91Ó°ÊÓ

A first-order linear differential equation is one that involves a derivative of one dependent variable (typically denoted as \( y \)) with respect to one independent variable (often denoted as \( x \)). The general form of such an equation is \( \frac{d y}{d x} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). In our specific problem, we have

• \( P(x) = -2x \)
• \( Q(x) = 2x \)

The goal in solving a first-order linear differential equation is to find the function \( y(x) \) that satisfies the equation for given initial or boundary conditions. Recognizing the type of differential equation is crucial as it helps in applying the correct solving technique, such as the use of an integrating factor, which we’ll explore next.

The integrating factor is a critical tool for solving first-order linear differential equations. It transforms the original equation into an easier form, often allowing for straightforward integration. The integrating factor \( \mu(x) \) is determined using the function \( P(x) \) from the standard form of the differential equation \( \frac{d y}{d x} + P(x)y = Q(x) \). The formula for the integrating factor is:\[ \mu(x) = e^{\int P(x) \, dx} \]In the exercise at hand, \( P(x) = -2x \), so calculating the integrating factor involves finding:

• \( \int -2x \, dx = -x^2 \)

Substituting back into our formula gives us:

• \( \mu(x) = e^{-x^2} \)

The integrating factor simplifies the original problem, and when multiplied through the differential equation, it allows for easy integration by enabling us to write the left-hand side as a derivative of the product \( y \mu(x) \). This simplification lets us proceed to integration, a process we'll delve into next.

Integration by substitution is a method often used to solve integrals that involve complex functions. It simplifies the process by transforming the variable of integration into another variable for easier computation.In the problem we tackled, after multiplying the differential equation by the integrating factor, we reached:\[ \frac{d}{dx}(ye^{-x^2}) = 2xe^{-x^2} \]For the integration on the right-hand side, we introduced a substitution. Let \( u = -x^2 \), then the differential \( du = -2x \, dx \) becomes useful for this substitution. Solving the integral:

• \( \int 2xe^{-x^2} \, dx = -e^{-x^2} + C \)

where \( C \) is the constant of integration. This substitution allows us to handle the integral \( 2xe^{-x^2} \, dx \) more readily, converting it into a form that aligns clearly with \( e^u \) derivatives. Once the integral is resolved, we revert back to the original variable to find the solution for \( y \), using the initial conditions to determine the constant \( C \) and thus arriving at the complete solution for the differential equation.