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}