x dy - 2[y + xy²(1 + lnx)] dx = 0x·dy/dx - 2y = 2xy²(1 + lnx)、两边除以xy²(1/y²)(dy/dx) - 2/(xy) = 2(1 + lnx)令z = 1/y、dz/dx = dz/dy·dy/dx = (- 1/y²)(dy/dx)、代入原式得- dz...