Problem 35 - Calculus: Graphical, Numerical, Algebraic

Question

Logistic Differential Equation Show that the solution of the differential equation \(\frac{d P}{d t}=k P(M-P) \quad\) is \(\quad P=\frac{M}{1+A e^{-M k t}}\) where \(A\) is a constant determined by an appropriate initial condition.

Step-by-Step Solution

Answer

The solution of the logistic differential equation \( \frac{d P}{d t}=k P(M-P) \) is \( P = \frac{M}{1 + A e^{-Mkt}} \) where \( A = e^C \) and \( C \) is a constant determined by an appropriate initial condition.

Step 1: Separation of Variables

First, rewrite the given logistic differential equation with variables separated: rewrite the equation \( \frac{d P}{d t}=k P(M-P) \) as \( \frac{d P}{P(M-P)}= k dt \). This means that all \( P \) terms are on one side of the equation and all \( t \) terms are on the other.

Step 2: Perform Integration

Next, integrate both sides of the equation. On left hand side, this is a complex fraction; perform the integral by performing a partial fraction decomposition into more manageable pieces: \( \frac{1}{P(M-P)} = \frac{1}{M}\left( \frac{1}{P} + \frac{1}{M-P} \right) \). The resulting integrals form is \( \int \frac{1}{P(M-P)} dP = k \int dt \). Solving these integrals gives us \( \frac{1}{M}ln | \frac{P}{M-P} | = kt + C \), where \( C \) is the constant of integration.

Step 3: Simplify the Equation

Use properties of logarithms to simplify and solve for \( P \). Start by multiply through by \( M \) to get \( ln | \frac{P}{M-P} | = Mkt + C \). Then exponentiate both sides to remove the natural log to get: \( | \frac{P}{M-P} | = e^{Mkt+C} \).

Step 4: Isolate \( P \) in the Equation

We can rewrite \( e^{Mkt+C} \) as \( A e^{Mkt} \) where \( A = e^C \). We then have \( | \frac{P}{M-P} | = A e^{Mkt} \). To remove the absolute value we have \( \frac{P}{M-P} = \pm A e^{Mkt} \). The positive is selected based on the context (as \( P \) represents population, it cannot be negative). Then cross multiply to yield \( P = \frac{M}{1 + A e^{-Mkt}} \), which is the required solution.

Key Concepts

Separation of Variables

The technique of separation of variables is essential when dealing with differential equations. It involves rearranging an equation to isolate all terms involving one variable on one side of the equation, and all terms involving the other variable on the opposite side. In the context of a logistic differential equation, this approach simplifies the integration process by allowing us to directly integrate with respect to each variable independently.

Consider the logistic differential equation \(\frac{d P}{d t}=k P(M-P)\). To apply separation of variables, we divide both sides by \(P(M-P)\) and multiply by \(dt\), resulting in \(\frac{d P}{P(M-P)}= k dt\). Here, all the \(P\) terms are on one side, and all the \(t\) terms are on the other, setting the stage for straightforward integration.

Partial Fraction Decomposition

Partial fraction decomposition is a method used in calculus to break down complex rational expressions into simpler ones that are easier to integrate. Specifically, when faced with a fraction where the numerator is of lower degree than the denominator, and the denominator can be factored into linear factors, we can decompose the complex fraction into a sum of simpler fractions.

In our logistic differential equation after applying separation of variables, we obtain \(\frac{1}{P(M-P)}\), a complex fraction that is challenging to integrate directly. We use partial fraction decomposition to express it as \(\frac{1}{M}\left( \frac{1}{P} + \frac{1}{M-P} \right)\). Each of these simpler fractions corresponds to a basic integral that we can evaluate using fundamental integration techniques.

Exponential Growth and Decay

Exponential growth and decay are phenomena observed in various natural and scientific contexts, and they are described by differential equations where the rate of change of a quantity is proportional to the quantity itself. For exponential growth, the rate is positive leading to an increase over time, while for decay, the rate is negative, indicating a reduction in the quantity over time.

In our logistic differential equation, the solution \(P = \frac{M}{1+A e^{-M k t}}\) characterizes a form of growth that initially behaves like exponential growth but levels off as it approaches the carrying capacity \(M\). The term \(A e^{-M k t}\) demonstrates exponential decay, which effectively reduces the value of the denominator over time, causing \(P\) to approach \(M\). The solution hence encapsulates both the explosive nature of exponential growth at the outset and the stabilizing effect of decay as the population nears its maximum sustainable size.