∵(2x+y-4)dx+(x+y-1)dy=0
==>(2x-4)dx+(y-1)dy+(ydx+xdy)=0
==>d(x^2-4x)+d(y^2/2-y)+d(xy)=0
==>x^2-4x+y^2/2-y+xy=C/2(C是任意常数)
==>2x^2-8x+y^2-2y+2xy=C
∴原方程的通解是2x^2-8x+y^2-2y+2xy=C.