ODE with Bessel decay This is related to my previous question, but it is probably less scary and an expert in using Mathematica could figure out an answer easily.
I would like to estimate the asymptotic behaviour of the solution of the following ODE
$$ w''(r)+\frac{1}{r}w'(r)-k^{2}w(r)=-g(r)$$
where $g$ is a smooth positive function satisfying
$$ g(r) = O\left(\frac{e^{-kr}}{\sqrt{r}}\right) \ \ \ \ \ r\to +\infty . $$ 
Assuming that $w(r)$ decays exponentially fast, is it true that $w(r)$ has exactly the same asymptotic decay as $g(r)$?
 A: I don't think this statement is correct. As a counter example, take
$$g(r)=\frac{e^{-k r} (8 k r-1)}{4 r^{3/2}}$$
which has the desired $r^{-1/2}e^{-kr}$ decay for $r\rightarrow\infty$. It changes sign for small $r$, but I would think that is irrelevant for the large-$r$ decay of $w(r)$. I then note that the function
$$w(r)=e^{-kr}\sqrt{r}$$
solves the differential equation, but decays more slowly than $g(r)$.
A: Maple 2018 solves the ODE under consideration:
dsolve(((D@@2)(w))(r)+(D(w))(r)/r-k^2*w(r) = -g(r))

$$w \left( r \right) ={{\ I}_{0}\left(kr\right)}{\it \_C2}+{{ K}_{0
}\left(kr\right)}{\it \_C1}-\int \!{{ K}_{0}\left(kr\right)}g
 \left( r \right) r\,{\rm d}r{{ I}_{0}\left(kr\right)}+$$ $$\int \!{
{ I}_{0}\left(kr\right)}g \left( r \right) r\,{\rm d}r{{ K}_{0}
\left(kr\right)}
 $$
The assumption $w(r)$  decays exponentially fast implies the constant ${\it \_C2} $ equals zero because of
asympt(BesselI(0, k*r), r, 2)

$$\frac 1 2\,{\frac {\sqrt {2}\sqrt {{k}^{-1}}{{\rm e}^{kr}}\sqrt {{r}^{-1}}}{
\sqrt {\pi}}}+O \left(  \left( {r}^{-1} \right) ^{3/2} \right)
 $$
and
asympt(BesselK(0, k*r), r, 1)

$$\frac 1 2\,{\frac {\sqrt {2}\sqrt {\pi}{{\rm e}^{-kr}}\sqrt {{r}^{-1}}}{
\sqrt {k}}}+O \left(  \left( {r}^{-1} \right) ^{3/2} \right)
 $$
Hope that would be useful.
Addition. The Mathematica 11.3 command
AsymptoticDSolveValue[ D[w[r], {r, 2}] +D[w[r], r]/r - k^2*w[r] == -Exp[-k*r]/Sqrt[r],w[r], 
{r, Infinity, 1}, Assumptions -> k > 0]

disproves the suggestion of the question. See the long output here https://www.dropbox.com/s/on0jz2kpb7bin7l/AS.pdf?dl=0 and pay your attention to the term $\frac{\sqrt{r}}{2 k} $.
A: This answer provides a slight modification of the excellent answer by Carlo Beenakker, to get $g>0$ on the entire interval $(0,\infty)$: if 
\begin{equation}
 w(r)=e^{-k r} \left(32 k \sqrt{r}-\frac{2}{\sqrt{r}}\right) 
\end{equation}
and 
\begin{equation}
 g(r)=\frac{e^{-k r} \left(128 k^2 r^2-16 k r+1\right)}{2 r^{5/2}} 
\end{equation}
for some $k\ne0$ and all real $r>0$, 
then $g>0$ on $(0,\infty)$, the equation $w''(r)+\frac1r\,w'(r)-k^2w(r)=-g(r)$ is satisfied for all real $r>0$, and $g(r) = O(e^{-kr}/\sqrt r)$ as $r\to\infty$, whereas $ w(r)\sim 32k e^{-k r} \sqrt r\ne O(e^{-kr}/\sqrt r)$ as $r\to\infty$. 
