Problem 68 - Calculus and its applications

Question

The validity of the WeberFechner Law has been the subject of great debate among psychologists. An alternative model, $$ \frac{d R}{d S}=k \cdot \frac{R}{S} $$ where \(k\) is a positive constant, has been proposed. Find the general solution of this equation. (This model has also been referred to as the Power Law of Stimulus-Response.)

Step-by-Step Solution

Answer

The general solution is \( R = C \cdot S^k \), where \( C \) is an arbitrary constant.

Step 1: Recognize the Differential Equation

The given differential equation is \( \frac{dR}{dS} = k \cdot \frac{R}{S} \). This is a first-order linear differential equation that can be solved by separating variables.

Step 2: Separate the Variables

Rearrange the terms of the equation to separate variables: \( \frac{dR}{R} = k \cdot \frac{dS}{S} \). This allows us to integrate each side with respect to its variable.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( R \) and the right side with respect to \( S \): \[ \int \frac{dR}{R} = \int k \cdot \frac{dS}{S} \]This results in the following integrals:\[ \ln |R| = k \cdot \ln |S| + C \] where \( C \) is the constant of integration.

Step 4: Solve for R

Exponentiate both sides to solve for \( R \):\[ e^{\ln |R|} = e^{k \ln |S| + C} \]This simplifies to:\[ |R| = e^C \cdot S^k \]Thus, \( R = C_1 \cdot S^k \), where \( C_1 = \pm e^C \) represents a constant.

Step 5: Write the General Solution

The general solution of the differential equation is:\[ R = C \cdot S^k \]where \( C \) is an arbitrary constant, and \( k \) is the given positive constant in the problem.

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation involves the derivative of the first degree. Essentially, it looks something like \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( y \) is the function we want to find. In our problem, \( \frac{dR}{dS} = k \cdot \frac{R}{S} \) is the first-order differential equation. Though it might seem complex, breaking it down into simpler terms helps us solve it effectively. Notice the equation can be rewritten in the form of a linear first-order differential equation by isolating \( R \) terms on one side and \( S \) terms on the other.

Separation of Variables

The technique of separation of variables comes in handy when solving differential equations. This method involves rearranging terms to separate variables on opposite sides of the equation. For the equation \( \frac{dR}{dS} = k \cdot \frac{R}{S} \), we move all terms involving \( R \) to one side and \( S \) to the other, resulting in \( \frac{dR}{R} = k \cdot \frac{dS}{S} \). By doing this, each side of the equation contains only one variable. It's a simple yet powerful way to make the equation easier to work with before integration.

Integration

Once we have the equation \( \frac{dR}{R} = k \cdot \frac{dS}{S} \) nicely split up, it's time to integrate both sides. Integration is about finding the antiderivative. For the left side, the integral of \( \frac{dR}{R} \) is \( \ln |R| \), while for the right side, \( \int k \cdot \frac{dS}{S} \) gives us \( k \cdot \ln |S| \). This process results in the relationship \( \ln |R| = k \cdot \ln |S| + C \), where \( C \) is the integration constant. Keep in mind that integration transforms our separated variables into a more expressive form that we can work with to find \( R \).

General Solution

To find the general solution from our integrated form \( \ln |R| = k \cdot \ln |S| + C \), we exponentiate both sides. This step's goal is to solve for \( R \). By doing this, we eliminate the natural logarithms, yielding \( |R| = e^C \cdot S^k \). The general solution formula becomes \( R = C_1 \cdot S^k \), where \( C_1 = \pm e^C \). Here, \( C_1 \) is an arbitrary constant, which can change based on initial conditions or specific values you may have. Ultimately, this general solution provides a family of functions that satisfies the original differential equation.