Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
jocelynevsq
Group Title
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)?
 one year ago
 one year ago
jocelynevsq Group Title
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)?
 one year ago
 one year ago

This Question is Open

hewsmike Group TitleBest ResponseYou've already chosen the best response.0
Moderately nasty. Setup :\[\frac{dP}{dt}=P(2\frac{P}{5000})=P\frac{(10^4P)}{5000}\]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 :\[1=A(10^{4}P)+B(P)=10^{4}AAP+BP = 10^{4}A+P(BA)\]equate like powers of P :\[A=10^{4}\]\[BA = 0\]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\[(10^{4}P)Ce^{2t}=P\]rearranged ( 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.
 one year ago

hewsmike Group TitleBest ResponseYou've already chosen the best response.0
Whoops ( darn the Equation editor lag .... ) :\[\frac{dP}{dt}=2P\frac{P^{2}}{5000}\]
 one year ago

hewsmike Group TitleBest ResponseYou've already chosen the best response.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 ).
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.