In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal? In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal?
Comprobe the orthogonality direct by integration.
Where,
$$x^2y''+xy'+\lambda y=0\tag1$$
with boundary condition:
$y(1)=0,\,\,\,y(b)=0$
My attempt
The solution of the ODE is:
$$y(x)=C_1\cos(\ln(x)\sqrt{\lambda})+C_2\sin(\ln(x)\sqrt{x})$$
Note the eigenvalues of $(1)$ are:
$$\lambda_k=\frac{k^2\pi^2}{ln(b)^2}$$
with $b>1$ and $k=0,1,2,...$
The minimun eigenvalue is $\lambda_0=0$ and 
$$\lim_{k\rightarrow\infty}\lambda_k=\infty$$
Moreover, the equation $(1)$ in Sturm-Liouville form is:
$$xy''+y'+\frac{\lambda}{x}y=0\tag2$$
Here, i'm stuck. Can someone help me?
 A: First of all, $\lambda \ne 0$ since it would result in a trivial solution (prove this yourself). So the eigenfunctions are
$$ y_k(x) = \sin\left(k\pi \frac{\ln x}{\ln b}\right) $$
up to a multiplicative constant, with $\lambda_k = \frac{k^2\pi^2}{\ln^2 b}$ and $k = 1,2,3,\dots$
Let's compare 2 distinct eigenfunctions $y_n$ and $y_m$, with corresponding eigenvalues $\lambda_n\ne\lambda_m$. By the Sturm-Liouville form, we have the following relations:
\begin{align} 
\frac{\lambda_n}{x}y_n &= -(x{y_n}')' \\
\frac{\lambda_m}{x}y_m &= -(x{y_m}')' 
\end{align}
Using integration by parts, we can show
\begin{align}
\int_1^b \frac{\lambda_n}{x}y_ny_m\ dx &= -\int_1^b (x{y_n}')'y_m\ dx = \int_1^b x{y_n}'{y_m}'\ dx \\
\int_1^b \frac{\lambda_m}{x}y_my_n\ dx &= -\int_1^b (x{y_m}')'y_n\ dx = \int_1^b x{y_m}'{y_n}'\ dx
\end{align}
It follows that
$$ \int_1^b \frac{\lambda_n}{x}y_ny_m\ dx = \int_1^b \frac{\lambda_m}{x}y_my_n\ dx \implies (\lambda_n-\lambda_m) \int_1^b \frac{1}{x}y_ny_m\ dx = 0 $$
But $\lambda_n\ne \lambda_m$, so the integral must be zero. This means the eigenfunctions $y_n$ and $y_m$ are orthogonal w.r.t the weighing function $w(x)=\frac{1}{x}$

The second part of the question asks you to show this by direct integration, i.e
$$ \int_1^b \frac{1}{x} \sin\left(n\pi\frac{\ln x}{\ln b}\right)\sin\left(m\pi\frac{\ln x}{\ln b}\right) dx = 0 $$
which can be done with the substitution $t = \ln x$
