[tex]y'=\dfrac{2y-x}{2x-y}[/tex]
Let [tex]y=xv[/tex], where [tex]v=v(x)[/tex], so that [tex]y'=xv'+v[/tex]. Then the ODE is
[tex]xv'+v=\dfrac{2xv-x}{2x-xv}[/tex]
[tex]xv'=\dfrac{2x(v-1)}{x(2-v)}-v[/tex]
[tex]xv'=\dfrac{2(v-1)}{2-v}-\dfrac{v(2-v)}{2-v}[/tex]
[tex]xv'=\dfrac{v^2-2}{2-v}[/tex]
This is separable, so you have
[tex]\dfrac{2-v}{v^2-2}\,\mathrm dv=\dfrac{\mathrm dx}x[/tex]
Integrate both sides, solving for [tex]v[/tex] if possible, then replace using [tex]v=\dfrac yx[/tex] and solve for [tex]y[/tex] if possible.
