Step 1: Rewrite the differential equation

Given:

$$(x - y) \, dy - (x + y) \, dx = 0$$

Rearranging:

$$(x - y) \, dy = (x + y) \, dx$$

Therefore:

$$\frac{dy}{dx} = \frac{x + y}{x - y}$$

Step 2: Show that the equation is homogeneous

Dividing numerator and denominator by $x$:

$$\frac{dy}{dx} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}}$$

This is of the form $F\left(\frac{y}{x}\right)$, which proves the equation is homogeneous.

Step 3: Solve using substitution

Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substituting:

$$v + x\frac{dv}{dx} = \frac{1 + v}{1 - v}$$ $$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v^2}{1 - v}$$

Separating variables:

$$\frac{1 - v}{1 + v^2} \, dv = \frac{dx}{x}$$

Step 4: Integrate both sides

$$\int \frac{1 - v}{1 + v^2} \, dv = \int \frac{dx}{x}$$ $$\int \frac{1}{1 + v^2} \, dv - \int \frac{v}{1 + v^2} \, dv = \ln|x| + C$$ $$\tan^{-1}(v) - \frac{1}{2}\ln(1 + v^2) = \ln|x| + C$$

Step 5: Substitute back

Replacing $v = \frac{y}{x}$:

$$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln\left(\frac{x^2 + y^2}{x^2}\right) = \ln|x| + C$$

Simplifying:

$$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln(x^2 + y^2) = C$$

∴ This is the general solution of the given differential equation.