Step 1: Rewrite the equation in standard form

$$(x^2 - y^2) \, dx + 2xy \, dy = 0$$ $$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$$

Step 2: Show that the equation is homogeneous

$$f(x, y) = \frac{y^2 - x^2}{2xy}$$ $$f(tx, ty) = \frac{(ty)^2 - (tx)^2}{2(tx)(ty)} = \frac{t^2(y^2 - x^2)}{2t^2xy} = \frac{y^2 - x^2}{2xy} = t^0 \cdot f(x, y)$$

Since $f(tx, ty) = t^0 \cdot f(x, y)$, the equation is homogeneous of degree 0.

Step 3: Apply substitution y = vx

Let $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

$$v + x\frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2x \cdot vx} = \frac{v^2 - 1}{2v}$$

Step 4: Separate variables and integrate

$$x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{v^2 + 1}{2v}$$ $$\frac{2v}{v^2 + 1} \, dv = -\frac{dx}{x}$$

Integrating:

$$\ln(v^2 + 1) = -\ln x + \ln C$$ $$\ln(v^2 + 1) + \ln x = \ln C$$ $$x(v^2 + 1) = C$$

Step 5: Substitute back v = y/x

$$x\left(\frac{y^2}{x^2} + 1\right) = C$$ $$\frac{x^2 + y^2}{x} = C$$

∴ $x^2 + y^2 = Cx$ (where $C$ is an arbitrary constant)