设y=xt,则t=y/x,y'=xt'+t代入原方程得xt'+t+t=1/t==>xt'=(1-2t2)/t==>tdt/(1-2t2)=dx/x==>d(1-2t2)/(1-2t2)=-4dx/x==>ln│1-2t2│=-4ln│x│+ln│C│ (C是积分常数)==>1-2t2=C/x^4==>(1-2(y/x)2)x^4=C==>x2(x2-2y2)=...