How to Solve the Logistic Growth Differential Equation Step-by-Step

📚 Understanding Logistic Growth

The logistic growth equation describes how a population's growth rate changes as it approaches its carrying capacity. Unlike exponential growth, which assumes unlimited resources, logistic growth accounts for the fact that resources are finite.

📜 A Brief History

The logistic growth model was first introduced by Pierre-François Verhulst in the 19th century. He used it to describe the self-limiting growth of populations. Although initially overlooked, it gained prominence in the 20th century with the rise of ecology and resource management.

🔑 Key Principles of the Logistic Growth Equation

The logistic growth differential equation is typically written as:

$\frac{dP}{dt} = rP(1 - \frac{P}{K})$

Where:

• 🌱 P(t) represents the population size at time t.
• 📈 r is the intrinsic growth rate (the rate at which the population would grow if resources were unlimited).
• 🏠 K is the carrying capacity (the maximum population size that the environment can sustain).

📝 Solving the Logistic Growth Differential Equation: Step-by-Step

Here’s how to solve the logistic growth differential equation:

1. ➗ Step 1: Separation of Variables

   Rearrange the equation to separate the variables P and t:

   $\frac{dP}{P(1 - \frac{P}{K})} = r dt$

2. ➕ Step 2: Partial Fraction Decomposition

   Decompose the left side into partial fractions:

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

   Solving for A and B, we find A = 1/K and B = 1/K.

   Therefore, the equation becomes:

   $\frac{1}{K}(\frac{1}{P} + \frac{1}{1 - \frac{P}{K}})dP = r dt$

3. ∫ Step 3: Integration

   Integrate both sides of the equation:

   $\int \frac{1}{K}(\frac{1}{P} + \frac{1}{1 - \frac{P}{K}})dP = \int r dt$

   This results in:

   $\ln|P| - \ln|1 - \frac{P}{K}| = rKt + C$

4. Simplify and Solve for P

   Using logarithm properties, simplify the left side:

   $\ln|\frac{P}{1 - \frac{P}{K}}| = rKt + C$

   Exponentiate both sides:

   $\frac{P}{1 - \frac{P}{K}} = e^{rKt + C} = e^{C}e^{rKt}$

   Let $A = e^{C}$, then:

   $\frac{P}{1 - \frac{P}{K}} = A e^{rKt}$

   Solve for P:

   $P = \frac{K A e^{rKt}}{K + A e^{rKt}}$

   Divide top and bottom by $Ae^{rKt}$

   $P(t) = \frac{K}{1 + \frac{K}{A}e^{-rKt}}$

5. 📍 Applying the Initial Condition

   If you know $P(0)$ (the initial population), you can find the value of A.

   $P(0) = \frac{KA}{K+A}$

   Solving for A:

   $A = \frac{KP(0)}{K-P(0)}$

   Plug this back into the solution for $P(t)$ to obtain the specific solution.

   $P(t) = \frac{K}{1+(\frac{K-P(0)}{P(0)})e^{-rt}}$

🌍 Real-World Examples

• 🦠 Bacterial Growth: Modeling the growth of bacteria in a petri dish with limited nutrients.
• 🐟 Fish Populations: Predicting the population size of fish in a lake, considering factors like food availability and predation.
• 🌲 Forestry: Managing forest resources by understanding tree growth rates and carrying capacities.

💡 Tips for Success

• 🔍 Practice: Work through plenty of example problems to solidify your understanding.
• 🧪 Understand the Concepts: Make sure you grasp the meaning of each variable and how they relate to each other.
• 📝 Check Your Work: Always double-check your calculations to avoid errors.

🏆 Conclusion

The logistic growth differential equation is a powerful tool for modeling real-world phenomena where growth is limited by resources. By understanding the steps involved in solving it, you can gain valuable insights into population dynamics and other important processes.