Asymptotic behavior of the solution I just started research in physics, but along the way, I want to show something like
$$\partial_x \beta(x) \sim -\frac{A}{x^2}[\alpha(x)+\beta(x)]\,\, as\,\, x\rightarrow\infty$$
where $\alpha(x)$ and $\beta(x)$ are non-negative monotonically decreasing much faster than $\frac{1}{x^n}$ (something like exponential decay or Gaussian) with a constant $A$.
After doing method of integrating factor, I am putting a boundary condition that $\beta(x)$ at infinity is equal to 0. Then I tried to bound the integral or show some asymptotic behavior, but it seems like my math is not sufficient enough.
Anyway, my goal is to show $\alpha$ is subdominant to $\beta$ at infinity or vice versa, how should I approach these kind of problem without losing generality (not assuming they are decaying like exp or gaus).
Any suggestion for how to show asymptotic behavior of beta would be appreciated! thank you
 A: Let $f=\frac{A}{x^2}$,
$$\beta_x=-f\beta-\alpha f$$
$$\frac{\partial}{\partial x}e^{\int f dx}\beta=-e^{\int f dx}\alpha f$$
We observe that
$$e^{\int f dx} = k e^{-\frac{A}{x}}$$
So
$$ke^{-\frac{A}{x}}\beta=C - \int{k \frac{A}{x^2}e^{-\frac{A}{x}}\alpha dx}$$
Let $I$ be the right hand integral. Since $\alpha < Dx^{-n}$,
$$0<I<kD\int{\frac{A}{x^{2+n}}e^{-\frac{A}{x}}dx}=\frac{kD}{A^{n+1}}\Gamma(n+1,\frac{A}{x})$$
That means,
$$\beta>\left(\frac{C}{k}-\frac{D}{A^{n+1}}\Gamma(n+1,\frac{A}{x})\right)e^{\frac{A}{x}}$$
where $\Gamma(n+1, u)$ is the incomplete Gamma function. Using the relation,
$$\Gamma(n+1,\frac{A}{x})=n!e^{-\frac{A}{x}}\sum_{m=0}^{n}\left(\frac{1}{m!}\left(\frac{A}{x}\right)^m\right)$$
Thus,
$$\beta > \frac{C}{k}e^{\frac{A}{x}} -\frac{n!D}{A^{n+1}} \sum_{m=0}^{n}\left(\frac{1}{m!}\left(\frac{A}{x}\right)^m\right)$$
Upon expanding the exponential term into a series, we note that the coefficients for the $x^{-j}$ terms when $0<j<n$ are nonzero. Strictly speaking then, $\beta$ cannot fall as $O(x^{-n})$, and decays more slowly than $\alpha$, but does tend towards $0$ as $x\rightarrow \infty$.
