Question

Population Growth In the study of population dynamics one of the most famous models for a growing but bounded population is the logistic equation

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{P}\left(a-bP\right)\mathbf{,}$

where and are positive constants. Although we will come back to this equation and solve it by an alternative method in Section 3.2, solve the DE this first time using the fact that it is a Bernoulli equation.

Short answer

The equation is$P=\frac{a{e}^{at}}{b{e}^{at}+ac}$

Step-by-step solution

Step 1:

We know that a Bernoulli's equation is of the form

$\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)y^n$

where n is any real number. We substitute u = y^1-n to reduce any Bernoulli's equation to a linear equation.

For the given exercise, we first rewrite the given DE in Bernoulli's equation form by simplifying as

$$\begin{gathered} \frac{dP}{dt}=P\left(a-bP\right) \\ =aP-b{P}^2 \\ \frac{dP}{dt}+\left(-a\right)P=-b{P}^2 \end{gathered}$$

Here, the variables P and t are used instead of y and x respectively with value of n = 2.

We substitute $u=P^{1-2}=P^{-1}orP=u^{-1}$to reduce the equation to a linear one. Then, using the chain rule, we have

$\frac{dP}{dt}=\frac{dP}{du}.\frac{du}{dt}=-u^{-2}\frac{du}{dt}$

Upon substitution into the given equation, we get

$$\begin{gathered} -u^{-2}\frac{du}{dt}-a{n}^{-1}=-b{\left(u^{-1}\right)}^2 \\ =-b{u}^{-2}\frac{du}{dt}+au=b \end{gathered}$$

We have reduced the Bernoulli's equation to a linear equation now. The integrating factor for this linear equation is

$e^{adt}=e^{at}$

Multiply the differential equation

We multiply the differential equation (1) by e^at to get

$e^{at}\frac{du}{dt}+au{e}^{at}=b{e}^{at}\Rightarrow \frac{d}{dt}{e}^{at}u=b{e}^{at}$

Upon integration, we get

$$\begin{gathered} \frac{d}{dt}\left[e^{at}u\right]dt=e^{at}bdt \\ {e}^{at}u=\frac{b}{a}{e}^{at}+c \\ u=\frac{b}{a}+c{e}^{-at} \end{gathered}$$

Step 3:Final proof

Since$u=P^{-1}=\frac{1}{P\text{'}},$the solution of the given equation will be

$\begin{array}{rcl}\frac{1}{P}& =& \frac{b}{a}+c{e}^{-at}\\ & =& \frac{b}{a}+\frac{c}{e^{at}}\\ & =& \frac{b}{a}+\frac{c}{e^{at}}\\ & =& \frac{b{e}^{at}+ac}{a{e}^{at}}\\ P& =& \frac{a{e}^{at}}{b{e}^{at}+ac}\end{array}$