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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 $r\to\infty$.