设y=xt,则dy=xdt+tdx于是,代入原方程得2xydx-(x^2+y^2)dy=0==>2x²tdx-(x²+x²t²)(xdt+tdx)=0==>2tdx-(1+t²)(xdt+tdx)=0==>t(1-t²)dx=x(1+t²)dt==>dx/x=(1+t²)dt/(t(1-t²...