= Solution
The homogeneous <Cauchy-Euler equation> has powers $x^r$ with $r(r-1)+2r-2=(r-1)(r+2)=0$. Boundedness at zero selects $x$, and boundedness at infinity selects $x^{-2}$. Put $G=Ax$ for $x<\zeta$ and $G=Bx^{-2}$ for $x>\zeta$. Continuity gives $B=A\zeta^3$. Since the operator is $L=(x^2\,d/dx)'-2$, the required delta-function normalization is $\zeta^2[G_x]_{\zeta^-}^{\zeta^+}=1$. Thus $-3A\zeta^2=1$, giving the <bounded Green kernel for a radial Euler operator>:
$$
\boxed{G(x,\zeta)=\begin{cases}-x/(3\zeta^2),&0<x<\zeta,\\-\zeta/(3x^2),&\zeta<x<\infty.\end{cases}}
$$
In the requested notation, $\boxed{a=1,\ b=-2,\ c=1,\ d=-2,\ k_-=k_+=-1/3}$. The two expressions agree at $x=\zeta$.
With ordinary measure $d\zeta$, the solution is
$$
\boxed{y(x)=\int_0^\infty G(x,\zeta)f(\zeta)d\zeta=-\frac1{3x^2}\int_0^x\zeta f(\zeta)d\zeta-\frac x3\int_x^\infty\frac{f(\zeta)}{\zeta^2}d\zeta.}
$$
For bounded forcing these integrals converge and give a bounded solution. It is unique because no nonzero combination of $x,x^{-2}$ is bounded at both ends. For a more general forcing, convergence and boundedness of the displayed integral must be checked.
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