Asymptotic behaviour for ODE Imagine I have an ordinary linear homogeneous second-order differential equation of the form:
$$
y''=f(x)y'+g(x)y,
$$
where the functions $f$ and $g$ have the regularity you wish, are bounded, and satisfy
$$
\lim_{x\rightarrow \infty}f(x)=a, \qquad \lim_{x\rightarrow \infty}g(x)=b,
$$
where $a$ and $b$ are real constants.
Heuristically, this tells that the asymptotics of a solution of our ODE will behave as the solution of the autonomous ODE:
$$
v''=av'+bv.
$$
For example, if we have a solution for the first equation, $y$, converging to zero, and $v$ is a solution for the second equation converging to zero (i.e., exponential decay), do we have
$$
y(x)=O(v(x)),\qquad v(x)=O(y(x))
$$
as $x \rightarrow \infty$?
EDIT:
To clarify my question, I will point out here an example suggested in one of the answers. Let $f(x)=e^{-x}+1$ ang $g(x)=0$. Then we have
\begin{equation}
y''(x)+(e^{-x}+1)y'(x)=0 \implies y(x) = c_1e^{e^{-x}}+c_2.
\end{equation}
And between all these solutions, take for example the ones converging to zero, so take $c_1=-c_2$.
I claim that asymptotically this solution and the solutions converging to zero of the approximation
\begin{equation}
y''(x)+y'(x)=0 \implies y(x) = \tilde{c}e^{-x}
\end{equation}
are the same. In this case, this is true, since
$$
\lim_{x \rightarrow \infty} \frac{c_1e^{e^{-x}}-c_1}{\tilde{c}e^{-x}}=c_1/\tilde{c}.
$$
Question:
Is this true in general? If so, is there any kind of result I can use to formalize this? Note that I assumed the solutions to converge to zero for simplicity, but I think the asymptotics would be the same between any two reasonable solutions (as far as we do not compare solutions with a finite limit and solutions going to infinity, for example).
I have tried to compute the error between the approximation and the first solution, and prove that this error has a faster decay than the solution itself, but it didn't work.
 A: You cannot have that much accuracy with that weak assumptions. Take $y(x)=e^{-x}$ solving $y''=y$, say. Now take $Y(x)=xe^{-x}$. Then $Y''=(x-2)e^{-x}=\frac{x-2}xY$ (assuming the domain $x\ge 1$, say), so they satisfy asymptotically the same equation but $Y$ is much larger near $\infty$. In general, just take any exponentially decaying solution of any equation with constant coefficients, multiply it by any function whose derivatives are o-small of its value as $x\to\infty$ and you will construct a similar example with pretty much any sub-exponential deviation. The life gets better if you impose some particular speed of convergence of $f$ and $g$ to their limiting values, but it is a separate story.
A: Counter example (I think):
Let $f(x)=e^{-x}+1$ and $g(x)=0$ so that
\begin{equation}
\lim_{x\rightarrow\infty} f(x) = 1.
\end{equation}
We have that
\begin{equation}
y''(x)+(e^{-x}+1)y'(x)=0 \implies y(x) = c_1e^{e^{-x}}+c_2.
\end{equation}
From your heuristic proposition we have:
\begin{equation}
y''(x)+y'(x)=0 \implies y(x) = c_1e^{-x}+c_2.
\end{equation}
Lastly, we have
\begin{equation}
\lim_{x\rightarrow\infty} c_1e^{e^{-x}}+c_2 = c_1+c_2 \neq c_2 = \lim_{x\rightarrow\infty}  c_1e^{-x}+c_2.
\end{equation}
which completes the counter example.
A: As far as I understand the question, it can be addressed as follows.
Define $\tilde f(x) := f(x) - a$ and $\tilde g(x) := g(x) - b$. Let $y(x)$ be the solution of $y'' = f(x)y'+g(x)y$ and $z$ be the solution of $z'' = az'+bz$. Define $e(x) = y(x)-z(x)$. Then $e' = y'-z'$ and $$e'' = \tilde f(x)y' + ay' + \tilde g(x) y + b y - az' - bz$$ that is $$e'' = ae' + b e + \tilde f(x)y' + \tilde g(x) y.$$
The solution $y$ converges to $z$ iff $e$ goes to zero. For that one, we need that

*

*the solution $z$ of $z'' = az'+bz$ is (exponentially) converging to zero (upd: $a$ and $b$ both negative)

*the extra term $\tilde f(x)y' + \tilde g(x) y$ converges to zero, e.g., $y$ and $y'$ are bounded.

I would say that the second item follows from the first under some assumptions on $f$ and $g$.
A: Even though I gave the bounty to the answer of @fedja, due to his clarifying counterexample, I think it would be useful to also write here what I have discovered about this topic.
As the counterexample shows, the statement is not true in general. But we can still say something if we have more information about the functions $f$ and $g$.
Assume $\Lambda=\{\lambda_1(x), \lambda_2(x)\}$ is a complete set of roots for the characteristic equation
$$
r^2-f(x)r-g(x)=0.
$$
Then if $\Lambda$ posses certain technical properties which I am not going to develop here (roughly speaking, we are asking for the asymptotics of $\lambda_1$ and $\lambda_2$ to be different enough), then the differential equation
$$
y''-f(x)y'-g(x)y=0
$$
has a fundamental system of solutions of the form
$$
y_1(x)=\exp \left[\int\gamma_1(x)\, dx \right], \qquad y_2(x)=\exp \left[\int\gamma_2(x)\, dx \right]
$$
where $\gamma_1, \gamma_2$ are functions satisfying
$$
\gamma_1(x)\sim \lambda_1(x), \qquad \gamma_2(x)\sim \lambda_2(x)
$$
as $x\rightarrow \infty$. The details of this statement and a proof can be found in the book "Asymptotics of Linear Differential Equations" by Lanstman, Theorem 10.29.
Example:
In the comments, the following example was given as a proof that my original statement was not correct, but we can see that it is easy to find the asymptotics of such an equation with the theorem just mentioned. Consider the ODE:
$$
y''=\frac{x-2}{x}y,
$$
and we look for the asymptotics of any solution approaching zero. In this case, the function $g(x)=\frac{x-2}{x}$ satisfies all the conditions of the theorem (if you do not believe me, check the book I cited). And the set
$$
\left\{\sqrt{\frac{x-2}{x}}, -\sqrt{\frac{x-2}{x}}\right\}
$$
is a complete set of roots for the corresponding characteristic equation. Since
$$
\sqrt{\frac{x-2}{x}}=1-\frac{1}{x}+o(\frac{1}{x^2})
$$
as $x\rightarrow \infty$, then we have that every solution approaching zero will behave as
$$
\exp \left[\int - \sqrt{\frac{x-2}{x}}\right]=\exp \left[\int -x+\frac{1}{x}+o(\frac{1}{x^2})\right]\sim  Cxe^{-x}.
$$
We can double-check this with the general explicit solutions of our equation that approach zero: $y(x)=Cxe^{-x}$. (The fact that we obtained exactly the solution of the equation is a coincidence, the theorem only gives the asymptotic behavior of the solutions, not their explicit form).
