(adsbygoogle = window.adsbygoogle || []).push({}); 1. A population grows accordin to the logistic law, with a limiting population of 5 x 10^8 individuals. When the population is low, it doubles every 40 minutes. What will the population be after 2 hours if initially it is (a) 10^8, and (b) 10^9?

P=P0e^a(t-t0)

N=a/b

p(t) = ap0/[bp0 + (a-bp0)e^-a(t-t0)]

I have actually solved the problem, but I want to confirm that I am doing it right so here is my solution, i only showed (a) since a and b are the same.

p=p0e^a(t-t0) (i used this since it gives you info about a low population)

2=1e^a(40-0)

2=e^40a

40a = ln 2

a = 1/40 ln 2

then i used N = a/b since N is given to be 5 x 10^8

5 x 10^8 = a/b

5 x 10^8 = (1/40 ln 2) / b

b = (1/40 ln 2)/(5 x 10^8)

then i used the logistic model

p(t) = ap0/[bp0 + (a-bp0)e^-a(t-t0)]

p(120)= (1/40 ln 2)(10^8)/[((1/40 ln 2)/(5x10^8)) (10^8) + (1/40 ln 2 - ((1/40 ln 2)/(5x10^8)) (10^8) )e^(-1/40 ln 2)(120-0)] (sorry i know this looks confusing bare with me)

P(120) = (10^8/40 ln 2) / [1/8 ln 2 + (1/40 ln 2- (1/8 ln 2))e^-3 ln 2)]

P(120) = (10^8/40 ln 2) / [1/8 ln 2 + (1/40 ln 2 - (1/8 ln 2))(1/8)] (e^-3 ln 2 = 1/8, is this correct??)

P(120) = (10^8/40 ln 2) / [1/8 ln 2 + 1/320 ln 2 - 1/64 ln 2]

then i got rid of all the ln 2's

P(120) = (10^8/40) / [1/8 + 1/320 - 1/64]

P(120) = (10^8/40) / 9/80

P(120) = (10^8/40)(80/9)

P(120) = (2 x 10^8)/9 = 22,222,222

i really hope this is right.

and i did the same thing for (b) but changed 10^8 to 10^9 and got 22,792,022

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Differential equations - logistic law problem

**Physics Forums | Science Articles, Homework Help, Discussion**