}y)(x)=\frac{d}{dx}\left(x\,\frac{dy}{dx}\right)-\frac{n^2}{x}\,y(x), \:0<x<1$$
with boundary conditions $$y(0)=y(1)=0$$ can be the form of $$G(x,t)= \begin{cases} \dfrac{1}{2n} \left(\dfrac{x}{t}\right)^n(1-t^{2n})& x\leq t\\ \dfrac{1}{2n} \left(\dfrac{t}{x}\right)^{\!n}(1-x^{2n}) & x>t \end{cases}. $$
Now using this, I would like to convert the following ODE to an integral equation for nonzero number $n$ (which I got stuck in) $$x^2y''+xy'+(\lambda x^2-n^2)y=0$$ $$y(0)=y(1)=0. $$ We also can write the above equation in $\mathcal{L}y(x)+\lambda xy=0$ relavant to $y(0)=y(1)=0$, so it may be correct to say $$\int_0^1G(x,s)(\mathcal{L}y(s)+\lambda sy(s))\,ds=0$$
So now it equals $\displaystyle y(x)+\lambda \int_0^xG(x,s)sy\,ds+\lambda \int_x^1G(x,s)sy\,ds=0$.
I'm confused here, since I assumed $\displaystyle y(x)=\int_0^1G(x,s)Ly(s)\,ds $ from this http://www.nada.kth.se/~annak/greens1d_odes.pdf
However from Zeyman, I think we have $\displaystyle Ly(x)=\int_0^1G(x,s)Ly(s)\,ds$, since $G(x,y)$ is a kernel. Anyways, I'm not sure how to solve this question; I appreciate any help.