30.9k views
0 votes
A disease is spreading throughout a community of 3,000 people at a rate (measured in number infected per day) proportional to the product of number of people infected and the number of people not yet infected, with constant of proportionality k = 0.004. Initially, 500 people are infected. Write an initial value problem.

1 Answer

3 votes

Answer:


I'(t)=12I-0.004I^2, I_o=500

Explanation:

Population of the Community=3000

Let the number of infected=I

The number of uninfected=3000-I

The rate at which disease is spreading is proportional to the product of number of people infected and the number of people not yet infected.


(dI)/(dt)\propto I(3000-I) \\(dI)/(dt)=k I(3000-I)\\(dI)/(dt)=0.004 I(3000-I)\\$Let I_o$=Initial Number of Infected=500\\Therefore, the initial value problem is given as:\\I'(t)=12I-0.004I^2, I_o=500

User Angle
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories