rabbit population

• February 12th 2009, 03:31 PM
razorfever
rabbit population
dN/dt = cN^1.01

(i) show that, if there are N(0) = 2 rabbits initially, and 16 rabbits after 3 months, then the population N(t) become unbounded at a finite time t
(ii) does the population become unbounded far any initial condition N(0) > 0, and any c>0, expain your answer

i solved the DE and got

N(t) = [(0.68t - 99.31) / -100]^(-100)
how do i show that the population becomes unbounded at a finite time t

can someone give me a detailed explanation of both parts??
• February 12th 2009, 11:28 PM
Rincewind
Quote:

Originally Posted by razorfever
dN/dt = cN^1.01

(i) show that, if there are N(0) = 2 rabbits initially, and 16 rabbits after 3 months, then the population N(t) become unbounded at a finite time t
(ii) does the population become unbounded far any initial condition N(0) > 0, and any c>0, expain your answer

i solved the DE and got

N(t) = [(0.68t - 99.31) / -100]^(-100)
how do i show that the population becomes unbounded at a finite time t

can someone give me a detailed explanation of both parts??

Another way to write your solution is

$N(t) = \left(\frac{100}{D - ct}\right)^{100}$

where you have determined $D \approx 99.31$ and $c \approx 0.68$.

Now you can see from this that when the denominator is zero then N become unbounded (division by zero) so this will happen at the critical time

$t_{critical} = \frac{D}{c}$

For part one the critical time is

$t_{critical} \approx \frac{99.31}{0.68} \approx 146$

For part 2, if N(0) > 0 then

$D = \frac{100}{N(0)^{0.01}}$

which is positive and if c is positive then also $t_{critical}$ will be positive.

Hope this helps