Question

suppose the population P(t) of Ongwediva town at any time t>0 satisfies the logic law
dP/dt=P/100-P^2/〖10〗^8
Where t is measured in years. Given that the population of this city was 100000 in 1980 determine the population as a function t>1980

Answer 1

$\frac{dP}{dt}=\frac{P}{100}-\frac{P^2}{10^8}\\ 10^8\frac{dP}{10^6P-P^2}=dt$

$10^8\int{\frac{dP}{10^6P-P^2}}=\int{dt}$

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

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

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

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

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