91Ó°ÊÓ

When dealing with first-order linear differential equations, using an integrating factor is a common technique for finding solutions. Let's break it down further. An integrating factor is a function that we multiply through the differential equation to help transform it into a form that is easier to solve. First, identify the standard form of the first-order linear differential equation, which is given by:

• \( y' + P(x)y = Q(x) \).

In the example equation, \( y' - 2y = e^{-2x} \), we identify \( P(x) = -2 \) and \( Q(x) = e^{-2x} \). Next, calculate the integrating factor \( \mu(x) \) using:

• \( \mu(x) = e^{\int P(x) \, dx} \).

For \( P(x) = -2 \), this integral calculates to \( \mu(x) = e^{-2x} \). Multiplying through the entire differential equation by this integrating factor allows us to proceed to find the solution with relative ease.

Solving a first-order linear differential equation involves transforming the equation using the integrating factor. Here's how it works:Once you have the integrating factor \( \mu(x) = e^{-2x} \), apply it to the differential equation. This multiplication transforms the left side of the equation into the derivative of a product:

• \( y' - 2y = e^{-2x} \) becomes \( e^{-2x}y' - 2e^{-2x}y = e^{-4x} \).
• This can be expressed as \( \frac{d}{dx}(e^{-2x}y) = e^{-4x} \).

Now, integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{-2x}y) \, dx = \int e^{-4x} \, dx \]The integration results in:

• Left side: \( e^{-2x}y \)
• Right side: \( -\frac{1}{4}e^{-4x} + C \), where \( C \) is the constant of integration.

Solve for \( y \) by isolating it:Multiply through by \( e^{2x} \) to get:\[ y = -\frac{1}{4}e^{-2x} + Ce^{2x} \]This solution represents the general form, accommodating any constant \( C \).

Validating the solution is a crucial final step in solving differential equations. After solving for \( y \), it's essential to verify that it satisfies the original differential equation. Here's how you do it:Start by differentiating the solution:

• Given \( y = -\frac{1}{4}e^{-2x} + Ce^{2x} \), differentiate to find \( y' \).
• The result is \( y' = \frac{1}{2}e^{-2x} + 2Ce^{2x} \).

Next, substitute both \( y \) and \( y' \) back into the original equation \( y' - 2y = e^{-2x} \):

• Plugging in gives: \( \frac{1}{2}e^{-2x} + 2Ce^{2x} - 2\left(-\frac{1}{4}e^{-2x} + Ce^{2x}\right) \).
• This simplifies to \( e^{-2x} \), thus verifying the solution is correct.

This verification ensures the computed solution is accurate and fully solves the original differential equation. Always validate as a best practice in mathematical problem-solving.