## jocelynevsq 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

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1. hewsmike

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

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

3. hewsmike

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 ).