Poor Bug, Will it ever succeed?

Assume at time $\displaystyle t$ , the bug is at a distance $\displaystyle x$ from the wall.
At the same time, the free end of the rubber band is at a distance $\displaystyle L+vt$ from the wall.

Since the band stretches uniformly,
The speed of stretching at the position of the bug is,
$v\frac{x}{L+vt}$

Thus, the relative velocity of the bug with respect to the wall is,
$v\frac{x}{L+vt} - u = \frac{dx}{dt}$

Solving the differential equation, and substituting $\displaystyle x=0$ , we get,

$t = \frac{L}{v}\left(e^{v/u}-1\right) \approx \boxed{95.427sec}$

$\displaystyle\dot x=\frac{3x}{15+3t}-1=\frac x{5+t}-1$

$\displaystyle\dot x-\frac x{5+t}=-1$

$\displaystyle\dot x+f(t)x=g(t)$

The integrating factor is $\large e^{\int f(t)\ \mathrm dt}=e^{\int-\frac{\mathrm dt}{5+t}}=e^{-\ln|t+5|}=\frac 1{t+5}$

Multiplying the whole equation by the integrating factor.

$\displaystyle\frac{\dot x}{t+5}-\frac x{(t+5)^2}=\frac{-1}{t+5}$

$\displaystyle\int\left[\frac{\dot x}{t+5}-\frac x{(t+5)^2}\right]\mathrm dt=\int\frac{-\ \mathrm dt}{t+5}$

$\displaystyle\frac{x}{t+5}=\int\frac{-\ \mathrm dt}{t+5}$

$\displaystyle\frac{x}{t+5}=-\ln(t+5)+C$

Putting $t=0$ and $x=15$ gives $C=3+\ln5$ .

$\displaystyle\frac{x}{t+5}=-\ln(t+5)+\ln5+3=\ln\frac{5e^3}{t+5}$

$\displaystyle x=(t+5)\ln\frac{5e^3}{t+5}$

Solving $x=0$ gives $t=95.43$ .

Let x be the distance of the bug from the wall. dx/dt = u + [x/(L+vt)]v

Consider the auxiliary variable z = x/(L+vt). Using this new variable leads to: d/dt[(L+vt)z]=u+vz

or

(L+vt)dz/dt + vz = u + vz

or

dz = u dt/ (L+vt)

Integrating both sides yields z(t) = (u/v) ln [1+vt/L]

By using the condition z=1 when the beg reaches the end of the rubber band, we can get the fllowing through elementary algebra:

t = (L/V) [ exp(v/u) -1] = 5 ( e^3 - 1) = 95.4276846 s

Consider coordinate $\psi$ measured along the band, with the starting fixed point $\psi = 0$ , and the end point $\psi = 1$ . In this system, all points on the system remains fixed (in terms of $\psi$ ) as the ropes stretches. A point $x =X$ on the the rope will have a coordinate $\phi(t)=\psi = \dfrac{X}{L +vt},$ and the speed of the bug is equivalent to $\phi'(t) = -\dfrac{u}{L+vt}$

We thus reduced the requirement of solving the Difference Equation.

So, $\phi(t) = -\dfrac{u}{v}\ln(L+vt) + \kappa$ where $\kappa$ is the constant of integration.

Putting limits to get the value of $\kappa$ , we know $\phi(0) = 1$ $\begin{aligned}1 &= -\dfrac u v \ln L + \kappa\\\kappa &= 1+\dfrac uv \ln L\\\\\Rightarrow \phi(t) &= -\dfrac{u}{v}\ln(L+vt) + 1+\dfrac uv \ln L\\&=\dfrac uv \ln\left(\dfrac{L}{L+vt} + 1\right)\end{aligned}$

Hence, to get time $T$ , $\begin{aligned}\phi(T) = 0 &= \dfrac uv \ln\left(\dfrac{L}{L+vT} +1\right)\\ \Rightarrow 1 &= \dfrac uv \ln \left(\dfrac {L+vT}L \right)\\\Large\Rightarrow T &= \dfrac Lv \left(e^{\dfrac vu} - 1\right)\end{aligned}$

$\displaystyle \Huge \therefore \boxed{T = \dfrac Lv \left(e^{\frac vu} - 1\right)}\Large\approx 95.4277$