Question

Separable differential equations: Suppose a population P = P(t) grows in such a way that its rate of growth $\frac{dP}{dt}$obeys the equation $\frac{dP}{dt}=P\left(100-P\right)$

This is called a differential equation because it is an equation that involves a derivative. In the series of steps that follow, you will find a function P(t) that behaves according to this differential equation.

Set the answer from the previous step equal to $\int 1dt=t+C_2$, and solve for P = P(t). Along the way you can combine unknown constants into new constants; for example, if you encounter C2 − C1, then you could just rename that constant C and proceed from there. At the end of your calculations

you should have$P\left(t\right)=\frac{100A{e}^{100t}}{1+A{e}^{100t}}$for some constant A.

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given information

Differential equation is$\frac{dP}{dt}=P\left(100-P\right)$

Step 2. Explanation

$\begin{array}{rcl}\frac{1}{100}\left(\ln \left|P\right|-\ln \left|100-P\right|\right)+C_1& =& t+C_2\\ \frac{1}{100}\ln \left|\frac{P}{100-P}\right|& =& t+C_2-C_1\\ \ln \left|\frac{P}{100-P}\right|& =& 100\left(t+C_2-C_1\right)\\ \left|\frac{P}{100-P}\right|& =& e^{100\left(t+C_2-C_1\right)}\\ P& =& \left(100-P\right)e^{100t}{e}^{100\left(C_2-C_1\right)}\\ P\left(1+e^{100t}{e}^{100\left(C_2-C_1\right)}\right)& =& 100e^{100t}{e}^{100\left(C_2-C_1\right)}\\ P\left(t\right)& =& \frac{100e^{100t}{e}^{100\left(C_2-C_1\right)}}{1+e^{100t}{e}^{100\left(C_2-C_1\right)}}\\ & =& \frac{100A{e}^{100t}}{1+A{e}^{100t}}\end{array}$