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