Differential approach to solve the 'ant on rubber rope' problem I have been trying to solve the Ant on Rubber Rope problem myself, but I hit some end. Maybe I made a mistake. 
The 'Ant on a rubber rope' problem describes an infinite elastic rubber band, whose end (denoted by $y$) is moving from the origin with a velocity $u$. On the rubber band there is an ant (at position $x$) which crawls much slower towards the end of the rubber band, with a velocity $v$ ($u>v$). 
I am concentrating on the relative position of the ant to the end of the rubber band 
$$\epsilon = \frac{y}{x}$$
So at the beginning the ant is at position $x_0$ and the end of the rubber band is $y_0$:
$$\epsilon_0 = \frac{x_0}{y_0}$$
This value would not change if the ant were stationary and not moving, as the ant's position would stretch relative to the rubber band. But the ant is moving, so after some time $\Delta t$ the end of the rubber band is at 
$$y_1 = y_0 + u \Delta t$$
and the ant's new position is
$$x_1 = x_0 + \epsilon_0\cdot u\cdot \Delta t + v\Delta t$$
The second term is because of the stretch of the rubber band, the third term is the motion of the ant itself. 
Now I calculate the difference $\epsilon_1 - \epsilon_0$ which I find to be
$$\Delta \epsilon = \epsilon_1 - \epsilon_0 = \frac{v\Delta t}{y_0 + u\Delta t}$$
and it follows
$$d\epsilon = \frac{v\,dt}{y_0 + u\, dt}$$
when I have not made a mistake. But if this is right, you can transform this equation to 
$$d\epsilon y_0 + u\,dt\,d\epsilon= v\,dt$$
where you can neglect the double differential term and you end up with 
$$\epsilon = \frac{v\,t}{y_0}$$
which I believe is not a correct solution. 
Where did I made the mistake?
 A: You seem to be confusing $t$, $\Delta t$, and $dt$ a bit.
Let $t$ be the current time, so at time $t$ the length of the band is $y = y_0 + ut$.
At time $t$, the ant is at position $x=\epsilon y$ moving at speed $v$ relative to the band. The change in these values during a time period $dt$ would be $dx = \epsilon\,dy + y\,d\epsilon = (\epsilon u + v)\,dt$ as $dy = u\,dt$ due to the band expanding. This is what you have derived, but expressed with $\Delta t$ instead of $dt$.
The speed of the ant relative to the band is $y\,d\epsilon$ which should be equal to $v\,dt$. You can solve this in terms of $x$, but as you seem to have realised, it is easier to do it in terms of $\epsilon$.
Since $y\,d\epsilon=v\,dt$, we get $d\epsilon/dt = v/y = v/(y_0+ut)$. You almost get to this, but since you have been using $\Delta t$ as both time $t$ and change in time $dt$, you end with the expression $\delta\epsilon=v\,\Delta t/(y_0+u\,\Delta t)$ and replace $\Delta t$ with $dt$, although one represents $t$.
Integrate $d\epsilon/dt = v/(y_0+ut)$ and set $\epsilon=0$ at $t=0$, and you get
$$
\epsilon = \frac{v}{u}\cdot\ln\left(1+\frac{ut}{y_0}\right).
$$
A: The culprit is in your use of the indexes zero and one like $x_0$ and $x_1$.
You found out the relative speed of the ant at the fixed time $t=0$, but... this should not be applied to all the relative speed at other times.
We should instead generalize your equations:
$$ \begin{array}{}
\text{The relative position:}& \epsilon_t &=& \frac{x_t}{y_t}\\
\text{The end position:}& y_{t+\Delta t} &=& y_t  +\hphantom{\epsilon_t} u \Delta t\\
\text{The ant position:}& x_{t+\Delta t} &=& x_t +\epsilon_tu \Delta t +v\Delta t   
\end{array}$$
The relative position is than changing as:
$$\Delta \epsilon_t = \epsilon_{t+\Delta t} - \epsilon_{t} = \frac{x_t +\epsilon_tu \Delta t +v\Delta t }{y_t  + u \Delta t} - \frac{x_t}{y_t} = \frac{ v\Delta t }{y_t  + u \Delta t} \neq \frac{v\Delta t}{y_0 + u\Delta t}  $$
So you will get 
$$ \frac{d \epsilon_t }{dt} =\lim_{\Delta t \to 0}  \frac{\Delta \epsilon_t}{ \Delta t} = \lim_{\Delta t \to 0}  \frac{v}{y_t + u\Delta t} =\frac{v}{y_t} = \frac{v}{y_0+ut} $$
This means that the relative speed $\frac{d \epsilon_t }{dt}$ of the ant is not constant $\frac{v}{y_0}$, but instead decreasing in time $\frac{v}{y_t}$ as the rope length $y_t$ is increasing in time.
This will eventually lead to:
$$ \epsilon_t = \epsilon_0  + \frac{v}{u} \log \left(1 + \frac{ut}{y_0} \right)$$
