The population P(t) of a species satisfies the logistic differential equation dp/dt=P{2-(P/5000)}, where initial population P(0)=3000 and t is the time in years. What is the limit P(t) as t tends to infinity?

1 Answer
Jan 17, 2018

P(t) = (30000e^(2t)) / (7 + 3e^(2t))

As t rarr oo we have:

lim_(t rarr oo) P(t) = 10000

Explanation:

We have:

(dP)/(dt) = P{2 - (P/5000)} and P(0)=3000

This is a First Order separable ODE, so we can manipulate the DE as follows:

(dP)/(dt) = P{ (10000 - P)/5000}

:. 5000/( P(1000-P) ) \ (dP)/(dt) = 1

So we can "separate the variables" to get

int \ 5000/( P(10000-P) ) \ dP = int \ dt

The RHS integral is trivial and we can use partial fraction decomposition for the LHS integral:

5000/( P(10000-P) ) -= A/P + B/(10000-P)
" " = (A(10000-P) + BP)/( P(10000-P) )

Leading to the identity:

5000 -= A(10000-P) + BP

Where A,B are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put P = 0 => 5000 = 10000A => A=1/2
Put P=10000 => 10000B=5000 => B = 1/2

So we can now write the seperated DE as:

1/2 int \ 1/P + 1/(10000-P) \ dP = int \ dt

And we can now integrate to get:

1/2 {ln|P| -ln |10000-P| } = t + C

We now use the initial condition P(0)=3000

1/2 {ln3000 -ln (10000-3000) } = 0+ C
:. C = 1/2 ln (3000/7000) = 1/2ln(3/7)

So we have the solution:

1/2 {ln|P| -ln |10000-P| } = t + 1/2ln(3/7)

:. ln|P/(10000-P)| -ln (3/7) = 2t

:. ln | (P/(10000-P)) / (3/7) | = 2t
:. ln | (7P)/(30000-3P) | = 2t

Noting that t is positive we then have:

(7P)/(30000-3P) = e^(2t)

:. 7P = e^(2t) (30000-3P)
:. 7P = 30000e^(2t)-3Pe^(2t)
:. 7P + 3Pe^(2t) = 30000e^(2t)
:. P(7 + 3e^(2t)) = 30000e^(2t)
:. P = (30000e^(2t)) / (7 + 3e^(2t))

We now want to examine the behaviour of P(t) as t rarr oo.

lim_(t rarr oo) P(t) = lim_(t rarr oo) (30000e^(2t)) / (7 + 3e^(2t)) \ e^(-2t)/e^(-2t)

" " = lim_(t rarr oo) (30000) / (7e^(-2t) + 3 )

" " = (30000) / (0 + 3 )

" " = 10000