RaySix

Post

at November 4th 2023, 2:09:20 pm.

Question:

A population is currently growing at a rate proportional to its size. The logistic growth equation for this population is given by:

\frac{dP}{dt} = kP \left(1- \frac{P}{M}\right)

Where:

$\frac{dP}{dt}$ represents the rate of change of the population with respect to time.
$P$ represents the population size.
$M$ represents the carrying capacity of the population.
$k$ represents the proportionality constant.

Given that the population size $P$ is initially 100 and the carrying capacity $M$ is 200, answer the following questions:

(a) Determine the differential equation that represents the given logistic growth equation.

(b) Solve the differential equation from part (a) to find the function $P(t)$ that models the population growth.

(c) Find the population size after 3 units of time, using the function $P(t)$ from part (b).

Answer:

(a) To determine the differential equation, we differentiate $P$ with respect to $t$ :

\frac{dP}{dt} = kP \left(1- \frac{P}{M}\right)

Therefore, the differential equation is:

\frac{dP}{dt} = kP \left(1- \frac{P}{M}\right)

(b)

\frac{dP}{P \left(1- \frac{P}{M}\right)} = k dt

Next, we decompose the fraction on the left side using partial fractions:

\frac{1}{P \left(1- \frac{P}{M}\right)} = \frac{A}{P} + \frac{B}{1-\frac{P}{M}}

Multiplying through by $P \left(1- \frac{P}{M}\right)$ , we get:

1 = A \left(1- \frac{P}{M}\right) + BP

Simplifying the equation, we have:

1 = A + \left(B - \frac{AB}{M}\right)P

Since the left side is a constant, the coefficients of $P$ on the right side must also be equal to a constant. Therefore, we have:

A = 1 \quad \text{and} \quad B - \frac{AB}{M} = 0

Now, substituting these values back into the fraction, we get:

\frac{1}{P \left(1- \frac{P}{M}\right)} = \frac{1}{P} - \frac{\frac{1}{M}}{1-\frac{P}{M}}

Substituting this back into the original equation, we have:

\frac{1}{P} - \frac{\frac{1}{M}}{1-\frac{P}{M}} = k dt

Integrating both sides, we obtain:

\ln|P| - \ln|M-P| = kt + C

Applying the logarithmic property $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$ , we can simplify the equation further:

\ln\left(\frac{P}{M-P}\right) = kt + C

Finally, we exponentiate both sides to obtain the population growth model $P(t)$ :

\frac{P}{M-P} = e^{kt+C}

After rearranging and solving for $P$ , we get:

P(t) = \frac{M}{1+\left(\frac{M}{P_0}-1\right)e^{-kt}}

where $P_0$ represents the initial population size.

(c) To find the population size after 3 units of time, we substitute $t = 3$ into the function $P(t)$ from part (b) with $P_0 = 100$ and $M = 200$ :

P(3) = \frac{200}{1+\left(\frac{200}{100}-1\right)e^{-k(3)}}

Simplifying further, we get:

P(3) = \frac{200}{1+e^{-3k}}

Thus, the population size after 3 units of time is given by $P(3) = \frac{200}{1+e^{-3k}}$ .