Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, solve each differential equation. $$ \frac{d y}{d x}+\frac{y}{x}=\frac{1}{x} $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{x + C}{|x|} \).

Step 1: Identify the Type of Differential Equation

This is a first-order linear differential equation in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \( P(x) = \frac{1}{x} \) and \( Q(x) = \frac{1}{x} \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) is calculated using the formula \( \mu(x) = e^{\int P(x) \, dx} \). For this problem, \( P(x) = \frac{1}{x} \), so:\[ \mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln |x|} = |x|. \]

Step 3: Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( |x| \):\[ |x| \frac{dy}{dx} + |x| \frac{y}{x} = |x| \cdot \frac{1}{x}. \]This simplifies to:\[ |x| \frac{dy}{dx} + y = 1. \]

Step 4: Rewrite as Exact Derivative

Notice that the left side is the derivative of the product of \( |x| \) and \( y \):\[ \frac{d}{dx}(|x|y) = 1. \]

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(|x|y) \, dx = \int 1 \, dx. \]This gives:\[ |x|y = x + C, \]where \( C \) is the constant of integration.

Step 6: Solve for \( y \)

Divide both sides by \(|x|\) to solve for \( y \):\[ y = \frac{x + C}{|x|}. \]

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a key concept in solving many real-world problems. It has the general form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

Here, the term \( \frac{dy}{dx} \) represents the derivative of the unknown function \( y \) with respect to \( x \), indicating that changes in one variable directly affect the other.

To identify a first-order linear differential equation, look for the following features:

• The equation involves the first derivative \( \frac{dy}{dx} \). Higher derivatives do not appear.
• The unknown function \( y \) is not raised to a power (it appears linearly).
• Functions \( P(x) \) and \( Q(x) \) are functions only of \( x \) and not of \( y \).

Integrating Factor

The integrating factor is a powerful tool used to solve a first-order linear differential equation. It transforms a difficult equation into one that is much easier to integrate.

To find the integrating factor, \( \mu(x) \), use the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

This function \( \mu(x) \) simplifies the original equation by converting it into an exact derivative, making it straightforward to solve.

Steps to use the integrating factor:
- Calculate \( \mu(x) \) using the formula.
- Multiply every term in the differential equation by \( \mu(x) \). This makes the left-hand side of the equation an exact derivative.
- Proceed to integrate both sides of the resulting equation.

Exact Derivative

An exact derivative is an expression derived from a function whose components form a complete derivative of another function.

When solving differential equations, especially first-order linear ones, converting an equation to the form of an exact derivative simplifies the integration process.

How it works:
- Multiply the integrating factor with the entire equation to rearrange it.
- This operation allows the left side of the equation to express the derivative of a single function with respect to \( x \).
- The challenge is reduced to merely integrating a known derivative, which is straightforward.

For instance, in the equation \( |x| \frac{dy}{dx} + y = 1 \), the left side \( |x| \frac{dy}{dx} + y \) can be rewritten as \( \frac{d}{dx} (|x|y) \). This transformation highlights the power of making a differential equation exact for easier integration.

Constant of Integration

The constant of integration, usually represented by \( C \), arises when you integrate a mathematical expression.

During indefinite integration, actual solutions can differ by a constant amount; hence, we include \( C \) to represent all possible antiderivatives.

Why it's important:
- Provides a general solution rather than a specific one.
- Especially vital in solving differential equations, as it accounts for all potential functions fitting the differential equation.
- Allows further conditions, like initial conditions, to pinpoint a unique solution.

For example, upon integrating \( \frac{d}{dx}(|x|y) = 1 \), you obtain \( |x|y = x + C \), indicating a family of solutions that represent the same derivative behavior.