How to solve logistic differential equations using separation of variables: a guide.

📚 Understanding Logistic Differential Equations

A logistic differential equation models population growth with a carrying capacity. Unlike simple exponential growth, it considers resource limitations. This results in a more realistic model for population dynamics.

📜 Historical Context

The logistic equation was popularized by Pierre François Verhulst in the 19th century to describe the self-limiting growth of a biological population. It addressed the limitations of the Malthusian growth model, which predicted unchecked exponential population increase.

🔑 Key Principles: Separation of Variables

The core idea behind solving logistic differential equations is using separation of variables. This technique allows us to rearrange the equation so each variable ($y$ and $t$ in this case) appears on only one side.

📝 Solving the Logistic Equation: A Step-by-Step Guide

Let's consider the standard logistic differential equation:

$\frac{dy}{dt} = ky(1 - \frac{y}{K})$

where:

• 🌱 $y(t)$ is the population at time $t$
• 📈 $k$ is the growth rate
• 🏠 $K$ is the carrying capacity

Here's how to solve it using separation of variables:

• ➕ Step 1: Separate the variables

  Rewrite the equation to get all $y$ terms on one side and all $t$ terms on the other:

  $\frac{dy}{y(1 - \frac{y}{K})} = k dt$

• ➗ Step 2: Integrate both sides

  Integrate both sides of the equation. The left side requires partial fraction decomposition:

  $\int \frac{dy}{y(1 - \frac{y}{K})} = \int k dt$

  After partial fraction decomposition, we get:

  $\int (\frac{1}{y} + \frac{1/K}{1 - \frac{y}{K}}) dy = \int k dt$

  Integrating both sides yields:

  $\ln|y| - \ln|1 - \frac{y}{K}| = kt + C$

• ➗ Step 3: Simplify and solve for $y(t)$

  Combine the logarithms and exponentiate:

  $\ln|\frac{y}{1 - \frac{y}{K}}| = kt + C$

  $\frac{y}{1 - \frac{y}{K}} = e^{kt + C} = Ae^{kt}$

  Where $A = e^C$. Now solve for $y$:

  $y = Ae^{kt}(1 - \frac{y}{K})$

  $y = Ae^{kt} - \frac{Ae^{kt}y}{K}$

  $y(1 + \frac{Ae^{kt}}{K}) = Ae^{kt}$

  $y(t) = \frac{Ae^{kt}}{1 + \frac{Ae^{kt}}{K}} = \frac{KAe^{kt}}{K + Ae^{kt}}$

• ✏️ Step 4: Apply Initial Conditions

  Use the initial condition $y(0) = y_0$ to find $A$:

  $y_0 = \frac{KA}{K + A}$

  $y_0(K + A) = KA$

  $y_0K + y_0A = KA$

  $A(K - y_0) = y_0K$

  $A = \frac{y_0K}{K - y_0}$

• ✅ Step 5: The Solution

  Substitute $A$ back into the equation for $y(t)$:

  $y(t) = \frac{K \frac{y_0K}{K - y_0} e^{kt}}{K + \frac{y_0K}{K - y_0} e^{kt}} = \frac{Ky_0 e^{kt}}{K - y_0 + y_0 e^{kt}}$

  $y(t) = \frac{Ky_0}{y_0 + (K - y_0)e^{-kt}}$

🌍 Real-World Examples

• 🐟 Fish Population: Modeling fish population growth in a lake, considering the lake's carrying capacity.
• 🦠 Bacterial Growth: Describing the growth of a bacterial colony in a petri dish, limited by available nutrients.
• 📈 Spread of Disease: Modeling the spread of a disease in a population, considering the number of susceptible individuals.
• 🛍️ Product Adoption: Modeling the adoption rate of a new product in a market, considering market saturation.

💡 Conclusion

Logistic differential equations provide a powerful tool for modeling real-world phenomena where growth is limited. By mastering separation of variables, you can solve these equations and gain insights into diverse applications. Keep practicing and exploring different scenarios!