Question

\(x\left(y^{\prime}-y\right)=e^{x}\).

Short answer

The general solution is \( y = e^{x}(\ln|x| + C) \).

Step-by-step solution

Identify the Equation Type

The given equation is \( x(y' - y) = e^x \). This is a first-order linear differential equation because it involves the first derivative \( y' \) of \( y \) and is linear in both \( y \) and \( y' \).

Simplify the Equation

Rewrite the expression: \( x(y' - y) = e^x \) as \( xy' - xy = e^x \). This format helps identify the function and derivative terms separately.

Rewrite in Standard Form

Express the equation in the standard linear form \( y' + P(x) y = Q(x) \) by dividing through by \( x \): \( y' - y = \frac{e^x}{x} \). This gives us \( P(x) = -1 \) and \( Q(x) = \frac{e^x}{x} \).

Find the Integrating Factor

The integrating factor, \( \mu(x) \), is calculated using \( e^{\int P(x) \, dx} \). Here, \( P(x) = -1 \), so the integrating factor is \( \mu(x) = e^{-x} \).

Multiply Through by Integrating Factor

Multiply the entire differential equation by the integrating factor: \( e^{-x}(y' - y) = e^{-x}\frac{e^x}{x} \). This simplifies to \( e^{-x}y' - e^{-x}y = \frac{1}{x} \).

Solve Left-Hand Side as Product Derivative

Recognize the left-hand side as the derivative of a product: \((e^{-x} y)' = \frac{1}{x}\). This helps in integrating the equation easily.

Integrate Both Sides

Integrate both sides with respect to \( x \). The left side gives \( e^{-x} y \) and the right side requires the integral \( \int \frac{1}{x} \, dx = \ln|x| + C \).

Solve for General Solution

After integrating, we have \( e^{-x} y = \ln|x| + C \). Solving for \( y \) gives \( y = e^{x}(\ln|x| + C) \). This is the general solution to the differential equation.

Key concepts

Integrating Factor in Differential Equations

To solve first-order linear differential equations, we often need a tool called the integrating factor. This is a strategic function that simplifies the process of integrating a differential equation.
It is particularly useful when the equation is in the form

• \( y' + P(x)y = Q(x) \)

In this setup, the integrating factor \( \mu(x) \) is given by the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

By multiplying through the entire equation by this factor, the equation transforms.
This simplification allows the left-hand side to be expressed as the derivative of a single function, often leading to an easier integration process. In our exercise, \( P(x) = -1 \), yielding an integrating factor \( \mu(x) = e^{-x} \). By introducing this, we transformed our equation into a form where the left-hand side could be rewritten as a product derivative, setting up easy integration of both sides.

Standard Form of Differential Equations

The standard form of a first-order linear differential equation is key for determining the correct integrating factor.
This standard format is expressed as

• \( y' + P(x)y = Q(x) \)

It presents the equation in terms of a single derivative term \( y' \) and allows us to clearly identify functions \( P(x) \) and \( Q(x) \). Having this form unravels the structure of the equation and makes it ready for systematic application of techniques such as integrating factors.
Our original problem \( x(y' - y) = e^x \) needs to be rearranged to this configuration.
Dividing each term by \( x \), we achieved the standard form: \( y' - y = \frac{e^x}{x} \). Through this, \( P(x) \) was identified as \(-1\) and \( Q(x) \) as \(\frac{e^x}{x}\), which are crucial for solving the equation efficiently.

General Solution of Differential Equations

The concept of the general solution is fundamental in the study of differential equations. Once an equation is neatly expressed using an integrating factor and brought to an integrable form, solving it yields what's called the general solution. This encompasses the complete set of solutions that satisfy the differential equation.
Generally, after performing the integration, we end up with a form like

• \( e^{-x}y = \ln|x| + C \)

By isolating \( y \), we finally derive the general solution encompassing all possible specific solutions. The constant \( C \) represents an unknown that can be adjusted to satisfy particular initial conditions.
For the equation in the exercise, solving the integrals provides:

• \( y = e^x(\ln|x| + C) \)

This expression gives insight into how solutions of differential equations are systematically unrolled through steps, involving algebra and calculus, representing a broad spectrum of potential solutions based on varying \( C \) values.