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A first-order linear differential equation is an equation that involves the first derivative of a function and is expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). In simpler terms, it's an equation in which the highest derivative is the first derivative.

- **Components Explained:** - \( P(x) \): A function of the variable \( x \) multiplying the variable \( y \). - \( Q(x) \): A function of the variable \( x \) alone on the right side of the equation.In the problem given, our differential equation \( \frac{dy}{dx} + y = 3 + e^{-x} \) fits this standard formula where \( P(x) = 1 \) and \( Q(x) = 3 + e^{-x} \). Recognizing this form is crucial as it indicates that we can apply specific methods like the integrating factor technique.

This kind of equation is commonly seen in modeling real-world processes such as population growth, electrical circuits, and anywhere involving a rate of change.

To solve a first-order linear differential equation, we often use what's called an "integrating factor." The integrating factor is a function that, when multiplied by every term of the differential equation, makes it easily integrable. For our equation, the integrating factor \( \mu(x) \) is given by:\[ \mu(x) = e^{\int P(x) \, dx} \]
- **Steps in Our Problem:** - Here, \( P(x) = 1 \), so \( \int P(x) \, dx = x \). - Therefore, \( \mu(x) = e^x \).By multiplying the entire differential equation by this integrating factor \( e^x \), we transform it into a format where the left side of the equation can be recognized as a derivative of a product.

This step essentially "unlocks" the equation, paving the way for integration and simplifying the solution process significantly.

The product rule in calculus is a method of finding the derivative of the product of two functions. It's expressed as:\[ \frac{d}{dx}(uv) = u'v + uv' \]In the context of solving our problem, recognizing the left side of the equation formed after applying the integrating factor becomes vital. After multiplying the equation by the integrating factor \( e^x \), the left side becomes:\[ e^x \frac{dy}{dx} + e^x y \]
- **Recognizing Product Rule:** - This expression can be rewritten using the product rule as \( \frac{d}{dx}(e^x y) \).Understanding and identifying this transformation allows us to integrate both sides of the equation smoothly in the next step. This recognition is crucial because it directly links the problem to a known derivative form, thus simplifying the integration process.

The initial condition is an additional piece of information about the value of the function at a specific point, helping us determine the constant of integration. Here, we have \( y(0) = -1 \), which we will use to find our constant \( C \) after solving the differential equation.- **Steps to Apply:** - First, we integrate our transformed equation to solve for \( y \), giving us a general solution involving \( C \). - In our example, \( y = 3 + xe^{-x} + Ce^{-x} \).Now, substitute \( x = 0 \) and \( y = -1 \) into this solution:- The equation becomes: - \( -1 = 3 + 0 \cdot e^0 + C \cdot e^0 \) - Simplifying gives \( C = -4 \).This allows us to write the specific solution \( y(x) = 3 + xe^{-x} - 4e^{-x} \), solving the initial value problem fully. Initial conditions nudge the general solution to fit a specific scenario practical in real-world applications.