Return to Question

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)?

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 $g(r)$?

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)?

Edit: The answers below suggest this is not true. Let me change a little bit the assumptions on $g(r)$: suppose that there exists $0<\alpha<1$ such that $$ g(r) = O\left(\frac{e^{-kr}}{r^{\alpha}\sqrt{r}}\right) \ \ \ \ \ r\to +\infty . $$ Can we now say that $w(r)=O\left(\frac{e^{-kr}}{\sqrt{r}}\right)$ ?

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)?

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)?

Edit: The answers below suggest this is not true. Let me change a little bit the assumptions on $g(r)$: suppose that there exists $0<\alpha<1$ such that $$ g(r) = O\left(\frac{e^{-kr}}{r^{\alpha}\sqrt{r}}\right) \ \ \ \ \ r\to +\infty . $$ Can we know say that $w(r)=O\left(\frac{e^{-kr}}{\sqrt{r}}\right)$ ?

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)?

Edit: The answers below suggest this is not true. Let me change a little bit the assumptions on $g(r)$: suppose that there exists $0<\alpha<1$ such that $$ g(r) = O\left(\frac{e^{-kr}}{r^{\alpha}\sqrt{r}}\right) \ \ \ \ \ r\to +\infty . $$ Can we now say that $w(r)=O\left(\frac{e^{-kr}}{\sqrt{r}}\right)$ ?