Respuesta :
Answer:
a) [tex] t = 2 *ln(\frac{82}{5}) =5.595[/tex]
b) [tex] t = 2 *ln(-\frac{820}{p_0 -820}) [/tex]
c) [tex] p_0 = 820-\frac{820}{e^6}[/tex]
Step-by-step explanation:
For this case we have the following differential equation:
[tex] \frac{dp}{dt}=\frac{1}{2} (p-820)[/tex]
And if we rewrite the expression we got:
[tex] \frac{dp}{p-820}= \frac{1}{2} dt[/tex]
If we integrate both sides we have:
[tex]ln|P-820|= \frac{1}{2}t +c[/tex]
Using exponential on both sides we got:
[tex] P= 820 + P_o e^{1/2t}[/tex]
Part a
For this case we know that p(0) = 770 so we have this:
[tex] 770 = 820 + P_o e^0[/tex]
[tex] P_o = -50[/tex]
So then our model would be given by:
[tex] P(t) = -50e^{1/2t} +820[/tex]
And if we want to find at which time the population would be extinct we have:
[tex] 0=-50 e^{1/2 t} +820[/tex]
[tex] \frac{820}{50} = e^{1/2 t}[/tex]
Using natural log on both sides we got:
[tex] ln(\frac{82}{5}) = \frac{1}{2}t[/tex]
And solving for t we got:
[tex] t = 2 *ln(\frac{82}{5}) =5.595[/tex]
Part b
For this case we know that p(0) = p0 so we have this:
[tex] p_0 = 820 + P_o e^0[/tex]
[tex] P_o = p_0 -820[/tex]
So then our model would be given by:
[tex] P(t) = (p_o -820)e^{1/2t} +820[/tex]
And if we want to find at which time the population would be extinct we have:
[tex] 0=(p_o -820)e^{1/2 t} +820[/tex]
[tex] -\frac{820}{p_0 -820} = e^{1/2 t}[/tex]
Using natural log on both sides we got:
[tex] ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t[/tex]
And solving for t we got:
[tex] t = 2 *ln(-\frac{820}{p_0 -820}) [/tex]
Part c
For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:
[tex] 12 = 2 *ln(\frac{820}{820-p_0}) [/tex]
[tex] 6 = ln (\frac{820}{820-p_0}) [/tex]
Using exponentials we got:
[tex] e^6 = \frac{820}{820-p_0}[/tex]
[tex] (820-p_0) e^6 = 820[/tex]
[tex] 820-p_0 = \frac{820}{e^6}[/tex]
[tex] p_0 = 820-\frac{820}{e^6}[/tex]
For the given case of increment of mice' population, we get following figures:
- After 5.59 months approx, the population of mice will extinct.
- The extinction time (in months) of population of mice when its given that [tex]p(0) = p_0[/tex] is given by:
- [tex]t = 2\ln(\dfrac{820}{820-p_0})[/tex]
- The initial population of mice for given conditions would be approx 818
What is differential equation?
An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations. The derivatives might be of any order, some terms might contain product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.
For the considered case, the population of mice with respect to time passed in months is given by the differential equation:
[tex]\dfrac{dp}{dt} = 0.5p - 410[/tex]
Taking same variable terms on same side, and then integrating, we get:
[tex]\dfrac{dp}{0.5p - 410} = dt\\\\\int \dfrac{dp}{0.5p - 410} = \int dt\\\\\dfrac{\ln(|0.5p - 410|)}{0.5} = t + C_1\\\ln(|0.5p - 410|) = 0.5t + 0.5C_1 = 0.5t + C[/tex]
where C₁ is integration constant.
Since it is specified that at time t = 0, the population p = 770, therefore, putting these values in the equation obtained above, we get:
[tex]\ln(|0.5p - 410|) = 0.5t + 0.5C_1 = 0.5t + C\\\\\ln(|0.5 \times 770 - 410|) = 0.5 \times 0 + C\\\\\ln(|-25|) = C\\C = \ln(25) \approx 3.22[/tex]
Therefore, we get the relation between p and t as:
[tex]\ln(|0.5p - 410|) = 0.5t + 0.5C_1 = 0.5t + C\\\ln(|0.5p - 410|) \approx 0.5t + 3.22\\\\|0.5p - 410| \approx e^{0.5t + 3.22}\\\text{Squaring both the sides}\\\\(0.5p - 410)^2 \approx e^{t+6.44}\\(p-820)^2 \approx 4e^{t+6.44}\\\\p^2 -1640p + 672400 \approx 4e^{t+6.44}[/tex]
Calculating the needed figures for each sub-parts of the problem:
a): The time at which the population becomes extinct.
Let it be t at which p becomes 0, then, from the equation obtained, we get:
[tex]p^2 -1640p + 672400 \approx 4e^{t+6.44}\\\text{At p = 0}\\672400 \approx 4e^{t+6.44}\\\\t \approx \ln{(\dfrac{672400}{4}) - 6.44 = \ln(168100) - 6.44 \approx 5.59 \text{\: (In months)}[/tex]
Thus, after 5.59 months approx, the population of mice will extinct.
b) Find the time of extinction if p(0) = p0, where 0 < p0 < 820
From the equation [tex]\ln(|0.5p - 410|) = 0.5t + C[/tex]
putting [tex]p = p_0[/tex] when t = 0, we get the value of C as:
[tex]\ln(|0.5p_0 - 410|) = C[/tex]
Thus, the equation becomes
[tex]\ln(|0.5p - 410|) = 0.5t + \ln(|0.5p_0 - 410|)[/tex]
At time of extension t months, p becomes 0, thus,
[tex]\ln(|0.5p - 410|) = 0.5t + \ln(|0.5p_0 - 410|)\\\text{At p = 0, we get}\\\\\ln(410)=0.5t + \ln(|0.5p_0 - 410|)\\\\t = 2\ln(\dfrac{410}{0.5p_0 - 410}) = 2\ln(\dfrac{820}{|p_0-820|})\\\\\text{Since 0 } < p_0 < 820, \text{ thus, we get }\\\\t = 2\ln(\dfrac{820}{820-p_0})[/tex]
Thus, the extinction time (in months) of population of mice when its given that [tex]p(0) = p_0[/tex] is given by:
[tex]t = 2\ln(\dfrac{820}{820-p_0})[/tex]
c) Find the initial population [tex]p_0[/tex] if the population is to become extinct in 1 year.
Putting t = 12 (since t is measured in months, and that 1 year = 12 months) in the equation obtained in the second part, we get the value of initial population as:
[tex]t = 2\ln(\dfrac{820}{820-p_0})\\\\12 = 2\ln(\dfrac{820}{820-p_0})\\e^{6} = \dfrac{820}{820-p_0}\\1 - \dfrac{p_0}{820} = \dfrac{1}{e^6}\\p_0 \approx 820(1 - \dfrac{1}{e^6}}) \approx 818[/tex]
Thus, the initial population of mice for given conditions would be approx 818
Therefore, for the given case of increment of mice' population, we get following figures:
- After 5.59 months approx, the population of mice will extinct.
- The extinction time (in months) of population of mice when its given that [tex]p(0) = p_0[/tex] is given by:
- [tex]t = 2\ln(\dfrac{820}{820-p_0})[/tex]
- The initial population of mice for given conditions would be approx 818
Learn more about differential equations here:
https://brainly.com/question/14744969
