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The integrating factor is a crucial tool when solving linear first-order differential equations, allowing us to transform the equation into a more easily solvable form.
Understanding when and how to use it requires recognizing that you're dealing with a linear first-order differential equation, typically of the form:
\[ y' + p(x)y = q(x) \]
where \(p(x)\) and \(q(x)\) are functions of \(x\).
The integrating factor, \( \mu(x) \), is given by the exponential of the integral of \(p(x)\):

• \( \mu(x) = e^{\int p(x)\, dx} \)

After determining this factor, you multiply through the entire differential equation, transforming it into:

• \( \frac{d}{dx}[\mu(x) y] = \mu(x) q(x) \)

This simplification allows us to integrate both sides with respect to \(x\), ultimately solving for \(y\).
In the original exercise, the integrating factor is \(e^{-2x}\), helping to isolate and solve for \(w = u'\).
Understanding how to choose and apply an integrating factor is essential for tackling these types of equations.

Linear first-order differential equations are the simplest form of differential equations, involving no powers or products of the unknown function and its derivative.
These equations have the general form:

• \( y' + p(x)y = q(x) \)

where \(y'\) represents the derivative of \(y\) with respect to \(x\), \(p(x)\) and \(q(x)\) are given functions of \(x\).
These equations describe various processes, like growth and decay, and are fundamental to understanding more complex differential systems.
The main objective is to solve for \(y\), using techniques such as identifying an integrating factor.By transforming the equation with the integrating factor, it becomes straightforward to derive the solution by integrating the transformed expression.
In the given exercise, the differential equation \( w' - 2w = xe^{-x} \) is a linear first-order differential equation, recognizable by its structure and the presence of the derivative \( w' \).
Solving this equation requires identifying and working with the integrating factor as a critical step.

The general solution of a differential equation encompasses all possible solutions, accounting for arbitrary constants.
These constants arise from integration during the solving process, providing a family of solutions rather than a single unique one.
In the context of linear differential equations, once the solution form is identified, integrating offers the general solution containing constants that fit various initial conditions.
The aim is to express the solution in terms of known functions, such as exponential or polynomial forms.
For the exercise at hand, we obtain a general solution:

• \( y = \frac{1}{3} x + \frac{4}{9} + c_2 e^{3x} + c_3 e^{x} \)

This expression not only solves the equation but continues to satisfy initial values set by specific conditions.
Understanding how to determine the general solution allows for adapting to a variety of real-world scenarios depicted by the equation.

Differentiation and integration are fundamental operations in calculus, each serving distinct purposes in analyzing functions.
The derivative measures how a function changes concerning its variable, reflecting rates of change or slopes of tangents.
Integration, on the other hand, is the inverse process of differentiation. It sums up infinitesimal changes over an interval, useful for finding areas under curves or accumulated quantities.
When solving differential equations, we frequently alternate between using derivatives to set up the problem and integrals to solve it.
Each step in the solution often involves integrating or differentiating to simplify expressions and find solutions.
In this exercise, we differentiate to transform the equation, and integrate to solve it, like in the step:

• \( \frac{d}{dx}[e^{-2x} w] = xe^{-3x} \)

Mastering these processes is essential for effectively tackling a wide array of mathematical problems.