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A first-order linear differential equation is one of the simplest types of differential equations that you'll encounter. It involves an unknown function and its first derivative but no higher derivatives. An equation in this form looks like \( \frac{dy}{dx} + P(x)y = Q(x) \). The terms \( P(x) \) and \( Q(x) \) are functions of \( x \), generally known beforehand. The goal is usually to find the function \( y \) that satisfies this equation.

In the given exercise, the differential equation \( \frac{d y}{d x} - 2xy = 2x \) is recognized as a first-order linear differential equation. This is because it can be rewritten as \( \frac{dy}{dx} + (-2x)y = 2x \), fitting the format exactly. Solving these equations involves methods that simplify the equation into a form where integration can be applied.

The initial value problem tells us that besides solving the equation, we have specific starting conditions. Here, it means that when \( x = 0 \), \( y \) must equal 3.

The integrating factor is a critical part of solving a first-order linear differential equation. It's a technique that allows us to simplify the equation and solve for the unknown function \( y \). The integrating factor is denoted as \( \mu(x) \) and calculated based on the function \( P(x) \).

The formula for finding the integrating factor is \( \mu(x) = e^{\int P(x) \, dx} \). In our exercise, since \( P(x) = -2x \), we compute \( \int -2x \, dx = -x^2 \). Consequently, the integrating factor becomes \( \mu(x) = e^{-x^2} \).

Multiplying through the entire differential equation by this integrating factor transforms the left hand side into the derivative of a product. In this case, \( e^{-x^2}\frac{d y}{d x} - 2x e^{-x^2} y = \frac{d}{dx}(e^{-x^2}y) \). This transformation prepares the equation for integration, a crucial step in the process.

The substitution method is a technique used in calculus to simplify complex integrals, and here it plays a key role in integrating the transformed differential equation. When dealing with the right side integral \( \int 2x e^{-x^2} \, dx \), substitution becomes handy.

You set \( u = -x^2 \), hence \( du = -2x \, dx \). This change of variable allows the integration to become more straightforward. The integral \( \int 2x e^{-x^2} \, dx \) simplifies to \( \int e^{u} (-du) \), which easily integrates to \( -e^{u} + C \), or \( -e^{-x^2} + C \) when reversing back the substitution.

This integral contributes directly to solving for \( y \) as part of the general solution \( e^{-x^2}y = -e^{-x^2} + C \). Further manipulation with the integrating factor clears out \( e^{-x^2} \), resulting ultimately in \( y = -1 + Ce^{x^2} \). Initial conditions allow determination of \( C \), solidifying the particular solution.