   Chapter 8.5, Problem 49E

Chapter
Section
Textbook Problem

# Epidemic Model A single infected individual enters a community of n susceptible individuals. Let x be the number of newly infected individuals at time t. The common epidemic model assumes that the disease spreads at a rate proportional to the product of the total number infected and the number not yet infected. So. d x / d t = k ( x + 1 ) ( n − x ) and you obtain ∫ 1 ( x + 1 ) ( n − x )   d x = ∫ k   d t Solve for x as a function of t.

To determine

To calculate: The value of x in the common epidemic model, dxdt=k(x+1)(nx) as a function of t.

Explanation

Given:

The common epidemic model is dxdt=k(x+1)(nx)

So, 1(x+1)(nx)dx=kdt

Formula used:

1udu=lnu+c

Calculation:

Consider the given integral,

1(x+1)(nx)dx=kdt

Let, 1(x+1)(nx)=Ax+1+Bnx

Then,

1(x+1)(nx)=A(nx)+B(x+1)(x+1)(nx)1=AnAx+Bx+B0x+1=(A+B)x+(An+B)

Now, compare the coefficients on both the sides,

A+B=0

And

An+B=1

Then,

A=1n+1

And

B=1n+1

Therefore,

1(x+1)(nx)dx=kdt[1(n+1)(x+1)+1(n+1)(nx)]dx=kt+C1n+1(1x+1+1nx)dx=kt+C1n+1(ln|x+1nx|)=kt+C

Since, x be the number of newly infected individuals at time t

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