[y+(x^2+y^2)^1/2]dx-xdy=0>dy/dx=y/x+(1+(y/x)^2)^(1/2)设z=y/x，则dy/dx=z+xdz/dx>z+xdz/dx=z+(1+z^2)^(1/2)>1/(1+z^2)^(1/2)dz=1/xdx>z+(1+z^2)^(1/2)=cx把y回代：>y+(x^2+y^2)^1/2=cx^2【OK？】
CCd170
