Put each equation in self -adjoint form
$x^{2}{y}''-x{y}'+\lambda y=0$
If I want to put it in the self-attached form of Sturm Liouvulle, I first do this:
$x{y}''-{y}'+\frac{\lambda }{x}y=0$
$e^{-\int \frac{dx}{x}}$
$e^{-ln(x)}=\frac{1}{x}$
$\frac{1}{x}\left [x{y}''-{y}'+\frac{\lambda }{x}y \right ]=0$
${y}''-\frac{1}{x}{y}'+\frac{\lambda }{x^{2}}y=0$
but I don't know if what I did is good, you can help me
If you want to multiply by $a$ to put in self-adjoint form, $$ x^2 a(x) y''(x)-x a(x)y'(x)+\lambda a(x)y=0, $$ then you must arrange it so that $(x^2 a)'=-xa$, or $$ 2x a + x^2 a'=-xa \\ x^2 a' = -3xa \\ \frac{a'}{a} = -\frac{3}{x} \\ a= Cx^{-3} $$ $C=1$ is good enough. Then the original equation becomes $$ \frac{1}{x}y''-\frac{1}{x^2}y'+\frac{\lambda}{x^3}y=0 \\ \left(\frac{1}{x}y'\right)'+\frac{\lambda}{x^3}y=0 $$
