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The logistic growth of a population.

Most populations do not grow exponentially, rather they follow a logistic model. Once the population has reached its carrying capacity, it will stabilize and the exponential curve will level off towards the carrying capacity, which is usually when a population has depleted most its natural resources.

The Logistic Equation

${\frac {dP}{dt}}=kP(1-{\frac {P}{K}})$

Where,

$P(t)$= the population after time t

$t$= time a population grows

$k$= relative growth rate coefficient

$K$= carrying capacity of the population; defined by ecologists as the maximum population size that a particular environment can sustain.

The Analytic Logistic Solution

This is a separable differential equation that can be derived through integration. The analytic solution is useful in analyzing the behavior of population models.

The equation is separable and to find the solution we integrate.

$\int {\frac {dP}{P(1-{\frac {P}{K}})}}=\int k\cdot dt$

Working on just the left side of the equation, the fraction in the denominator is eliminated by multiplying the variable K, and then the fraction is split in 2.

${\frac {1}{P(1-{\frac {P}{K}})}}\cdot {\frac {K}{K}}={\frac {K}{P(K-P)}}$

${\frac {K}{P(K-P)}}={\frac {1}{P}}+{\frac {1}{K-P}}$

The partial fraction is then integrated more easily.

$\int ({\frac {1}{P}}+{\frac {1}{K-P}})dP=\int k\cdot dt$

After integrating and using U substitution, we get

$-\ln \left\vert P\right\vert +\ln \left\vert K-P\right\vert =-kt-C$

$\ln \left\vert {\frac {K-P}{P}}\right\vert =-kt-C$

Exponentiate both sides to get rid of the natural log. This is the equation that remains:

$\left\vert {\frac {K-P}{P}}\right\vert =e^{-kt-C}$

Get rid of the absolute value and split the $e$ into 2 parts.

${\frac {K-P}{P}}=\pm e^{-C}e^{-kt}$

Let $A=\pm e^{-C}$ and get

${\frac {K-P}{P}}=Ae^{-kt}$

Solve for $P$ to get the explicit solution to the logistic equation as

$P(t)={\frac {K}{1+Ae^{-kt}}}$; where $A={\frac {K-P_{0}}{P_{0}}}$ and $P_{0}=$ the initial population at time 0.