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)