Answered: A virus infect with contact and is most…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A virus infect with contact and is most infectious at winter. If you get infected, you remains a carrier for an unlimited time. In an isolated population with P persons the rate of infection by time (t =months after the first of january) is propotional with the product of :

1: the number y(t) infected
2: the number not infected
3: 1 + cos((pi*t)/6)

1/10 of the population is infected 1.january

Make a differential equation that satisfies y(t), solve it, and explain every step

if a fifth of the population is infected a month later, how many are infected a year later?

Expert Solution

Step 1

$(a)$ Let total population is $P$ and infected people is $y (t)$ .

Then

$\frac{d y}{d t} \propto y (t) . . . . (1)$

$\frac{d y}{d t} \propto [P - y (t)] . . . . (2)$

$\frac{d y}{d t} \propto [1 + \cos (\frac{πt}{6})] . . . . (3)$

$\Rightarrow \frac{d y}{d t} = k y (t) [p - y (t)] [1 + \cos (\frac{πt}{6})] \frac{d y}{y (t) [p - y (t)]} = k [1 + \cos (\frac{πt}{6})] d t \int \frac{d y}{p y (t) - y^{2} (t)} = k \int [1 + \cos (\frac{πt}{6})] d t \frac{d y}{{(\frac{p}{2})}^{2} - [y^{2} - 2 (\frac{P}{2}) y + {(\frac{P}{2})}^{2}]} = k [t + \frac{\sin (\frac{πt}{6})}{π / 6}] + C$

Step 2

By solving this integration we get

$\frac{1}{P} \log (\frac{y}{P - y}) = k [t + \frac{6}{π} \sin (\frac{πt}{6})] + C . . . . . (4) A s a t t = 0, y (t) = \frac{P}{10} \Rightarrow \frac{1}{P} \log (\frac{P / 10}{9 P / 10}) = k (0) + C \Rightarrow C = \frac{- 2 \log 3}{P} \frac{1}{P} \log (\frac{y}{p - y}) = k [t + \frac{6}{π} \sin (\frac{πt}{6})] - \frac{2 \log 3}{9} . . . . . (5) \frac{1}{P} \log (\frac{y}{p - y} + \log 9) = k [t + \frac{6}{π} \sin (\frac{πt}{6})] \Rightarrow \frac{9 y}{P - y} = e^{k [t + \frac{6}{π} \sin (\frac{πt}{6})] P} \Rightarrow 9 y = P e^{k [t + \frac{6}{π} \sin (\frac{πt}{6})] P} - y e^{^{k [t + \frac{6}{π} \sin (\frac{πt}{6})] P}} \Rightarrow y (t) = \frac{P e^{^{k [t + \frac{6}{π} \sin (\frac{πt}{6})] P}}}{9 + e^{k [t + \frac{6}{π} \sin (\frac{πt}{6})] P}} . . . . (6)$

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