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

(D) Given differential equation: $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$
Rearranging the terms: $x dy = (y + \sqrt{x^{2} + y^{2}}) dx$
$\frac{dy}{dx} = \frac{y + \sqrt{x^{2} + y^{2}}}{x} = F(x, y)$
To check for homogeneity,replace $x$ with $\lambda x$ and $y$ with $\lambda y$:
$F(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^{2} + (\lambda y)^{2}}}{\lambda x} = \frac{\lambda (y + \sqrt{x^{2} + y^{2}})}{\lambda x} = \lambda^{0} F(x, y)$
Since $F(\lambda x, \lambda y) = \lambda^{0} F(x, y)$,the equation is homogeneous.
Substitute $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$:
$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^{2} + v^{2}x^{2}}}{x} = v + \sqrt{1 + v^{2}}$
$x \frac{dv}{dx} = \sqrt{1 + v^{2}}$
$\frac{dv}{\sqrt{1 + v^{2}}} = \frac{dx}{x}$
Integrating both sides:
$\int \frac{dv}{\sqrt{1 + v^{2}}} = \int \frac{dx}{x}$
$\log |v + \sqrt{1 + v^{2}}| = \log |x| + \log C$
$v + \sqrt{1 + v^{2}} = Cx$
Substituting $v = \frac{y}{x}$:
$\frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} = Cx$
$\frac{y + \sqrt{x^{2} + y^{2}}}{x} = Cx$
$y + \sqrt{x^{2} + y^{2}} = Cx^{2}$