91Ó°ÊÓ

First-order linear differential equations are a pivotal topic in calculus. They often appear in the form \( y' = f(x) \), where the derivative \( y' \) is expressed in terms of \( x \). The goal is to find the function \( y \) that satisfies this relationship.
To determine the general solution, you need to integrate the right-hand side of the equation. In our exercise, \( y' = 2e^{-x} + x \) implies integrating each component separately. Calculating these integrals gives us separate pieces that combine to form the solution.
Specifically, for \( abla y = 2e^{-x} + x \):

• The term \( 2e^{-x} \) integrates to \( -2e^{-x} \)
• The term \( x \) integrates to \( \frac{x^2}{2} \)

These integrals together form \( y = -2e^{-x} + \frac{x^2}{2} + C \). Here, \( C \) is a constant of integration, representing a family of solutions on any vertical shift along the y-axis.
Thus, the general solution encompasses this entire family of curves, shifting up or down depending on \( C \). This flexibility captures all potential functions that solve the differential equation, serving as a foundation for exploring specific solutions.

Integration is the process of finding the antiderivative of a function. It's essential for solving differential equations, as it helps us find the general solution. For first-order linear differential equations like \( y' = 2e^{-x} + x \), integration of the right side is required.
To achieve this, you break down the equation into simpler parts:

• First, focus on \( \int 2e^{-x} \, dx \), where the antiderivative is \( -2e^{-x} \). This results from using the formula for integrating exponential functions, \( \int e^{ax} \, dx = \frac{1}{a}e^{ax} + C \).
• Next, take \( \int x \, dx \). Here, the antiderivative is \( \frac{x^2}{2} \), following the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n = 1 \).

Combining these, you get the general solution: \( y = -2e^{-x} + \frac{x^2}{2} + C \). Through mastering these basic integration techniques, you gain the ability to solve a variety of similar equations. Remember that different functions may require different integration methods, but these basics are a great starting point.

Particular solutions of a differential equation are specific functions derived from the general solution by choosing particular values of the constant \( C \). They're exact cases of the function \( y \) satisfying the equation \( y' = f(x) \).
In our solution, we have the general form \( y = -2e^{-x} + \frac{x^2}{2} + C \). Choosing different values for \( C \) results in distinct particular solutions:

• When \( C = 0 \), the particular solution is \( y = -2e^{-x} + \frac{x^2}{2} \).
• When \( C = 1 \), it becomes \( y = -2e^{-x} + \frac{x^2}{2} + 1 \).
• If \( C = -1 \), then \( y = -2e^{-x} + \frac{x^2}{2} - 1 \).

These particular solutions correspond to specific cases or conditions where an equation applies.
For example, you might use a particular solution to model real-world situations where initial conditions are known.
This process of customizing solutions to fit specific scenarios is key in applying mathematics to practical problems. Understanding the difference between the general solution and particular solutions helps you grasp how differential equations model dynamic systems.