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Part 1
Suppose that the population​ P(t) of a country satisfies the differential equation StartFraction dP Over dt EndFraction equals kP left parenthesis 800 minus Upper P right parenthesis
 with k constant. Its population in 1960 was 200 million and was then growing at the rate of 3 million per year. Predict this​ country's population for the year 2020.

The given differential equation for the population \( P(t) \) is:

\[
\frac{dP}{dt} = kP(800 - P)
\]

Where:
- \( P(t) \) is the population (in millions) at time \( t \),
- \( k \) is a constant,
- \( \frac{dP}{dt} \) is the rate of change of population.

We are given the following initial conditions:
- In 1960, the population was 200 million: \( P(0) = 200 \).
- The rate of population growth in 1960 was 3 million per year: \( \frac{dP}{dt}(0) = 3 \).

We are asked to predict the population in 2020, i.e., 60 years later, so \( t = 60 \).

### Step 1: Set up the differential equation and solve for \( k \)
The differential equation is separable. Rewrite it as:

\[
\frac{dP}{P(800 - P)} = k \, dt
\]

We now solve for \( k \) using the initial condition \( P(0) = 200 \) and \( \frac{dP}{dt}(0) = 3 \).

### Step 2: Solve the separable differential equation
Let me solve the differential equation explicitly and find \( k \) based on the initial conditions.The general solution to the differential equation is:

\[
P(t) = \frac{800}{C_1 e^{-800kt} + 1}
\]

Where \( C_1 \) is the constant to be determined using the initial conditions.

### Step 3: Use initial conditions to find \( C_1 \) and \( k \)
Given:
- \( P(0) = 200 \), which will help us find \( C_1 \),
- \( \frac{dP}{dt}(0) = 3 \), which will allow us to solve for \( k \).

Let me now substitute \( P(0) = 200 \) into the solution to determine \( C_1 \), and then use \( \frac{dP}{dt}(0) = 3 \) to find \( k \).Using the initial condition \( P(0) = 200 \), we find that \( C_1 = 3 \).

Now, let's use the other condition, \( \frac{dP}{dt}(0) = 3 \), to find the value of \( k \). This involves differentiating the solution for \( P(t) \) and applying this condition at \( t = 0 \). I will compute this next.Using the condition \( \frac{dP}{dt}(0) = 3 \), we find that \( k = \frac{1}{40000} \).

### Step 4: Predict the population in 2020
Now that we know \( C_1 = 3 \) and \( k = \frac{1}{40000} \), we can substitute these values into the solution for \( P(t) \) and find the population at \( t = 60 \) (corresponding to the year 2020). Let's calculate that next.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like me to manually explain the rest of the process or clarify any steps so far?

Here are some follow-up questions:
1. How do initial conditions affect the population model?
2. What does the constant \( k \) represent in this logistic population model?
3. What assumptions are implicit in the logistic growth model used?
4. How might changes in \( k \) affect the long-term population prediction?
5. What alternative models could be used for predicting population growth?

**Tip:** In logistic models, the population tends to stabilize at the carrying capacity over time.

Suppose that the population P(t) of a country satisfies the differential equation dP/dt = kP(800 - P) with k constant. Its population in 1960 was 200 million and was then growing at the rate of 3 million per year. Predict this country's population for the year 2020.

Solve a logistic differential equation to predict a country's population in 2020. Using initial conditions, we explore how to calculate constants and predict future population trends. This detailed solution involves separation of variables and the logistic growth model.

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