Question

The population x(t) of a certain city satisfies the logistic law dx/dt = (x÷100)- (x^2 ÷ 10^8)
where t is measured in years. Given that the population of the city is 100000 in 1980, determine the population at any time t >1980 . Also find the population in the year 2000.

Answer 1

$\frac{dx}{dt}=\frac{x}{100}-\frac{x^2}{10^8}$

$10^8\frac{dx}{10^6x-x^2}=dt$

$10^8\int{\frac{dx}{10^6x-x^2}}=\int{dt}$

$100ln\frac{x}{x-10^6}=t+C_1$

$\frac{x}{x-10^6}=Ce^{\frac{t}{100}}$

Given that in the year 1980, t=0 and $x(0)=10^5$ we get $C=-\frac{1}{9}$

Then, $\frac{x}{x-10^6}=-\frac{1}{9}e^{\frac{t}{100}}$

$x(t)=\frac{10^6e^{\frac{t}{100}}}{e^{\frac{t}{100}}+9}$

in the year 2000, t=20

$x(20)=\frac{10^6e^{\frac{20}{100}}}{e^{\frac{20}{100}}+9}=119,494.6$