Solve the differential equation: (x² - y²) dx + 2xy dy = 0

$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$, which is homogeneous. Put $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. $v + x\frac{dv}{dx} = \frac{v^2 - 1}{2v} \Rightarrow x\frac{dv}{dx} = -\frac{1+v^2}{2v}$ $\int \frac{2v}{1+v^2}dv = -\int \frac{dx}{x} \Rightarrow \log(1+v^2) = -\log x + \log c$ $x(1+v^2) = c \Rightarrow x\left(1 + \frac{y^2}{x^2}\right) = c$ Hence $x^2 + y^2 = cx$.