Question

The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence of other factors, the population doubles each week. There are 200,000mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 20,000 mosquitoes perday. Set up a differential equation for the population of mosquitoes and make sure to solve for theproportionality constant using the information given. Determine the population of mosquitoes in thearea at any time.

Answer 1

Answer:

The population of mosquitoes in the area at any time t is:

$P(t)=201977.31-1977.31\times 2^{t}$

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=k\ dt$

Integrate the above expression as follows:

$\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}$

$\Rightarrow P=P_{0}e^{kt}$

It is provided that the population doubles every day.

Compute the value of k as follows:

$2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)$

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

$\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000$

The integrating factor for this is:

$e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}$

Then,

$P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}$

The initial population is 200,000.

Compute the value of C as follows:

$P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31$

Now substitute C in P (t),

$P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}$

Answer 2

This question is based on the concept of differential equation. Thus, the population of mosquitoes in the area at any time t is:$P(t) = 201977.31 - 1977.31 \times 2^t$.

Given:

There are 200,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 20,000 mosquitoes per day.

According to the question,

The rate of growth of mosquitoes can be expressed as:

$\dfrac{dP}{dt} = kP\\\dfrac{dP}{p} = k \;dt$

Now,integrating the above expression as follows:

$\int \dfrac{dP}{p} =\int k \;dt\\\\ln P = kt + C\\\\P =C e^{kt}\\\\P = P_0 \,e^{kt}$

Thus, It is given that the population doubles every day.  Now,compute the value of k as follows:

$2 = 1 \times e^{k \times 1}\\k = ln(2)$

It is also given that every day 20,000 mosquitoes are eaten.  The rate of growth per week can be expressed as:

$\dfrac{dP}{dt} = ln(2) P-14000\\\dfrac{dP}{dt} - ln(2) P= 14000$

Hence the integrating factor for this is,

$e^ {\int ln(2) dP }= e^ {\int ln(2)\int dt }= e^ {\int ln(2)t}$

Then,

$P(t) = e^{-ln(2)t} = \int e^{-ln(2)t} \, -14000dt\\\\ = -14000 \int e^{-ln(2)t} dt\\\\= -14000 \, \dfrac{e^{-ln(2)t}}{-ln(2) }+ C \\\\P(t) = (e^{-ln(2)t)} \times [ -14000 \times \dfrac{e^{-ln(2)t} }{-ln(2)} +C\\\\= \dfrac{14000}{ln(2)} +C \, e^{-ln(2) t$

Thus,the initial population is 200,000.  Compute the value of C as follows:

$P(t) = \dfrac{140000}{ln(2)} + C \, e^{-ln(2)t}\\\\200000 = \dfrac{140000}{ln(2)} + C \, e^{-ln(2)0}\\\\C = 200000 - \dfrac{140000}{ln(2)} \\\\C = -1977.31$

Now substituting C in P (t),

$P(t) = \dfrac{14000}{ln(2)} +C \, e^{-ln(2) t}\\\\P(t) = 201977.31 - 1977.31 \times 2^t$

Thus, the population of mosquitoes in the area at any time t is:$P(t) = 201977.31 - 1977.31 \times 2^t$.

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