= Solution
Match the coefficient of $y'$ with the <derivative> of the coefficient of $y''$. For <Sturm-Liouville form> this requires
$$
(rx^2)'=-rx,\qquad x^2r'+3xr=0.
$$
A positive <integrating factor> is therefore $r(x)=x^{-3}$, with any positive constant multiple equally valid. Direct calculation gives
$$
\boxed{r=x^{-3},\qquad\widehat Ly=(x^{-1}y')'+x^{-3}y.}
$$
This uses the plus-divergence sign convention of the question. The usual minus-divergence <Sturm-Liouville operator> is its negative; no extra sign is inserted into the requested <eigenvalue> equation.
Because $r$ never vanishes on the <interval>, multiplying the original <eigenvalue> equation by $r$ leaves the requested weighted <eigenvalue> problem unchanged:
$$
\boxed{\widehat\lambda_k=-k^2\pi^2,\qquad\widehat y_k=x\sin(k\pi\log x).}
$$
The weight in the <inner product> is essential. The substitution $t=\log x$ gives
$$
\int_1^e r\widehat y_j\widehat y_k\,dx
=\int_0^1\sin(j\pi t)\sin(k\pi t)\,dt
=\begin{cases}\tfrac12,&j=k,\\0,&j\ne k.\end{cases}
$$
Thus <orthogonality of Sturm-Liouville eigenfunctions> holds for the <weighted inner product> with weight $x^{-3}$, and the squared <norm> is $N_k=1/2$.
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