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