Show that the differential equation $(x-y) \frac{dy}{dx} = x+2y$ is homogeneous and solve it.

(N/A) The given differential equation can be expressed as
$\frac{dy}{dx} = \frac{x+2y}{x-y}$ ............$(1)$
Let $F(x, y) = \frac{x+2y}{x-y}$.
Now $F(\lambda x, \lambda y) = \frac{\lambda(x+2y)}{\lambda(x-y)} = \lambda^0 \cdot F(x, y)$.
Therefore,$F(x, y)$ is a homogeneous function of degree zero. So,the given differential equation is a homogeneous differential equation.
Alternatively,
$\frac{dy}{dx} = \frac{1+2(y/x)}{1-(y/x)} = g(y/x)$ .............$(2)$
Since the $R.H.S.$ is a function of $y/x$,it is a homogeneous function of degree zero.
To solve it,we make the substitution $y = vx$ ...........$(3)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$ ...........$(4)$
Substituting $(3)$ and $(4)$ into $(1)$:
$v + x \frac{dv}{dx} = \frac{1+2v}{1-v}$
$x \frac{dv}{dx} = \frac{1+2v}{1-v} - v = \frac{1+2v-v+v^2}{1-v} = \frac{v^2+v+1}{1-v}$
$\frac{v-1}{v^2+v+1} dv = -\frac{dx}{x}$
Integrating both sides:
$\int \frac{v-1}{v^2+v+1} dv = -\int \frac{dx}{x}$
$\frac{1}{2} \int \frac{2v+1-3}{v^2+v+1} dv = -\log|x| + C_1$
$\frac{1}{2} \log|v^2+v+1| - \frac{3}{2} \int \frac{1}{(v+1/2)^2 + (\sqrt{3}/2)^2} dv = -\log|x| + C_1$
$\frac{1}{2} \log|v^2+v+1| - \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right) = -\log|x| + C_1$
$\frac{1}{2} \log|v^2+v+1| + \log|x| = \sqrt{3} \tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right) + C_1$
$\frac{1}{2} \log|x^2(v^2+v+1)| = \sqrt{3} \tan^{-1}\left(\frac{2v+1}{\sqrt{3}}\right) + C_1$
Substituting $v = y/x$:
$\log|x^2+xy+y^2| = 2\sqrt{3} \tan^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right) + C$.