Suppose that the population​ P(t) of a country satisfies the differential equation StartFraction dP Over dt EndFraction equals kP left parenthesis 400 minus Upper P right parenthesis
dP
dt=kP(400−P) with k constant. Its population in 1960 was 200 million and was then growing at the rate of 11 million per year. Predict this​ country's population for the year 2010.

The given differential equation for population growth is:

\[
\frac{dP}{dt} = kP(400 - P)
\]

Where:
- \( P(t) \) is the population at time \( t \),
- \( k \) is a constant,
- \( 400 \) represents a carrying capacity (in millions).

### Step 1: Setup the initial conditions
- At \( t = 0 \) (which represents the year 1960), \( P(0) = 200 \) million.
- The rate of population growth in 1960 was \( \frac{dP}{dt} = 11 \) million per year.

Using these initial conditions, we can determine the value of \( k \).

### Step 2: Find \( k \)
From the differential equation, when \( t = 0 \), \( P = 200 \):

\[
\frac{dP}{dt} = k \cdot 200 \cdot (400 - 200)
\]

\[
11 = k \cdot 200 \cdot 200
\]

\[
11 = 40000k
\]

\[
k = \frac{11}{40000} = 0.000275
\]

### Step 3: Solve the differential equation
The equation is separable. We rearrange it to separate the variables \( P \) and \( t \):

\[
\frac{dP}{P(400 - P)} = k dt
\]

To solve this, we use partial fraction decomposition on the left-hand side:

\[
\frac{1}{P(400 - P)} = \frac{A}{P} + \frac{B}{400 - P}
\]

Solving this decomposition, we get:

\[
\frac{1}{P(400 - P)} = \frac{1}{400} \left( \frac{1}{P} + \frac{1}{400 - P} \right)
\]

Thus, the equation becomes:

\[
\frac{1}{400} \left( \frac{1}{P} + \frac{1}{400 - P} \right) dP = k dt
\]

Integrating both sides:

\[
\frac{1}{400} \left( \ln |P| - \ln |400 - P| \right) = kt + C
\]

Simplifying:

\[
\ln \left( \frac{P}{400 - P} \right) = 400kt + C'
\]

Where \( C' \) is a constant of integration. Exponentiating both sides:

\[
\frac{P}{400 - P} = e^{400kt + C'}
\]

\[
\frac{P}{400 - P}

Suppose that the population P(t) of a country satisfies the differential equation dP/dt = kP(400 - P) with k constant. Its population in 1960 was 200 million and was then growing at the rate of 11 million per year. Predict this country's population for the year 2010.

Explore how to predict population growth using a differential equation model. This example covers population dynamics, separation of variables, and partial fraction decomposition to estimate the population of a country from 1960 to 2010. Ideal for advanced calculus students or undergraduate math courses.

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