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Chapter 10: Problem 31

A model that describes the relationship between the price and the weekly sales of a product might have a form such as $$ \frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right), $$ where \(y\) is the volume of sales and \(p\) is the price per unit. That is, at any time, the rate of decrease of sales with respect to price is directly proportional to the sales level and inversely proportional to the sales price plus a constant. Solve this differential equation. (Figure 6 shows several typical solutions.)

Short Answer

Expert verified
The solution is \( y = K (p+3)^{-1/2} \).

Step by step solution

01

Identify the given differential equation

The given differential equation is \ \( \frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right) \).
02

Rewrite the equation to separate variables

Rewrite the equation: \ \( \frac{dy}{y} = -\frac{1}{2} \frac{dp}{p+3} \).
03

Integrate both sides

Integrate both sides of the separated equation: \ \( \int \frac{dy}{y} = \int -\frac{1}{2} \frac{dp}{p+3} \).
04

Perform the integrals

Perform the integration: \ \( \ln|y| = -\frac{1}{2} \ln|p+3| + C \), where \( C \) is the integration constant.
05

Simplify the solution

Exponentiate both sides to isolate \( y \): \ \( y = e^{C} (p+3)^{-1/2} \). Rename \( e^C \) to \( K \), a constant: \ \( y = K (p+3)^{-1/2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
Separation of variables is a powerful technique for solving differential equations. This method is especially useful when a function and its derivative appear in a product form.
In our exercise, the given differential equation is: \[ \frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right) \] First, rewrite the equation to separate the variables (terms involving y on one side, and terms involving p on the other): \[ \frac{dy}{y} = -\frac{1}{2} \frac{dp}{p+3} \] By separating the variables through algebraic manipulation, we make it easier to integrate and solve the differential equation.
This step transforms our original problem into a form that can be integrated on both sides.
Integration
Once the variables are separated, the next step is integration. This allows us to find a solution in terms of the original variables, y and p.
The separated equation is: \[ \int \frac{dy}{y} = \int -\frac{1}{2} \frac{dp}{p+3} \] Integrating both sides, we obtain: \[ \ln|y| = -\frac{1}{2} \ln|p+3| + C \] where C is the constant of integration.
To isolate y, we exponentiate both sides, resulting in: \[ y = e^{C} (p+3)^{-1/2} \] By renaming \( e^C \) to K (another constant), we simplify the expression: \[ y = K (p+3)^{-1/2} \] This step-by-step transformation through integration helps reveal the solution to our differential equation.
Initial Value Problem
An initial value problem (IVP) involves solving a differential equation subject to a specific initial condition. This condition allows us to determine the constant of integration, C, which depends on the initial state of the system. For example, suppose we know that when \( p = p_0, y = y_0 \). Using our general solution, we would substitute these values into: \[ y_0 = K (p_0 + 3)^{-1/2} \] Solving this for K: \[ K = y_0 (p_0 + 3)^{1/2} \] With K determined, we get the specific solution that fits the given initial condition. This process makes our general solution tailored to the initial value problem.
Exponential Functions
Exponential functions are a common component in solutions to differential equations. They often arise when integrating functions that involve natural logarithms.
In our example, the integration step produces natural logarithms: \[ \ln|y| = -\frac{1}{2} \ln|p+3| + C \] To solve for y, we exponentiate both sides: \[ y = e^{C} (p+3)^{-1/2} \] Exponential functions like \( e^C \) simplify into constants, making them easier to handle in differential equation solutions.
Understanding how these functions work is key to mastering the method of solving differential equations.

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Most popular questions from this chapter

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