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vagabundo [1.1K]
3 years ago
12

One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P

who have contracted the virusand the fraction who have not contracted the virus.a.Write out a differential equation that is satisfied by P.b.Solve the differential equation.c.Ina small communityof 100,000 inhabitants. At 8:00 am on the initial day of reporting, 100peoplehad contractedthe virus. By 8:00 am on the third day,200more people had contacted the virus.At what time will 90% of the population havecontracted the virus
Mathematics
1 answer:
Darya [45]3 years ago
8 0

Answer:

The differential equation for the model is

\frac{dP}{dt}=kP(1-P)

The model for P is

P(t)=\frac{1}{1-0.99e^{t/447}}

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

\frac{dP}{dt}=kP(1-P)

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}

P(0)=  \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)=  \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447

Then the model for the population infected at time t is:

P(t)=\frac{1}{1-0.99e^{t/447}}

Now, we can calculate t for P(t)=90,000

P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

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