Show that the differential equation $(x-y) dy - (x+y) dx = 0$ is a homogeneous equation and find its solution.

$(x-y) dy - (x+y) dx = 0$
$\Rightarrow \frac{dy}{dx} = \frac{x+y}{x-y}$
Let $F(x, y) = \frac{x+y}{x-y}$.
$\therefore F(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y} = \frac{x+y}{x-y} = \lambda^{0} F(x, y)$.
Thus,the given differential equation is a homogeneous equation. To solve it,we use the substitution $y = vx$.
$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting the values of $y$ and $\frac{dy}{dx}$ in the equation,we get:
$v + x \frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1+v}{1-v}$.
$x \frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v - v(1-v)}{1-v} = \frac{1+v^2}{1-v}$.
$\Rightarrow \frac{1-v}{1+v^2} dv = \frac{dx}{x}$.
Integrating both sides:
$\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{1}{x} dx$.
$\tan^{-1}(v) - \frac{1}{2} \log(1+v^2) = \log|x| + C$.
Substituting $v = \frac{y}{x}$:
$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left(1 + \frac{y^2}{x^2}\right) = \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left(\frac{x^2+y^2}{x^2}\right) = \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} [\log(x^2+y^2) - \log(x^2)] = \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^2+y^2) + \log|x| = \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2} \log(x^2+y^2) + C$.