**The logistic equation is given as:**

**\[ \dfrac{dP}{dt} = 0.05P + 0.0005(P)^2 \]**

**Where time $t$ is measured in the weeks.**

**What is the carrying capacity?****What is the value of k?**

This question aims to explain the carrying capacity **K** and value of relative growth rate coefficient** k** of the logistic equation that is given as:

\[ \dfrac{dP}{dt} = 0.05P + 0.0005(P)^2 \]

Logistic Differential Equations are used for modeling the growth of populations and other systems that have an exponentially increasing or decreasing function. A logistic differential equation is an ordinary differential equation that generates a logistic function.

The logistic population growth model is given as:

\[ \dfrac{dP}{dt} = kP(1 – \dfrac{P}{k} ) \]

Where:

t is the time taken by the population to grow.

k is the relative growth rate coefficient.

K is the carrying capacity of the logistic equation.

P is the population after the time t.

The carrying capacity K is the limiting value of the given population as time approaches infinity. The population must always tend towards the carrying capacity K. The relative growth rate coefficient k determines the rate at which the population is growing.

**Expert Answer:**

The general logistic equation for a population is given as:

\[ \dfrac{dP}{dt} = kP(1 – \dfrac{P}{k} ) \]

The logistic differential equation for the said population is given as:

\[ \dfrac{dP}{dt} = 0.05P + 0.0005(P)^2 \]

In order to calculate the carrying capacity $K$ and relative growth rate coefficient $k$, let’s modify the given logistic equation.

\[ \dfrac{dP}{dt} = 0.05P(1 + 0.01P ) \]

\[ \dfrac{dP}{dt} = 0.05P(1 + \dfrac{P}{100} ) \]

Now, compare it with the general logistic equation.

The value of the carrying capacity $K$ is given as:

\[ K = 100 \]

The value of the relative growth coefficient $k$ is given as:

\[ k = 0.05 \]

**Alternative Solution:**

Comparing both values that the equation gives,

The value of carrying capacity $K$ is:

\[ K = 100 \]

The value of relative growth coefficient is:

\[ k = 0.05 \]

**Example:**

Suppose that a population develops according to the logistic equation given:

\[ \dfrac{dP}{dt} = 0.08P – 0.0008(P)^2 \] where t is measured in weeks.

(a) What is the carrying capacity?

(b) What is the value of k?

The logistic equation given for the population is:

\[ \dfrac{dP}{dt} = 0.08P – 0.0008(P)^2 \]

Where time is measured in weeks.

The logistic equation for any population is defined as:

\[ \dfrac{dP}{dt} = kP(1 – \dfrac{P}{k} ) \]

Where $k$ is the relative growth coefficient and $K$ is the carrying capacity of the population.

In order to calculate the values of the carrying capacity and relative growth coefficients, let’s modify the given logistic equation for the population.

\[ \dfrac{dP}{dt} = 0.08P – 0.0008(P)^2 ) \]

\[ \dfrac{dP}{dt} = 0.08P( 1 – 0.01P ) \]

\[ \dfrac{dP}{dt} = 0.08P( 1 – \dfrac{P}{100} ) \]

Comparing the equation gives us:

\[ K = 100 \]

\[ k = 0.08 \]

**Therefore, the value of carrying capacity $K$ is $100$ and the value of relative growth coefficient $k$ is $0.08$.**