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The solution decays like $r^{2-n}$ if $n \geq 3$. Setting $u=r^{\frac{1-n}{2}} v$, then $$v''-\frac{(n-1)(n-3)}{4r^2} v +g(r)v=0$$ with $g=4q$. This gives the result if $n=3$ since $rg(r) \in L^1$ and then all solutions of the above equation behave like $1,r$ as $r \to \infty$ so that the decaying solution behaves like $r^{-1}$. If $n >3$ one obtains the result by using the Liouville transformation for the equation satified by $v$, as in Chapter 6 of the book by Olver, Asymptotics and special functions (with some care because the functions $\psi_{f,g}$ does not belong to $L^1$). I can write down the details if somebody is interested but I do not see a short way as above.