Continuous case of the "ant on a rubber band" problem Consider a thin and inﬁnitely stretchable rubber band held taut along an x-axis with its left endpoint ﬁxed at $x = 0$ and its right endpoint initially at $x = c, c > 0$. At time $t = 0$ the band starts to stretch uniformly and smoothly in such a way that the left endpoint remains stationary at $x = 0$ while the right endpoint moves away from the left endpoint with constant speed $v > 0$. A small ant leaves the left endpoint at time $t = 0$ and walks steadily and smoothly along the band towards the target point at a constant speed $\alpha > 0$ relative to the point on the band where the ant is at each moment. Will the ant ever reach the right endpoint?
Now I've seen these types of problems in my calculus class back in the day.
the place to start is to find $\frac{dy}{dt}$ and integrate. I think in this case $\frac{dy}{dt}$ will be the rate the rubber band moves.
Next we might find $\frac{dz}{dt}$ which will be the speed of the bug?
I'm guessing this will not involve a natural log but be a much simpler integrand but I'm not sure how to start.
 A: You don't need DiffEq to show this really.  Allow $ \lambda : [0,1] \to [0,c] $ be a connected path interval along the interval $ [0,c] $ (i.e. rubber band) and let $ \varphi : [0,c] \to [0,d] $ be a differentiable homeomorphism of the interval to a second interval.  Since $ \varphi $ is a homeomorphism it preserves open sets of the interval and the connectedness of the space and since it's differentiable it preserves the smoothness (i.e. the derivatives, i.e. the ability to calculate velocity of the ant).
Then $ \varphi \circ \lambda $ is such a path on $ [0,d] $.
A: For simplicity consider that the mechanical behavior of the rubber is linear. Consider the position of the points on the initial configuration given by $X \in [0,c]$. At time $t>0$ the right end will move to the position $x_c=c+vt$. Because of the assumption of linearity, the position each point $X$  at time $t>0$ is $x=vt/(c+vt)X$ (EDIT: linear $x(X)$ for $t>0$ knowing that the point $x=c+vt$ extends by $vt$ and the point $x=0$ does not extend at all). With this expression is easy to compute the velocity at any point by differentiation. The velocity of the ant is constant with respect to the velocity of the point $x$. The rest is nearly trivial.
A: The left endpoint is at $c+vt$.  If the ant is at $x(t)$ the speed of the ant is $\alpha+\frac x{c+vt}v$ because it is walking at $\alpha$ and the expansion is moving it at $\frac x{c+vt}$.  This gives the differential equation
$$\frac {dx}{dt}=\alpha+\frac x{c+vt}v$$
I could only get Alpha to solve this by interchanging $t$ and $v$.  When I change it back I get
$$x(t) = (\alpha (c + t v) \log(c + t v))/v + k_1 (c + t v)$$
From the fact that $x(0)=0$ we get
$$0=\frac {\alpha c \log c}v+k_1c\\k_1=-\frac{\alpha \log c}{v}$$
So we have to solve
$$c+vt= \frac \alpha v(c + t v) \log(c + t v)) -\frac{\alpha \log c}{v}(c + t v)\\
1+\frac{\alpha \log c}{v}= \frac \alpha v \log(c + t v))$$
for $t$, which is routine algebra.
