Arnold ODE Problem Problem 1 of Section 1.2.4 of Arnold's ODE book asks

Can the integral curves of a smooth (continuously differentiable) equation $\frac{dx}{dt} = v(x)$ approach each other faster than exponentially as $t\rightarrow \infty$?

It says that the answer is no when one of the curves corresponds to an equilibrium position but is yes otherwise.  
I interpret this to mean that for two integral curves $x_1, x_2$ defined for all values of $t$ larger than some constant, $\lim_{t\rightarrow \infty} e^t |x_1(t) - x_2(t)| = 0$.
But I can't think of an example of a ODE with solutions like that.  I tried the following ODEs


*

*$dx/dt = 1$.  This had solutions that maintained constant distance from each other

*$dx/dt = x^2$. The solutions to this ODE approach 0 for t large but their difference is on the order of $1/t^2$.  

*$dx/dt = -e^t$.
 A: Separate the variables in the original equation:
$$
\frac{dx}{v(x)}=dt
$$
$$
\int_{x(t_0)}^{x(t)}\frac{dx}{v(x)}= \int_{t_0}^{t} dt
$$
Denoting by $w(x)$ the antiderivative of $1/v(x)$, we  can write
$$
w(x(t))-w(x(t_0))= t-t_0.
$$
Let, for simplicity, $t_0=0$; denote $x_0= x(0)$. If we suppose that $w(x)$ has an inverse $w^{-1}(t)$, then the solution to the original equation can be written as
$$
x(t)= w^{-1}(t+w(x_0)).
$$
Let us have two initial values $x_{10}$ and $x_{20}$. The solutions to the corresponding initial value problems are
$$
x_1(t)=w^{-1}(t+w(x_{10})),\quad
x_2(t)=w^{-1}(t+w(x_{20}))
$$
Denote $C_1=w(x_{10})$, $C_2=w(x_{20})$ and write down the difference
$$\tag{1}
x_1(t)-x_2(t)=w^{-1}(t+C_1)-w^{-1}(t+C_2).
$$
We need the expression (1) to decrease faster than any exponent, i.e.
$$\tag{2}
\forall k\in\mathbb R\quad
\lim_{t\to+\infty} \left( w^{-1}(t+C_1)-w^{-1}(t+C_2) \right) e^{kt}= 0.
$$
When searching for possible $w^{-1}$ functions, the first thing that comes to mind is $w^{-1}(t)=e^{-t^2}$. And yes, it satisfies (2):
$$
\lim_{t\to+\infty} \left( e^{-(t+C_1)^2}-e^{-(t+C_2)^2} \right) e^{kt}= 
\lim_{t\to+\infty} e^{-t^2+kt}
 \left( e^{-2tC_1-C_1^2}-e^{-2tC_2-C_2^2} \right)=0.
$$
Now we reconstruct the function $v(x)$ participating in the original equation. Since $w(x)=\sqrt{-\ln x}$ and $w'(x)=1/v(x)$,
$$
w'(x)= \frac{-1}{2\sqrt{-\ln x}\cdot x}
$$
$$
v(x)=-2x\sqrt{-\ln x}.
$$
Finally, we got the equation
$$
\dot x= -2x\sqrt{-\ln x},\quad x\in(0,1).
$$
