Problem 37 - Calculus: Early Transcendentals

Question

(a) Show that the substitution \( z = 1/P \) transforms the logistic differential equation \( P' = kP(1 - P/M) \) into the linear differential equation \( z' + kz = \frac {k}{M} \) (b) Solve the linear differential equation in part (a) and thus obtain an expression for \( P(t). \) Compare with Equation 9.4.7.

Step-by-Step Solution

Answer

The substitution and transformation show the linear form; solving yields \( P(t) = \frac{1}{\frac{1}{M} + Ce^{-kt}} \).

Step 1: Substitute z for 1/P

We start by taking the substitution given in the problem: Let \( z = \frac{1}{P} \). From this relationship, differentiate both sides with respect to \( t \): \( z' = \frac{d}{dt} \left( \frac{1}{P} \right) = -\frac{1}{P^2} \cdot P' \).

Step 2: Express P' in terms of z and z'

Since \( z' = -\frac{1}{P^2} \cdot P' \), we know \( P = \frac{1}{z} \) and hence \( P^2 = \frac{1}{z^2} \). Solving for \( P' \), we have:\( P' = -z^2 z' \).

Step 3: Substitute P' into logistic equation

The given logistic differential equation is \( P' = kP(1 - \frac{P}{M}) \). Substitute \( P = \frac{1}{z} \) and \( P' = -z^2 z' \): \( -z^2 z' = k \cdot \frac{1}{z} \left( 1 - \frac{1}{zM} \right) \).

Step 4: Simplify to the linear form

Simplifying the right side of the equation: \[ k \left( \frac{1}{z} - \frac{1}{z^2 M} \right) = k \frac{1}{z} - \frac{k}{z^2 M} \]Thus, the equation becomes\( -z^2 z' = k \frac{1}{z} - \frac{k}{z^2 M} \) Multiply through by \( z^2 \) to obtain:\( -z' = kz - \frac{k}{M} \)or equivalently,\( z' + kz = \frac{k}{M} \).

Step 5: Recognize the standard linear differential form

The transformed equation \( z' + kz = \frac{k}{M} \) is a first-order linear differential equation of the form \( y' + py = q \) where \( p = k \) and \( q = \frac{k}{M} \).

Step 6: Solve the linear differential equation

The integrating factor \( \mu(t) \) for \( z' + kz = \frac{k}{M} \) is \( e^{\int k dt} = e^{kt} \).Multiply through by \( \mu(t) \):\( e^{kt}z' + ke^{kt}z = \frac{k}{M}e^{kt} \).This can be rewritten as \( \frac{d}{dt}(e^{kt}z) = \frac{k}{M}e^{kt} \).Integrate both sides:\( e^{kt}z = \frac{k}{M} \int e^{kt} dt + C \).Solve the integral: \( \int e^{kt} dt = \frac{1}{k} e^{kt} \).Substitute back:\( e^{kt}z = \frac{k}{M} \cdot \frac{1}{k} e^{kt} + C \)\( \Rightarrow e^{kt}z = \frac{1}{M} e^{kt} + C \).Solving for \( z \):\( z = \frac{1}{M} + Ce^{-kt} \).

Step 7: Express P in terms of t and compare with Equation 9.4.7

Since \( z = \frac{1}{P} \), we have \( \frac{1}{P} = \frac{1}{M} + Ce^{-kt} \).Rearrange to solve for \( P \):\( P = \frac{1}{\frac{1}{M} + Ce^{-kt}} \).This matches the standard logistic equation solution, providing the expression for \( P(t) \) as a function of time. The comparison with Equation 9.4.7 shows that they indeed represent the same solution form.

Key Concepts

Substitution method in differential equations

The substitution method is a powerful technique used in solving complex differential equations by simplifying them into a more familiar form. In the case of the logistic differential equation, we're initially dealing with the non-linear equation \( P' = kP(1 - P/M) \). To simplify this, we use the substitution \( z = 1/P \). This transformation allows us to rewrite the equation into a linear form.

This works because when we substitute \( z = 1/P \), we can express the derivative \( z' \) as \( z' = -\frac{1}{P^2} \cdot P' \). Thus, we link \( P \) and \( z \) with a differential relationship. This alteration changes a non-linear equation into a linear one, thereby making it more tractable. The substitution method essentially reframes the problem, providing a new perspective that might simplify both the interpretation and the solving process of the differential equation.

Linear differential equations

Linear differential equations are often easier to solve than non-linear equations due to their straightforward properties. An equation is considered linear if it can be written in the form \( y' + py = q \), where \( p \) and \( q \) are functions of the independent variable, commonly time \( t \). The significance in our transformed equation, \( z' + kz = \frac{k}{M} \), is that it fits this classic linear template.

The linearity lies in the structure, permitting techniques such as finding integrating factors or superposition, which aren't feasible for non-linear equations. By identifying our model as a linear equation, we've simplified the path to solving it. Linear equations are incredibly versatile and can describe a wide range of physical phenomena, making understanding their properties so vital for scientists and mathematicians alike.

Solving differential equations

When solving differential equations, especially first-order linear types, we often use an integrating factor. This method finds a particular solution by multiplying through the differential equation by a function that simplifies integration. For the equation \( z' + kz = \frac{k}{M} \), the integrating factor is \( \mu(t) = e^{kt} \).

By applying this integrating factor, the equation is manipulated to be easily solvable by integration, taking the form \( \frac{d}{dt}(e^{kt}z) = \frac{k}{M}e^{kt} \). Integration leads us to solve \( e^{kt}z = \frac{1}{M}e^{kt} + C \), where \( C \) represents the integration constant determined by initial conditions. Solving for \( z \), and then substituting back \( z = 1/P \), we derive \( P(t) \). This exercise demonstrates the power of methods like integrating factors and strategic substitutions in transforming and solving what initially appear to be daunting differential equations. Understanding these techniques makes solving such equations approachable and methodical.