Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 3 / g

Table of contents
- Solution g

Solution

Dividing the

I

equation by the

S

\frac{d I}{d S} = - 1 + \frac{γ}{βS} .

Therefore

I + S - \frac{γ}{β} lo g S = I (0) + S (0) - \frac{γ}{β} lo g S (0)

is a first integral. Define the basic reproduction number by

R_{0} = βS (0) / γ

. At the removal peak,

S = S (0) / R_{0}

, and hence

R_{peak} = \frac{S ( 0 )}{R _{0}} lo g R_{0},

and, using

S + I + R = N

,

I_{peak} = I (0) + S (0) - \frac{S ( 0 )}{R _{0}} 1 + lo g R_{0} .

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