Question

The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 600,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 80,000 mosquitoes/day. Determine the population of 3, 04 mosquitoes in the area at any time. (Note that the variable represents days.) udy

Answer 1

Let p (t) be the population of mosquitoes in the area with 't't in weeks, and k be the rate at that they double we see in absence of other factors there p_1 = kp thus the population grows like p (t) = p_0 e^kt where p_0 = 600000 is the initial population we may find that the rate k since we are given that the population double each week thus z = p (t + 1)/p (t) = e^k (t + 1)/e^kt = > e^k = > 2 = e K = > k = in 2 1/ weeks now that we have the rate at which the population grows, we may introduce the prerecord to the made o now by p ^1 = in 2p - 80000 mosquitoes/ day * 7 day/week p^21 = in 2 p - 560,000 the solution to this ODE is given by p (t) = e^t/n2 (p_0 - 560,000/in2) + (560,000/in2) p (t) = (600,000 in2 - 560,000)/in 2 2^t + (560,000/IN2)