∵x^y-2x+y=0 ==>e^(ylnx)-2x+y=0
==>e^(ylnx)d(ylnx)-2dx+dy=0
==>x^y(lnxdy+ydx/x)-2dx+dy=0
==>(x^ylnx+1)dy+(yx^(y-1))dx=0
∴x^y-2x+y=0的微分是(x^ylnx+1)dy+(yx^(y-1))dx=0