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 can first rearrange and separate the terms. Reorganizing gives: \[ 2 dy = (6x + \ln y - \ln(xy)) dx \] \[ 2 dy = (6x + \ln y - (\ln x + \ln y)) dx \] \[ 2 dy = (6x - \ln x) dx \] Now, we can separate variables by dividing both sides by 2 and integrating: \[ dy = \left(3x - \frac{1}{2} \ln x\right) dx \] Integrate both sides: \[ y = \int \left(3x - \frac{1}{2} \ln x\right) dx \] Calculating the integral: \[ y = \frac{3}{2} x^2 - \frac{1}{2} \left( x \ln x - x \right) + C \] Simplifying gives: \[ y = \frac{3}{2} x^2 - \frac{1}{2} x \ln x + \frac{1}{2} x + C \] Thus, the solution to the differential equation is: \[ y = \frac{3}{2} x^2 - \frac{1}{2} x \ln x + \frac{1}{2} x + C \]