y' = (2y-x)/(2x-y) 微分方程?

dy/dx=(2y-x)/(2x-y)dy/dx=(2y/x-1)/(2-y/x)令y/x=u,则u+xu'=(2u-1)/(2-u)[(2-u)/(u²-1)]du=dx/x两边积分得∫[(2-u)/(u²-1)]du=∫dx/x1/2∫[1/(u-1)-3/(u+1)]du=lnx1/2[ln(u-1)-3ln(u+1)]=lnx+C1y/x-1=Cx&...