Question

1.12. the time rate of change of an alligator population p in a swamp is proportional to the square of p. the swamp contained a dozen alligators in 1988, and two dozen in 1998. when will there be four dozen alligators in the swamp? what happens thereafter?

Answer 1

Given

p'(t) = kp²

p(0) = 12; p(10) = 24

Find

a) p(t)

b) t such that p(t) = 48

c) the behavior of p(t) after the time of part b

Solution

a) The differential equation is separable, so can be solved by separating the variables and integrating.

$\displaystyle\frac{d}{dt}p(t)=k\cdot p(t)^{2}\\\\\int{p^{-2}}\,dp=\int{k}\,dt\\\\-p^{-1}=kt+C\\\\p=\frac{-1}{kt+C}$

Plugging in the given boundary conditions, we can solve for k and C to find

$p(t)=\dfrac{240}{20-t}$

b) The population doubles when the time to t=20 is cut in half. The first doubling occurred in 10 years; the second one will occur in half that time, 5 years. There will be 48 alligators in the swamp in 2003.

c) The population doubles again in half the time of the previous doubling, so is predicted to be infinite in 2008.