Assume that the bug's distance from the wall when the band is not stretched is f(t) and the length of the rubber band (when it is stretched) is r(t). Let the distance from the wall to the bug (crawling along the band as it is stretched) be b(t). Then for an infinitesimal period dt, we have
b(t+dt) = b(t)r(t+dt)/r(t) + df(t).
Setting r(t+dt) = r(t) + r'(t)dt and df(t) = f'(t)dt, rearranging,
(b(t+dt)-b(t))/dt = b(t)r'(t)/r(t) + f'(t).
Taking the limit as dt approaches 0,
b'(t) = b(t)r'(t)/r(t) + f'(t).
The solution to this differential equation with boundary condition b(0)=0, is
b(T) = r(T)(Integral t=0->t=T f'(t)/r(t) dt),
If the bug reaches the end of the band at time T, so that b(T)=r(T), then we have
Integral t=0->t=T f'(t)/r(t) dt = 1.
The solution in T of this equation (when one exists) gives the answer for the general case.
Let f(t) = 0.00001t, and r(t) = t***. Then we have
Integral t=0->t=T 0.00001/(t***) dt = 1,
or, carrying out the integral,
0.00001 ln(T***) = 1.
So, T = e
100000 - 1 seconds.