3 Logistic population dynamics first order ode

Will use this section to solve some logistic odes. The logistic ode is given by

\begin{align} \frac {dP}{dt} & =rP\left ( 1-\frac {P}{K}\right ) \tag {1}\\ p\left ( 0\right ) & =p_{0}\nonumber \end{align}

Where in the above \(P\left ( t\right ) \) is the population at time \(t\) and \(r\) is the intrinsic growth rate, which is rate of birth minus rate of death of current population. \(K\) is the carrying capacity or maximum population also called equilibrium population that can be supported.

The solution to (1) is

\[ P\left ( t\right ) =\frac {p_{0}ke^{rt}}{k+p_{0}\left ( e^{rt}-1\right ) }\]

3.1 Problem 1

This is problem 2, page 56, from book Elementary differential equations, 8th edition by Earl Rainville.

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.

The solution to the ode (it is basic separable ode)

\[ P\left ( t\right ) =\frac {p_{0}ke^{rt}}{k+p_{0}\left ( e^{rt}-1\right ) }\]

But we are given that \(p_{0}=1000,k=10000\), hence the above becomes

\begin{align*} P\left ( t\right ) & =\frac {1000\left ( 10000\right ) e^{rt}}{10000+1000\left ( e^{rt}-1\right ) }\\ & =\frac {10000e^{rt}}{10+e^{rt}-1}\\ & =\frac {10000e^{rt}}{9+e^{rt}}\end{align*}

The only thing left is to find \(r\). Since we are told that after 1hr, the population is \(2000\), then this information is used to find \(r\). Using unit of time as hr, the above becomes at \(t=1\) as

\[ 2000=\frac {10000e^{r}}{9+e^{r}}\]

Rearranging

\begin{align*} 18000+2000e^{r} & =10000e^{r}\\ 18000 & =8000e^{r}\\ \frac {9}{4} & =e^{r}\end{align*}

Hence

\begin{align*} r & =\ln \left ( \frac {9}{4}\right ) \\ & =0.8109 \end{align*}

Therefore the solution now becomes

\begin{align*} P\left ( t\right ) & =\frac {10000e^{\ln \left ( \frac {9}{4}\right ) t}}{9+e^{\ln \left ( \frac {9}{4}\right ) t}}\\ & =\frac {10000\left ( \frac {9}{4}\right ) ^{t}}{9+\left ( \frac {9}{4}\right ) ^{t}}\end{align*}

Since the growth rate \(r\) used the unit of time as \(1\) hr, in the following plot of the above solution, the x axis will have unit of 1 hr. Here is a plot of the above solution

sol = 10000*(9/4)^t/(9 + (9/4)^t); 
p = Plot[{sol, 10000}, {t, 0, 9}, GridLines -> Automatic, 
GridLinesStyle -> LightGray, PlotStyle -> {Red, Dashed}, 
AxesLabel -> {"Time in units of hr", "population"}]