Respuesta :
Answer:
[tex]I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}[/tex]
Step-by-step explanation:
The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:
[tex]\frac{dI}{dt}=a*I*(1000-I)[/tex]
Rearranging and integrating
[tex]\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t[/tex]
At the initial breakout (t=0) there was one trooper infected (I=1)
[tex]-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755[/tex]
In two days (t=2) there were 5 troopers infected
[tex]-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725[/tex]
Rearranging, we can model the number of infected troops (I) as
[tex]-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}[/tex]
