The equation is rewritten as: \[x \log x \frac{dy}{dx} + y = 2 \log x \quad \Rightarrow \quad \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x \log x}.\] This is a first-order linear differential equation in the form \[\frac{dy}{dx} + P(x)y = Q(x).\] Final Answer: \( \boxed{{First-order linear differential equation}} \)