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 one year 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)?
 one year 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)?

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hewsmike
 one year ago
Best ResponseYou've already chosen the best response.0Moderately 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.

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

hewsmike
 one year ago
Best ResponseYou've already chosen the best response.0And 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 ).
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