x+yy'=2y, y'=2-x/y.
Let y=vx, y'=v+xv', v+xv'=2-1/v, v2+xvv'=2v-1, (v-1)2=-xvv'.
Separating the variables:
vdv/dx/(v-1)2=-dx/x, (v/(v-1)2)dv=-dx/x,
Let u=v-1, du=dv, v=u+1 so ∫(v/(v-1)2)dv=∫[(u+1)/u2]du=∫du/u+∫du/u2.
So, integrating:
ln|u|-1/u=ln|A/x|, where A is a constant,
ln|v-1|-1/(v-1)=ln(A/x), substituting u=v-1,
ln|y/x-1|-1/(y/x-1)=ln(A/x), substituting v=y/x, rewriting:
ln|(y-x)/x|-x/(y-x)=ln(A/x), ln|y-x|-x/(y-x)=ln(A)=C.
CHECK
ln|y-x|-x/(y-x)=C.
Differentiating:
(1/(y-x))(y'-1)-(y-xy')/(y-x)2=0,
y'-1-(y-xy')/(y-x)=0,
(y'-1)(y-x)-y+xy'=0,
yy'-y+x-y=0, x+yy'=2y.