设t=y/x,则dy/dx=t+xdt/dx∵y²+x²dy/dx=xydy/dx ==>(y/x)²=(1-y/x)dydx==>t²=(1-t)(t+xdt/dx)==>t(2t-1)=x(1-t)dt/dx==>(1-t)dt/[t(2t-1)]=dx/x==>ln|x|=∫[1/(2t-1)-1/t]dt==>ln|x|=(1/2)ln|2t...