### Home > CALC > Chapter 11 > Lesson 11.2.2 > Problem11-60

11-60.

The population $P$ of a certain species of fish grows at a rate of $\frac { d P } { d t } = 0.01 P ( 100 - P )$ fish per year. At time $t = 0$, the population was $50$ fish. What is the population after two years?

$\frac{100dP}{P(100-P)}=dt$
$\int\frac{100dP}{P(100-P)}=\int dt$
When $t = 0$, $P = 50$. What is the value of $C$?
Use your value of $C$ in your equation to determine the value of $P$ when $t = 2$.