设y=xu则y'=u+xu'代入原方程得：[xu-x(x^2+u^2x^2)]-x(u+xu')=0即x+u^2x+u'=0-xdx=du/(1+u^2)积分：-x^2/2+C=arctanuu=tan(c-x^2/2)y=xu=xtan(c-x^2/2)
