= Solution
Set $s=\log x$ and $Y(s)=y(e^s)$, so $0\leq s\leq1$. By the <chain rule>, the equation becomes $Y''+2Y'+(1+\lambda)Y=0$. The substitution $Y=e^{-s}v$ reduces it further to
$$
v''+\lambda v=0,\qquad v(0)=v(1)=0.
$$
Multiplication by $\overline v$ and <integration by parts> gives $\lambda\int_0^1|v|^2=\int_0^1|v'|^2>0$ for a nonzero solution, ruling out zero and negative <eigenvalues>. For $\lambda=k^2>0$, the first boundary condition gives $v=A\sin(ks)$, and the second gives $k=n\pi$, $n=1,2,\ldots$. Thus \b[all eigenpairs are]
$$
\boxed{\lambda_n=n^2\pi^2,\qquad y_n(x)=\frac1x\sin(n\pi\log x),\quad n\geq1,}
$$
with arbitrary nonzero scalar multiples of each <eigenfunction>. These are the complete list since all solutions of the constant-coefficient <differential equation> have been considered; the transformed sine system is also complete in the corresponding weighted <Hilbert space>.
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