Step 1: Rearrange the given differential equation

Given:

$$y \, dx + x \log\left(\frac{y}{x}\right) dy - 2x \, dy = 0$$

Rearranging:

$$y \, dx = 2x \, dy - x \log\left(\frac{y}{x}\right) dy$$ $$y \, dx = x\left[2 - \log\left(\frac{y}{x}\right)\right] dy$$ $$\frac{dx}{dy} = \frac{x}{y}\left[2 - \log\left(\frac{y}{x}\right)\right]$$

Step 2: Show that the equation is homogeneous

The right-hand side

$$\frac{x}{y}\left[2 - \log\left(\frac{y}{x}\right)\right]$$

is a function of $\frac{x}{y}$ only.

Hence, the given differential equation is homogeneous.

Step 3: Apply substitution x = vy

Let $x = vy$, so

$$\frac{dx}{dy} = v + y\frac{dv}{dy}$$

Substituting:

$$v + y\frac{dv}{dy} = v\left[2 - \log\left(\frac{1}{v}\right)\right]$$ $$v + y\frac{dv}{dy} = v[2 + \log v]$$ $$y\frac{dv}{dy} = v(1 + \log v)$$

Step 4: Separate variables and integrate

$$\frac{dv}{v(1 + \log v)} = \frac{dy}{y}$$

Let $1 + \log v = t$, then $dt = \frac{dv}{v}$

$$\int\frac{dt}{t} = \int\frac{dy}{y}$$ $$\log|t| = \log|y| + \log C$$ $$1 + \log v = Cy$$

Step 5: Substitute back to get the solution

Since $x = vy$, we have $v = \frac{x}{y}$

$$1 + \log\left(\frac{x}{y}\right) = Cy$$ $$\log\left(\frac{x}{y}\right) = Cy - 1$$ $$\frac{x}{y} = e^{Cy - 1}$$ $$x = y \cdot e^{Cy - 1}$$

Or:

$$\boxed{\log\left(\frac{y}{x}\right) + Cy = 1}$$