Answer to Question #106149 in Differential Equations for simran

Question

The population x(t) of a certain city satisfies the logistic law

2
8
10( )
1
100
1
x x
dt
dx
= −
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

1 STEP: Solve the differential equation

$$\frac{dx}{dt}=\frac{x}{100}-\frac{x^2}{10^8}\longrightarrow\\[0.3cm] \frac{dx}{\displaystyle\left(\frac{x}{100}-\frac{x^2}{10^8}\right)}=dt\longrightarrow\frac{10^8dx}{x(10^6-x)}=dt\longrightarrow\\[0.3cm] 10^8\cdot\int\frac{1}{10^6}\cdot\frac{(10^6-x)+x}{x(10^6-x)}dx=\int dt\longrightarrow\\[0.3cm] 10^2\cdot\int\left(\frac{1}{x}+\frac{1}{10^6-x}\right)dx=t\longrightarrow\\[0.3cm] \ln|x|-\ln|10^6-x|=\frac{t}{100}+\ln|C|\longrightarrow\\[0.3cm] \ln\left|\frac{x}{10^6-x}\right|=\frac{t}{100}+\ln|C|\longrightarrow\\[0.3cm] \frac{x}{10^6-x}=C\cdot e^{t/100}\longrightarrow\\[0.3cm] x=10^6\cdot C\cdot e^{t/100}-x\cdot C\cdot e^{t/100}\longrightarrow\\[0.3cm] x\cdot\left(1+C\cdot e^{t/100}\right)=10^6\cdot C\cdot e^{t/100}\longrightarrow\\[0.3cm] x(t)=\frac{10^6\cdot C\cdot e^{t/100}}{1+C\cdot e^{t/100}}\longrightarrow\\[0.3cm] \boxed{x(t)=\frac{10^6}{1+\frac{1}{C}\cdot e^{-t/100}}}$$

2 STEP: To determine the constant, we use the initial condition

$$x(1980)=100000\equiv 10^5=\frac{10^6}{1+\frac{1}{C}\cdot e^{-1980/100}}\longrightarrow\\[0.3cm] 1+\frac{1}{C}\cdot e^{-1980/100}=10\longrightarrow\frac{1}{C}\cdot e^{-1980/100}=9\longrightarrow\\[0.3cm] \boxed{\frac{1}{C}=9\cdot e^{1980/100}}$$

Conclusion,

$$x(t)=\frac{10^6}{1+9\cdot e^{1980/100}\cdot e^{-t/100}}\longrightarrow\\[0.3cm] \boxed{x(t)=\frac{10^6}{1+9\cdot e^{(1980-t)/100}}-\text{the population at any time}\,\,t>1980 }$$

3 STEP: We use the given formula to calculate the population in 2000

$$x(2000)=\frac{10^6}{1+9\cdot e^{(1980-2000)/100}}=\frac{10^6}{1+9\cdot e^{-0.2}}\longrightarrow\\[0.3cm] \boxed{x(2000)\approx 119494.63-\text{the population in 2000}}$$

ANSWER

$$x(t)=\frac{10^6}{1+9\cdot e^{(1980-t)/100}}-\text{the population at any time}\,\,t>1980\\[0.3cm] x(2000)\approx 119494.63-\text{the population in 2000}$$