 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 $G=B{x}^{-2}$ for $x>\zeta$. Continuity gives $B=A{\zeta }^3$. Since the operator is $L=(x^{2}d/dx{)}^{\prime }-2$, 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: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 isFor 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.