91Ó°ÊÓ

When solving first-order linear differential equations, the concept of an integrating factor is crucial. This method transforms the equation into a form that is much easier to handle. Given a differential equation in the form \( y' + P(x) y = Q(x) \), we find the integrating factor, \( \mu(x) \). This factor is given by \( \mu(x) = e^{\int P(x) \, dx} \), where \( P(x) \) is a function of \( x \) derived from the equation.

For the differential equation \( y' - y = e^x \), we identify \( P(x) = -1 \). Subsequently, the integrating factor becomes \( \mu(x) = e^{\int -1 \, dx} = e^{-x} \).

By multiplying every term of the differential equation by \( e^{-x} \), we can simplify and effectively combine terms involving \( y \). This multiplication step transforms the left-hand side into the derivative of \( e^{-x} y \), enabling easier integration on both sides.

The general solution of a differential equation provides a family of possible solutions, capturing all specific instances with a constant that can be determined if further conditions are provided. After obtaining the integrating factor for the differential equation \( y' - y = e^x \), the equation evolves to \( e^{-x} y' - e^{-x} y = 1 \).

With this simplification in place, recognize that the left-hand side is the derivative of a product: \( \frac{d}{dx}(e^{-x} y) = 1 \).

When both sides are integrated, the result is \( e^{-x} y = x + C \), where \( C \) is an integration constant and denotes any real number. Isolate \( y \) by multiplying through by \( e^{x} \), and we derive the general solution: \( y = e^{x}(x + C) \).

This expression outlines an infinite set of solutions depending on the value of \( C \). Specific solutions can be determined if initial or boundary conditions are given.

Verification is a key step to ensure that the derived solution actually satisfies the original differential equation. For our solution \( y = e^{x}(x + C) \), determining its derivative \( y' \) is necessary. Compute \( y' \) as follows: differentiate \( y = e^{x}(x + C) \) to obtain \( y' = e^x(x + C) + e^x \).

Plug \( y \) and \( y' \) back into the original equation \( y' = e^x + y \). Upon substitution, the equality \( e^x(x + C) + e^x = e^x + e^x(x + C) \) holds true. Both sides are identical, confirming that the solution satisfies the differential equation.

Through this process, verification ensures that the solution is neither a computational artifact nor an algebraic coincidence, but rather a true solution to the problem. This verification technique is a standard practice anytime solutions are claimed for differential equations, ensuring reliability of the results presented.