91Ó°ÊÓ

The integrating factor is a crucial concept when solving linear first-order differential equations. It allows us to transform a non-exact equation into an exact one, making it easier to solve. In the context of the original exercise, the equation is not immediately solvable, so we need to find an integrating factor.

To find the integrating factor, we use the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

In the given equation, \( P(x) \) is \( -\frac{1}{x} \). Calculating the integral of \( P(x) \), we get:

• \( \int -\frac{1}{x} \, dx = -\ln |x| \)
• \( \mu(x) = e^{-\ln |x|} = \frac{1}{x} \)

With this integrating factor, we multiply it throughout the equation, converting the left-hand side into the derivative of a single variable. This step is vital as it simplifies the equation considerably and sets up for finding the solution.

Integration by parts is a method used to integrate products of functions, based on the product rule for differentiation. In the process of solving our differential equation, we reached the integral \( \int \frac{2 \ln x}{x} \, dx \). This integral involves a logarithmic function—a classic candidate for integration by parts.

Here's a reminder of the integration by parts formula:

• \( \int u \, dv = uv - \int v \, du \)

In this context, let's choose:

• \( u = \ln x \), therefore \( du = \frac{1}{x} \, dx \)
• \( dv = \frac{2}{x} \, dx \), thus \( v = 2 \int \frac{1}{x} \, dx = 2 \ln x \)

Substituting back, we perform the integration as follows:

• \( \int \frac{2 \ln x}{x} \, dx = 2 \ln x \cdot \ln x - \int 2 \ln x \, \frac{1}{x} \, dx \)
• This simplifies to: \( 2(\ln x)^2 - 2 \int \ln x \, \cdot 1 \, dx \)

Finally, complete the calculation to obtain a result that plugs back into the solution for the general equation.

The general solution to a differential equation includes all possible solutions and often involves an arbitrary constant that enables the capture of specific conditions. In this problem, the result of our procedure allows us to say:

\( y(x) = x \left(2(\ln x)^2 - 2 \ln x + C\right) \)

• Here, \( C \) represents an arbitrary constant, introduced during the integration process.
• This constant is integral because it allows the solution to fit with any initial or boundary conditions presented later.

The general solution provides the most comprehensive form of the solution to the differential equation and can be adapted to various specific scenarios by substituting different values for \( C \). This flexibility is essential when modeling real-world situations that might have different starting points or conditions.