show directly That the eigenfunction of the given Sturm-Liouville problem are orthogonal without explicitly solving for therm , and state the orthogonality conditions
$x{y}''+{y}'+\lambda xy=0$
${y}'(0)=0$
$y(1)=0$
You can help me with this problem, my doubt is that I want to solve this equation by the Cauchy euler, but ${y}'$ does not have the $x$ to be able to use this method, I do not have any change of variable, I am confused on how to raise this problem , I know, I would appreciate any input regarding this
A Sturm-Liouville form of your equation is $$ -(xy')' = \lambda xy $$
Your problem is singular at $x=0$, which means there is no a priori guarantee that solutions will have a finite value or a finite derivative at $x=0$. But, assuming you have two solutions $y_1,y_2$ of the problem that you stated with different corresponding eigenvalues $\lambda_1,\lambda_2$, then $$ (\lambda_1-\lambda_2)\int_0^1 xy_1(x)y_2(x) \\ = \int_0^1 -(xy_1')'y_2+y_1(xy_2')' dx \\ = \int_0^1 \frac{d}{dx}(-xy_1'y_2+xy_1y_2')dx \\ = -xy_1'y_2+xy_1y_2'|_0^1 =0. $$ Because $\lambda_1\ne\lambda_2$, then $\int_0^1 xy_1(x)y_2(x)dx=0$ must hold. So the two solutions will be orthogonal with respect to the weight function $x$.