Solve the given differential equation. \( 2 d y+\ln x y d x=(6 x+\ln y) d x \) \( y=\square \)

To solve the differential equation \( 2 dy + \ln(xy)dx = (6x + \ln y)dx \), we first rearrange the terms: \[ 2dy = (6x + \ln y - \ln(xy))dx \] This simplifies to: \[ 2dy = (6x + \ln y - (\ln x + \ln y))dx \] Thus, we have: \[ 2dy = (6x - \ln x)dx \] Now, we can separate variables: \[ \frac{dy}{dx} = \frac{6x - \ln x}{2} \] Next, integrate both sides with respect to \(x\): \[ y = \int \frac{6x - \ln x}{2} dx = \int 3x dx - \frac{1}{2} \int \ln x dx \] The first integral gives \( \frac{3x^2}{2} \), and for the integral of \(\ln x\), we can use integration by parts, letting \(u = \ln x\) and \(dv = dx\). This yields: \[ \int \ln x \, dx = x \ln x - x + C\] Combining everything, we have: \[ y = \frac{3x^2}{2} - \frac{1}{2}(x \ln x - x) + C \] Which simplifies to: \[ y = \frac{3x^2}{2} - \frac{x \ln x}{2} + \frac{x}{2} + C \] So, the solution is: \[ y = \frac{3x^2}{2} - \frac{x \ln x}{2} + \frac{x}{2} + C \]