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.

Use partial fractions to show that the integral on the left side of this equation is equal to$\frac{1}{100}\left(\ln \left|P\right|-\ln \left|100-P\right|\right)+C_1$

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

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

Integrate both sides.

$\begin{array}{rcl}\int \frac{dP}{P\left(100-P\right)}& =& \int \frac{dP}{100P}-\frac{dP}{100\left(100-P\right)}\\ & =& \frac{1}{100}\left(\ln \left|P\right|-\ln \left|100-P\right|\right)+C_1\\ & & \end{array}$