= Solution
Substitution of $y=x^p$ into the homogeneous <Euler-Cauchy equation> gives $(p(p-1)+p-1)x^{p-2}=(p^2-1)x^{p-2}$, so $\boxed{p=1,-1}$. Use the solution $u(x)=x$ that vanishes at zero and $v(x)=1/x$ that decays at infinity. Their <Wronskian> is $uv'-u'v=-2/x$. The <half-line Green function for the first cylindrical harmonic> is therefore
$$
\boxed{G(x,s)=\begin{cases}-x/2,&0<x\leq s,\\-s^2/(2x),&0<s\leq x.\end{cases}}
$$
It is continuous at $x=s$, its first <derivative> has jump $1/2-(-1/2)=1$, and it solves the homogeneous equation elsewhere. Thus $L_xG=\delta(x-s)$ for integration with ordinary measure $ds$. The <boundary conditions> hold for each fixed $s>0$.
Integrating against the forcing gives
$$
y(x)=-\frac1{2x}\int_0^xs^2e^{-s}ds-\frac x2\int_x^\infty e^{-s}ds.
$$
Since the two <integrals> are $2-(x^2+2x+2)e^{-x}$ and $e^{-x}$, respectively,
$$
\boxed{y(x)=\frac{(x+1)e^{-x}-1}{x}\quad(x>0),\qquad y(0)=0.}
$$
A <Taylor expansion> gives $y=-x/2+O(x^2)$ near zero, while $y\sim-1/x$ at infinity. This verifies the two endpoint limits; direct differentiation verifies $Ly=e^{-x}$ on $x>0$. The differential expression itself is singular at zero, so the endpoint condition is imposed by continuity. No nonzero combination of $x$ and $1/x$ satisfies both homogeneous endpoint conditions, proving <uniqueness>.
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