Suppose that an insect population numbers 1000 initially, and that its death rate is 0 (so none of the insects are dying) and the birth rate is 0.0002P.

1. Construct the differential equation with initial condition.

2. Solve the differential equation.

3. What happens to the population at t --> 5?

We are given the following information about an insect population:

- Initial population: \( P(0) = 1000 \)
- Death rate: \( 0 \) (no insects are dying)
- Birth rate: \( 0.0002P \), where \( P \) is the population at time \( t \).

### 1. Constructing the Differential Equation

Let \( P(t) \) represent the insect population at time \( t \).

Since the birth rate is \( 0.0002P \) and there is no death rate, the rate of change of the population \( \frac{dP}{dt} \) can be expressed as:

\[
\frac{dP}{dt} = \text{birth rate} - \text{death rate} = 0.0002P - 0 = 0.0002P
\]

The differential equation with the initial condition \( P(0) = 1000 \) is:

\[
\frac{dP}{dt} = 0.0002P, \quad P(0) = 1000
\]

### 2. Solving the Differential Equation

This is a separable differential equation. We can solve it by separating the variables and integrating:

\[
\frac{dP}{P} = 0.0002 \, dt
\]

Integrating both sides:

\[
\int \frac{1}{P} \, dP = \int 0.0002 \, dt
\]

The integrals yield:

\[
\ln |P| = 0.0002t + C
\]

Exponentiating both sides to solve for \( P \):

\[
P = e^{0.0002t + C} = e^C \cdot e^{0.0002t}
\]

Let \( e^C \) be a constant \( A \), so:

\[
P(t) = A \cdot e^{0.0002t}
\]

Using the initial condition \( P(0) = 1000 \):

\[
1000 = A \cdot e^{0.0002 \cdot 0} = A \cdot 1
\]

Thus, \( A = 1000 \), and the solution to the differential equation is:

\[
P(t) = 1000 \cdot e^{0.0002t}
\]

### 3. What Happens to the Population as \( t \to 5 \)?

To determine the population at \( t = 5 \), we substitute \( t = 5 \) into the solution \( P(t) = 1000 \cdot e^{0.0002t} \):

\[
P(5) = 1000 \cdot e^{0.0002 \cdot 5}
\]

Let me calculate this.At \( t = 5 \), the population is approximately \( 1001 \) insects. This shows that the population increases slightly due to the small birth rate.

### Summary:
1. **Differential Equation**: \( \frac{dP}{dt} = 0.0002P \), with \( P(0) = 1000 \).
2. **Solution**: \( P(t) = 1000 \cdot e^{0.0002t} \).
3. **Population at \( t = 5 \)**: Approximately 1001 insects, indicating slow population growth.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the solution change if the birth rate was dependent on both population and time?
2. What happens to the population as \( t \to \infty \)?
3. How would introducing a death rate affect the differential equation and its solution?
4. What would the population be after 10 years under the same conditions?
5. How can this model be applied to real-world situations involving population dynamics?

**Tip**: When solving separable differential equations, always check if the initial conditions have been applied correctly to find the particular solution.

Suppose that an insect population numbers 1000 initially, and that its death rate is 0 (so none of the insects are dying) and the birth rate is 0.0002P. 1. Construct the differential equation with initial condition. 2. Solve the differential equation. 3. What happens to the population at t --> 5?

Learn how to model an insect population growth using a differential equation where the birth rate is 0.0002P. This problem involves solving a separable differential equation and applying exponential growth concepts. Suitable for Grades 11-12 and undergraduate students.

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