Differential Equations: Homogeneous First Order Differential Equation

$$(x - y)\,dy = (x + y)\,dx \implies \frac{dy}{dx} = \frac{x + y}{x - y}$$

Let $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$: $$v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$$ $$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v(1 - v)}{1 - v} = \frac{1 + v^2}{1 - v}$$

$$\frac{1 - v}{1 + v^2} \, dv = \frac{dx}{x} \implies \int \frac{1}{1 + v^2} \, dv - \frac{1}{2} \int \frac{2v}{1 + v^2} \, dv = \int \frac{dx}{x}$$ $$\arctan v - \frac{1}{2}\ln(1 + v^2) = \ln|x| + C$$ $$\arctan\left(\frac{y}{x}\right) = \ln|x| + \frac{1}{2}\ln\left(1 + \frac{y^2}{x^2}\right) + C = \ln\sqrt{x^2 + y^2} + C$$