Answer :
Answer:
Step-by-step explanation:
Assuming that the differential equation is
[tex]\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)[/tex].
We need to solve it and obtain an expression for [tex]P(t)[/tex] in order to complete the exercise.
First of all, this is an example of the logistic equation, which has the general form
[tex]\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)[/tex].
In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that [tex]k=0.04[/tex] and [tex]K=500[/tex] and the initial condition [tex]P(0)=100[/tex].
Notice that this equation is separable, then
[tex] \frac{dP}{P(1-P/K)} = kdt[/tex].
Now, intagrating in both sides of the equation
[tex] \int\frac{dP}{P(1-P/K)} = \int kdt = kt +C[/tex].
In order to calculate the integral in the left hand side we make a partial fraction decomposition:
[tex]\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P} [/tex].
So,
[tex] \int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|[/tex].
We have obtained that:
[tex] -\ln\left| \frac{K-P}{P}\right| = kt +C[/tex]
which is equivalent to
[tex] \ln\left| \frac{K-P}{P}\right|= -kt -C[/tex]
Taking exponentials in both hands:
[tex] \left| \frac{K-P}{P}\right| = e^{-kt -C} [/tex]
Hence,
[tex] \frac{K-P(t)}{P(t)} = Ae^{-kt}[/tex].
The next step is to substitute the given values in the statement of the problem:
[tex] \frac{500-P(t)}{P(t)} = Ae^{-0.04t}[/tex].
We calculate the value of [tex]A[/tex] using the initial condition [tex]P(0)=100[/tex], substituting [tex]t=0[/tex]:
[tex] \frac{500-100}{100} = A}[/tex] and [tex]A=4[/tex].
So,
[tex] \frac{500-P(t)}{P(t)} = 4e^{-0.04t}[/tex].
Finally, as we want the value of [tex]t[/tex] such that [tex]P(t)=200[/tex], we substitute this last value into the above equation. Thus,
[tex] \frac{500-200}{200} = 4e^{-0.04t}[/tex].
This is equivalent to [tex]\frac{3}{8} = e^{-0.04t}[/tex]. Taking logarithms we get [tex] \ln\frac{3}{8} = -0.04t[/tex]. Then,
[tex] t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325[/tex].
So, the population of rats will be 200 after 25 months.