Answer :
Answer: The differential equation can be written as;
dN/dt = k(P-N)
dN/dt = k(6100000 - N)
Step-by-step explanation:
Since, the number N of those people who have heard a certain rumor is proportional to the number of those who have not yet heard the rumor.
The rate of change of number of those who have heard the rumour is = dN/dt
dN/dt = k(P-N)
dN/dt = k(6100000 - N)
Where P is the population of the city = P = 6100000
k is the proportionality constant.
(P-N) = number of those that have not heard the rumour
The logistic differential equation that models this situation is:
[tex]\frac{df}{dx} = 1.0516\left(1 - \frac{f}{6100000}\right)f[/tex]
The number of people that hear the rumor has a limit, which is the carrying capacity, which is why the logistic differential equation is used.
It is given by:
[tex]\frac{df}{dx} = r\left(1 - \frac{f}{K}\right)f[/tex]
And the solution is:
[tex]f(x) = \frac{Kf(0)e^{rx}}{K + f(0)(e^{rx} - 1)}[/tex]
In which:
- K is the carrying capacity, which is the population size.
- r is the growth rate.
- f(0) is the initial value.
In this problem: [tex]K = 6100000, f(0) = 5600, f(5) = 0.15(6100000)[/tex].
This is applied into the solution to find r.
[tex]f(x) = \frac{Kf(0)e^{rx}}{K + f(0)(e^{rx} - 1)}[/tex]
[tex]0.15(6100000) = \frac{6100000(5600)e^{5r}}{6100000 + 5600(e^{5r} - 1)}[/tex]
[tex]\frac{5600e^{5r}}{6100000 + 5600(e^{5r} - 1)} = 0.15[/tex]
[tex]5600e^{5r} = 915000 + 840(e^{5r} - 1)[/tex]
[tex]4760e^{5r} = 914160[/tex]
[tex]e^{5r} = \frac{914160}{4760}[/tex]
[tex]e^{5r} = 192.05[/tex]
[tex]\ln{e^{5r}} = \ln{192.05}[/tex]
[tex]5r = \ln{192.05}[/tex]
[tex]r = \frac{\ln{192.05}}{5}[/tex]
[tex]r = 1.0516[/tex]
Hence, the differential equation is:
[tex]\frac{df}{dx} = 1.0516\left(1 - \frac{f}{6100000}\right)f[/tex]
A similar problem is given at https://brainly.com/question/13229117