Differential Equations: Homogeneous Differential Equations

$$2xy \, dy = -(x^2 - y^2) \, dx \implies \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} = \frac{(y/x)^2 - 1}{2(y/x)}$$ Both numerator and denominator are homogeneous of degree 2.

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

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

Substitute $x = 1, y = 1$: $$1^2 + 1^2 = C(1) \implies C = 2$$ Thus, the particular solution is: $$x^2 + y^2 = 2x \quad \text{or} \quad (x - 1)^2 + y^2 = 1$$