xdy-ydx=√(x²+y²)dxxdy=[√(x²+y²)+y]dxdy/dx=√[1+(y/x)²]+y/x设y/x=uu+xdu/dx=√(1+u²)+udu/√(1+u²)=dx/xarctanu=lnx+C即arctan(y/x)=lnx+C答案是：(x^2+y^2)^(1/2)-y=c...不好意思，我错了这一步积分我错了du/√(1+u²)=dx/x两边积分ln(u+√1+u²)=lnx+lnC1u+√(1+u²)=C1x即y/x+√(1+(y/x)²)=C1x[y+√(x²+y²)]/x=C1x[y+√(x²+y²)]/x²=C1分子有理化y-√(x²+y²)=C1√(x²+y²)-y=C