= Solution
Multiplying by $y$ and integrating with the zero endpoint values first gives $\lambda\int_1^by^2dx/x=\int_1^bx(y')^2dx>0$ for a nonzero real <eigenfunction>; hence $\lambda>0$. Using the printed <change of variables> $\xi=\lambda\ln x$, write $y(x)=Y(\xi)$. Then $(xy')'=\lambda^2Y_{\xi\xi}/x$, so $Y_{\xi\xi}+Y/\lambda=0$. The first endpoint gives $Y=A\sin(\xi/\sqrt\lambda)$, and the other gives $\sin(\sqrt\lambda\ln b)=0$. Thus
$$
\boxed{\lambda_n=\left(\frac{n\pi}{\ln b}\right)^2,\qquad y_n(x)=A_n\sin\left(\frac{n\pi\ln x}{\ln b}\right),\quad n=1,2,\ldots.}
$$
This is the <logarithmic change of variable in a Sturm-Liouville problem>. The measure $dx/x=d(\ln x)$ gives squared <weighted norm> $A_n^2\ln b/2$, so an orthonormal choice is $A_n=\sqrt{2/\ln b}$.
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