Suppose that a community contains 15,000 people who are susceptible to Michaud’s

Question

Suppose that a community contains 15,000 people who are susceptible to Michaud’s syndrome, a contagious disease. At time t = 0, the number N(t) of people who have developed Michaud’s syndrome is 5000 and is increasing at the rate of 500 per day. Assume that N'(t) is proportional to the product of the numbers of those who have caught the disease and of those who have not. How long will it take for another 5000 people to develop Michaud syndrome? PLEASE PROVIDE A DETAILED ANSWER.

Logistic Function:
(dP/dt)= P(a-bP)

Thanks!!

Explanation / Answer

N(t = 0) = 5000 dN/dt (t = 0) = 500 per day dN/dt = k*N*(15000 -N) "proportional to product of number of those who caught the disease(N) and those who have not (15000 - N) given all these, dP/dt (t = 0) = 500 per day = k*P(t = 0) * (15000 - P(t=0)) =k*(5000)*(15000 - 5000) k = 1.0e-5 dP/ [P*(15000 - P)] = kdt 1/[P*(15000 - P)] = A/P + B/(15000 - P) 1 = A*(15000 - P) + B*P = (B-A)*P + P*15000 A*15000 =1 B-A = 0 A =1/15000 B = A = 1/15000 k*dt = dP/ [P*(15000 - P)] = A*dP/P + B*dP/(15000 - P) = A*[dP/P +dP/(15000 - P)] (k/A)*dt = dP/P + dP/(15000 - P) (k/A)*t = ln(P/P0) - ln(15000 - P)/(15000 - P0) let P = 10000 (5000 + "another 5000") P0 = 5000 k/A = (1e-5 *15000) = 0.15 0.15*t = ln(10000/5000) - ln[(15000 - 10000)/(15000 - 5000)] =1.38629436 t= 9.24 days

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