Question

A bacterial population is known to have a logistic growth pattern with initial population 1000 and an equilibrium population of 10,000. A count shows that at the end of 1 hr there are 2000 bacteria present. Determine the population as a function of time.

Answer 1

$dN/dt=aN(k-N)$ is the logistic growth equation, where dN/dt is rate of growth of bacteria population, a is an undetermined constant and k is the saturation population or equilibrium population.

Solving the above equation we get;

$dN/N(k-N)=adt=(1/k)dN(1/N+1/(k-N))$

Integrating both sides we get;

$\implies \int dN(1/N+1/(k-N))=k\int adt$

$ln(N/(k-N))=(ka)t+c$

Using the given conditions :

$k=10000; N(0)=1000; N(1)=2000$

We get;

$c=ln(1000/(10000-1000))=-ln9=-2.197$

and

$ln(2000/(10000-2000))=ka+c$

$\implies ka=(-ln4-c)=-1.386+2.197=0.811$

Thus, the population as a function of time is given by;

$ln(N/(10000-N)))=0.811t-2.197$