Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

jocelynevsq

  • 3 years ago

The population P(t) of a species satisfies the logistic differential equation dP/dt= P(2-(P/5000)) where the initial population 3,000 and t is the time in years. What is the lim t>infinity P(t)?

  • This Question is Open
  1. hewsmike
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Moderately nasty. Setup :\[\frac{dP}{dt}=P(2-\frac{P}{5000})=P\frac{(10^4-P)}{5000}\]invert both sides :\[dt/dP=\frac{5000}{P(10^{4}-P)}\]we will want to integrate both sides with respect to P, but before we do that we want to express :\[\frac{1}{P(10^{4}-P)}\]in partial fractions :\[\frac{A}{P}+\frac{B}{(10^{4}-P)}\]for A and B to be determined as follows.\[\frac{1}{P(10^{4}-P)}=\frac{A}{P}+\frac{B}{(10^{4}-P)}\]multiply both sides by those denominators :\[1=A(10^{4}-P)+B(P)=10^{4}A-AP+BP = 10^{4}A+P(B-A)\]equate like powers of P :\[A=10^{-4}\]\[B-A = 0\]so\[B=A = 10^{-4}\]and\[dt/dP=\frac{5000\times10^{-4}}{P}+\frac{5000\times10^{-4}}{(10^{4}-P)}=\frac{1}{2P}+\frac{1}{2(10^{4}-P)}\]now integrate with respect to P:\[t=\frac{1}{2}\int_{}{}\frac{dP}{P}+\frac{1}{2}\int_{}{}\frac{dP}{(10^{4}-P)}=\frac{1}{2}[ln(P)-ln(10^{4}-P)]+c\]where c is some constant to be determined, but I can write it 'conveniently' without loss of generality as (-1/2)ln(C), giving :\[2t=ln(\frac{P}{C(10^{4}-P)})\]so\[e^{2t} = \frac{P}{C(10^{4}-P)}\]let's find C now using initial conditions :\[P(0)=3000\]\[e^{0} = 1= \frac{3000}{C(10^{4}-3000)}\]\[C=\frac{3000}{7000}=3/7\]now\[(10^{4}-P)Ce^{2t}=P\]re-arranged ( several steps skipped ):\[P=\frac{C10^{4}e^{2t}}{(1+Ce^{2t})}\]so as t -> infinity :\[P\rightarrow \frac{C 10^{4}e^{2t}}{Ce^{2t}}=10^{4}\]interesting result. Independent of C or if you like independent of the initial population ( provided it wasn't zero ), and the limit ( the fact that you have one at all ) is due to the quadratic term in the original differential equation, that you could write as :\[\frac{dP}{dt}=2P -10^{4}P^{2}\]so instead of increasing without bound in exponential fashion there is a check on the growth ( that's a negative coefficient on the quadratic ) that pulls it up.

  2. hewsmike
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Whoops ( darn the Equation editor lag .... ) :\[\frac{dP}{dt}=2P-\frac{P^{2}}{5000}\]

  3. hewsmike
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And it's also interesting that the ultimate ( steady state ) population depends only upon the ratios of the two coefficients in the differential equation :\[\frac{2}{(1/5000)}=10^{4}\]so if you want a bigger final population make 'em grow quicker ( choose a number larger than 2 ) and/or inhibit them less ( pick a number smaller than 1/5000 ).

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy