Step-by-Step Solution

The given differential equation is: \(x \log x \frac{dy}{dx} + y = 2 \log x\)

To determine the type of differential equation, we can rewrite it in the standard form for a first-order linear differential equation:

\(\frac{dy}{dx} + P(x)y = Q(x)\)

Divide the given equation by \(x \log x\):

\(\frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2 \log x}{x \log x}\)

\(\frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2}{x}\)

Here, \(P(x) = \frac{1}{x \log x}\) and \(Q(x) = \frac{2}{x}\). Since the equation can be written in the form \(\frac{dy}{dx} + P(x)y = Q(x)\), it is a first-order linear differential equation.