Question

The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population, P, of the rats is expected to grow according to the differential equation dP dt equals the product of 0.04 times P and the quantity 1 minus P over 500 , where P t is measured in months. When does the population of the rats reach 200? Give your answer to the nearest month. (

Answer 1

Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$.

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$.

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 $k=0.04$ and $K=500$ and the initial condition $P(0)=100$.

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$.

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$.

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$.

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$.

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$.

We calculate the value of $A$ using the initial condition $P(0)=100$, substituting $t=0$:

$\frac{500-100}{100} = A}$ and $A=4$.

So,

$\frac{500-P(t)}{P(t)} = 4e^{-0.04t}$.

Finally, as we want the value of $t$ such that $P(t)=200$, we substitute this last value into the above equation. Thus,

$\frac{500-200}{200} = 4e^{-0.04t}$.

This is equivalent to $\frac{3}{8} = e^{-0.04t}$. Taking logarithms we get $\ln\frac{3}{8} = -0.04t$. Then,

$t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325$.

So, the population of rats will be 200 after 25 months.