∵xy'-y-√(y-x)=0 ==>y'-y/x-√(y/x-1)=0∴设y=xt,则y'=xt'+t代入方程得xt'-√(t-1)=0 ==>dt/√(t-1)=dx/x==>ln(t+√(t-1))=ln│x│+ln│C│ (C是积分常数)==>t+√(t-1)=Cx==>y/x+√(y/x-1)=Cx==>y+√(y-x)=Cx故原...