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When approaching first-order linear differential equations, the integrating factor method is a popular and systematic approach. It allows us to simplify and solve these equations more conveniently. Let's break it down into steps.Linear differential equations typically have the form: \[ y' + P(x)y = Q(x) \] The "integrating factor" is the term used to help transform the equation into a form that is easier to solve. It is calculated using the expression: \[ e^{\int P(x) \, dx} \]In our exercise, the equation \( y' - y = 2 \) means \( P(x) = -1 \), and the integrating factor becomes \( e^{-x} \).Applying the integrating factor, we multiply it throughout the original equation. This step transforms the left-hand side into the derivative of a product, allowing the equation to be integrable. By employing this method, we simplify the process and create a direct path toward finding the solution.

Solving the differential equation using the integrating factor method involves a few sequential steps.After identifying the appropriate integrating factor, we apply it to transform the original equation: \[ e^{-x} y' - e^{-x} y = 2e^{-x} \]Next, we recognize that the left-hand side is the derivative of \( e^{-x}y \). Thus, the equation reshapes into the following form:\[ \frac{d}{dx}(e^{-x}y) = 2e^{-x} \]The strategy at this juncture is to tackle it by integrating both sides of the equation with respect to \( x \). As we perform integration, we find that:\[ e^{-x}y = -2e^{-x} + C \]Here, \( C \) represents the constant of integration, an essential component to account for indefinite integrals. We subsequently solve for \( y \) by multiplying through by \( e^{x} \):\[ y = -2 + Ce^{x} \]This is our general solution to the first-order linear differential equation.

Verifying the accuracy of the solution to a differential equation is just as crucial as finding the solution itself.Once you have determined a solution like \( y = -2 + Ce^{x} \), it is indispensable to check if it satisfies the original differential equation: \( y' - y = 2 \).To perform this check, differentiate \( y \) to determine \( y' \):\[ y' = Ce^{x} \]Substitute \( y \) and \( y' \) back into the original equation:\[ Ce^{x} - (-2 + Ce^{x}) \] Simplifying, you find:\[ 2 = 2 \]When both sides of the equation are equal, this verifies that the solution is correct. This step not only confirms accuracy but also provides confidence in understanding and applying the method correctly. Remember, always endeavor to check your solution to ensure it aligns perfectly with the given differential equation.