Show that the differential equation $(x^{2}-y^{2}) dx + 2xy dy = 0$ is a homogeneous equation and find its solution.

(N/A) The given differential equation is:
$(x^{2}-y^{2}) dx + 2xy dy = 0$
$\Rightarrow \frac{dy}{dx} = \frac{-(x^{2}-y^{2})}{2xy} = F(x, y)$ ..............$(1)$
To check for homogeneity,we evaluate $F(\lambda x, \lambda y)$:
$F(\lambda x, \lambda y) = \frac{-((\lambda x)^{2}-(\lambda y)^{2})}{2(\lambda x)(\lambda y)} = \frac{-\lambda^{2}(x^{2}-y^{2})}{\lambda^{2}(2xy)} = \lambda^{0} F(x, y)$
Since $F(\lambda x, \lambda y) = \lambda^{0} F(x, y)$,the equation is homogeneous.
To solve,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into $(1)$:
$v + x \frac{dv}{dx} = \frac{-(x^{2} - (vx)^{2})}{2x(vx)} = \frac{-(x^{2} - v^{2}x^{2})}{2vx^{2}} = \frac{v^{2}-1}{2v}$
$x \frac{dv}{dx} = \frac{v^{2}-1}{2v} - v = \frac{v^{2}-1-2v^{2}}{2v} = \frac{-(1+v^{2})}{2v}$
Separating variables:
$\frac{2v}{1+v^{2}} dv = -\frac{dx}{x}$
Integrating both sides:
$\int \frac{2v}{1+v^{2}} dv = -\int \frac{1}{x} dx$
$\ln(1+v^{2}) = -\ln|x| + \ln|C| = \ln|\frac{C}{x}|$
$1+v^{2} = \frac{C}{x}$
Substituting $v = \frac{y}{x}$:
$1 + \frac{y^{2}}{x^{2}} = \frac{C}{x}$
$\frac{x^{2}+y^{2}}{x^{2}} = \frac{C}{x}$
$x^{2}+y^{2} = Cx$