= Solution
Expand the <radial modified Helmholtz equation> as $Ly=-y''-2y'/x+y$. Setting $y=z/x$ gives $Ly=(-z''+z)/x$, so
$$
y(x)=\frac{Ae^x+Be^{-x}}x.
$$
Boundedness at zero requires $A+B=0$, whereas boundedness at infinity requires $A=0$. The respective adapted <homogeneous solutions> are therefore
$$
\boxed{u(x)=\frac{\sinh x}{x}\quad\text{at zero},\qquad v(x)=\frac{e^{-x}}x\quad\text{at infinity}.}
$$
The first has limit one at zero. Their <Wronskian> is $uv'-u'v=-x^{-2}$. Since the highest-derivative coefficient of $L$ is $-1$, the <Green function> must have first-derivative jump $-1$. The <radial modified Helmholtz Green function on the half-line> is
$$
\boxed{G(x,\xi)=\frac\xi x\sinh(\min(x,\xi))e^{-\max(x,\xi)},\qquad x,\xi>0.}
$$
It is continuous at the source, and its <derivative> jump is $-1$, giving $L_xG=\delta(x-\xi)$. For fixed source it tends to $\xi e^{-\xi}$ as $x\downarrow0$ and decays at infinity. This kernel is for the original operator and ordinary measure $d\xi$; multiplying the equation by $x^2$ would change the source normalization.
For the compact source, split the <integral> at $\xi=x$ when $x<R$. The needed antiderivatives are
$$
\int\xi\sinh\xi\,d\xi=\xi\cosh\xi-\sinh\xi,\qquad\int\xi e^{-\xi}d\xi=-(\xi+1)e^{-\xi}.
$$
Thus
$$
y(x)=\frac{e^{-x}}x\int_0^x\xi\sinh\xi\,d\xi+\frac{\sinh x}{x}\int_x^R\xi e^{-\xi}d\xi\quad(x<R),
$$
while for $x>R$ the result is $e^{-x}x^{-1}\int_0^R\xi\sinh\xi\,d\xi$. Simplifying gives
$$
\boxed{y(x)=\begin{cases}\displaystyle1-(R+1)e^{-R}\frac{\sinh x}{x},&0\leq x\leq R,\\\displaystyle(R\cosh R-\sinh R)\frac{e^{-x}}x,&x\geq R.\end{cases}}
$$
The continuous value at zero is $1-(R+1)e^{-R}$, and $y'(0)=0$. Both $y$ and $y'$ match at $R$, so no extra <Dirac delta> appears there. The only <homogeneous solution> bounded at both ends is zero, establishing uniqueness.
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