A population P obeys the logistic model. It satisfies the equation dP dt

The differential equation given is a logistic model for population growth:

$\frac{dP}{dt} = \frac{9}{1100} P (11 - P)$

To find $P(79)$, we first need to solve this differential equation with the initial condition $P(0) = 4$.

Step 1: Separate the variables

We start by separating the variables $P$ and $t$:

$\frac{dP}{P(11 - P)} = \frac{9}{1100} dt$

Step 2: Integrate both sides

The next step is to integrate both sides:

$\int \frac{dP}{P(11 - P)} = \frac{9}{1100} \int dt$

To integrate the left side, we can use partial fraction decomposition:

$\frac{1}{P(11 - P)} = \frac{A}{P} + \frac{B}{11 - P}$

Solving for $A$ and $B$:

$1 = A(11 - P) + BP$

This simplifies to:

$1 = 11A - AP + BP$

Matching coefficients, we get:

$A + B = 0 \quad \text{and} \quad 11A = 1$

So, $A = \frac{1}{11}$ and $B = -\frac{1}{11}$. Therefore:

$\frac{1}{P(11 - P)} = \frac{1}{11} \left(\frac{1}{P} + \frac{1}{11 - P}\right)$

Now, integrate both sides:

$\frac{1}{11} \left(\ln |P| - \ln |11 - P|\right) = \frac{9}{1100} t + C$

Step 3: Simplify the equation

Simplify the equation by combining the logarithms:

$\frac{1}{11} \ln \left|\frac{P}{11 - P}\right| = \frac{9}{1100} t + C$

Multiply both sides by 11:

$\ln \left|\frac{P}{11 - P}\right| = \frac{99}{1100} t + 11C$

Let $K = e^{11C}$, so:

$\frac{P}{11 - P} = K e^{\frac{99}{1100} t}$

Step 4: Solve for P(t)

Now solve for $P$:

$P = \frac{11K e^{\frac{99}{1100} t}}{1 + K e^{\frac{99}{1100} t}}$

Use the initial condition $P(0) = 4$ to solve for $K$:

$4 = \frac{11K}{1 + K}$

This gives:

$4 + 4K = 11K \quad \Rightarrow \quad 4 = 7K \quad \Rightarrow \quad K = \frac{4}{7}$

Thus, the equation for $P(t)$ is:

$P(t) = \frac{\frac{44}{7} e^{\frac{99}{1100} t}}{1 + \frac{4}{7} e^{\frac{99}{1100} t}}$

Step 5: Find $P(79)$

Finally, plug in $t = 79$:

$P(79) = \frac{\frac{44}{7} e^{\frac{99}{1100} \times 79}}{1 + \frac{4}{7} e^{\frac{99}{1100} \times 79}}$

Calculate the exponential term and simplify to get the final value of $P(79)$.

Let's calculate this numerically.The population at $t = 79$ is approximately $P(79) \approx 10.98$.

Would you like any more details or explanations?

Here are some related questions you might be interested in:

1. How is the logistic model used in real-world population studies?
2. What assumptions are made in the logistic growth model?
3. How does the initial population affect the growth curve in logistic models?
4. Can the logistic growth model be applied to populations with carrying capacities that change over time?
5. What is the significance of the parameter $\frac{9}{1100}$ in this equation?
6. How would the solution change if the initial population $P(0)$ was different?
7. How does the carrying capacity influence the long-term behavior of the population?
8. What are some limitations of using the logistic growth model for predicting real populations?

Tip: When solving differential equations, always double-check the boundary conditions, as they determine the specific solution to the equation.