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

(N/A) The given differential equation is $x^{2} \frac{dy}{dx} = x^{2} + xy - 2y^{2}$.
$\frac{dy}{dx} = \frac{x^{2} + xy - 2y^{2}}{x^{2}}$.
Let $F(x, y) = \frac{x^{2} + xy - 2y^{2}}{x^{2}}$.
$F(\lambda x, \lambda y) = \frac{(\lambda x)^{2} + (\lambda x)(\lambda y) - 2(\lambda y)^{2}}{(\lambda x)^{2}} = \frac{\lambda^{2}(x^{2} + xy - 2y^{2})}{\lambda^{2}x^{2}} = \lambda^{0} F(x, y)$.
Since $F(\lambda x, \lambda y) = \lambda^{0} F(x, y)$,the given differential equation is a homogeneous equation.
To solve it,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = \frac{x^{2} + x(vx) - 2(vx)^{2}}{x^{2}} = 1 + v - 2v^{2}$.
$x \frac{dv}{dx} = 1 - 2v^{2}$.
$\frac{dv}{1 - 2v^{2}} = \frac{dx}{x}$.
$\frac{1}{2} \int \frac{dv}{(\frac{1}{\sqrt{2}})^{2} - v^{2}} = \int \frac{dx}{x}$.
Using the formula $\int \frac{dx}{a^{2} - x^{2}} = \frac{1}{2a} \log |\frac{a+x}{a-x}| + C$:
$\frac{1}{2} \cdot \frac{1}{2(1/\sqrt{2})} \log |\frac{1/\sqrt{2} + v}{1/\sqrt{2} - v}| = \log |x| + C$.
$\frac{1}{2\sqrt{2}} \log |\frac{1 + \sqrt{2}v}{1 - \sqrt{2}v}| = \log |x| + C$.
Substituting $v = \frac{y}{x}$:
$\frac{1}{2\sqrt{2}} \log |\frac{x + \sqrt{2}y}{x - \sqrt{2}y}| = \log |x| + C$.