Respuesta :
Answer:
The solution to the corresponding initial-value problem is given by
[tex]N(t)=\frac{20000}{1+99e^{-0.7033t}}[/tex].
Step-by-step explanation:
Differential equations can be used to represent the size of a population as it varies over time.
The logistics equation is a differential equation that models population growth and is given by
[tex]\frac{dP}{dt} =P(r-\frac{r}{K}p )[/tex]
where [tex]K[/tex] is the carrying capacity, it is the maximum population that the environment can support.
With constants relabeled, the nonlinear equation is the same as
[tex]\frac{dP}{dt} =P(a-bP)[/tex]
The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution.
Consider the logistic differential equation subject to an initial population of [tex]P_0[/tex] with carrying capacity [tex]K[/tex] and growth rate [tex]r[/tex]. The solution to the corresponding initial-value problem is given by
[tex]P(t)=\dfrac{P_0Ke^{rt}}{(K-P_0)+P_0e^{rt}}[/tex]
or
[tex]P(t)=\dfrac{aP_0}{bP_0+(a-bP_0)e^{-at}}[/tex]
From here we get:
[tex]P(t)=\frac{aP_0}{bP_0} =\frac{a}{b}[/tex] as [tex]t\rightarrow \infty[/tex]
From the information given:
- The number [tex]N(t)[/tex] of people in a community who are exposed to a particular advertisement is governed by the logistic equation.
- [tex]N(0) = 200[/tex]
- [tex]N(1) = 400[/tex]
- The limiting number of people in the community who will see the advertisement is 20,000.
So, the logistic equation is
[tex]N(t)=\dfrac{a200}{200b+(a-200b)e^{-at}}[/tex]
Since the limiting number is 20,000, we get that
[tex]\lim_{t \to \infty} N(t)=\frac{a}{b}=20000[/tex]
Next, we divide by [tex]b[/tex] in the numerator and the denominator.
[tex]N(t)=\dfrac{200\frac{a}{b} }{200+(\frac{a}{b}-200)e^{-at}}[/tex]
[tex]N(t)=\dfrac{200\cdot 20000 }{200+(20000-200)e^{-at}}[/tex]
[tex]N(t)=\frac{20000}{1+99e^{-at}}[/tex]
To find the value of the constant [tex]a[/tex] we use the initial condition [tex]N(1) = 400[/tex].
[tex]N(1)=\frac{20000}{1+99e^{-a}}=400\\\\\frac{20000}{1+99e^{-a}}\left(1+99e^{-a}\right)=400\left(1+99e^{-a}\right)\\\\20000=400\left(1+99e^{-a}\right)\\\\\frac{400\left(1+99e^{-a}\right)}{400}=\frac{20000}{400}\\\\1+99e^{-a}=50\\\\99e^{-a}=49\\\\e^{-a}=\frac{49}{99}\\\\\ln \left(e^{-a}\right)=\ln \left(\frac{49}{99}\right)\\\\a=-\ln \left(\frac{49}{99}\right)\approx0.7033[/tex]
The solution to the corresponding initial-value problem is given by
[tex]N(t)=\frac{20000}{1+99e^{-0.7033t}}[/tex]
