Differential Equations: Homogeneous Differential Equation

$2xy \, dy = (y^2 - x^2) \, dx \implies \frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ Both numerator and denominator are homogeneous of degree 2.

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

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

Substituting $x = 1, y = 1$: $1^2 + 1^2 = C(1) \implies C = 2$ Thus, the particular solution is: $x^2 + y^2 = 2x \iff x^2 - 2x + y^2 = 0 \iff (x - 1)^2 + y^2 = 1$.