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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 $u{v}^{\prime }-u^{\prime }v=-2/x$. The half-line Green function for the first cylindrical harmonic is thereforeIt is continuous at $x=s$, its first derivative has jump $1/2-(-1/2)=1$, and it solves the homogeneous equation elsewhere. Thus $L_{x}G=\delta (x-s)$ for integration with ordinary measure $ds$. The boundary conditions hold for each fixed $s>0$.

Integrating against the forcing givesSince the two integrals are $2-(x^2+2x+2)e^{-x}$ and $e^{-x}$, respectively,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.