Question

The population in a town satisfies the logistic law.

dx/dt=x/100-x²/10⁸

where t is the number of years after the 2019 population census. if the population x was 100,000 in 2019.determine

1. population x as a function of time
2. The year the population will be double

Answer 1

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

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

$$\int\dfrac{100dx}{x}+\int\dfrac{100dx}{10^6-x}=\int dt$$

$$100\ln|x|-100\ln|x-10^6|=t+100\ln C$$

$$\ln\dfrac{x}{|x-10^6|}=0.01t+\ln C$$

$$\dfrac{x}{|x-10^6|}=Ce^{0.01t}$$

$x(0)=100000$

$$\dfrac{10^5}{|10^5-10^6|}=Ce^{0.01(0)}$$

$$C=\dfrac{1}{9}$$

$$\dfrac{x}{|x-10^6|}=\dfrac{1}{9}e^{0.01t}$$

1. If $x<1000000$

$$9x=(10^6-x)e^{0.01t}$$

$$x(t)=\dfrac{10^6}{1+9e^{-0.01t}}$$

2,

$x=2(100000)=2\times10^5$

$$\dfrac{2\times10^5}{10^6-2\times10^5}=\dfrac{1}{9}e^{0.01t}$$

$$18=8e^{0.01t}$$

$$0.01t=\ln 2.25$$

$$t=100\ln2.25\ years$$

$$t=81.1\ years$$