∵(x+y)dy+(x-y)dx=0
==>(1+y/x)dy+(1-y/x)dx=0
设y=xt,则dy=tdx+xdt
∴(x+y)dy+(x-y)dx=0
==>(1+t)(tdx+xdt)+(1-t)dx=0
==>(t²+1)dx+x(t+1)dt=0
==>dx/x+(t+1)/(t²+1)dt=0
==>ln|x|+∫t/(t²+1)dt+∫1/(t²+1)dt=ln|C| (C是积分常数)
==>ln|x|+1/2∫d(t²+1)/(t²+1)+arctant=ln|C|
==>ln|x|+1/2ln(t²+1)+arctant=ln|C|
==>x√(t²+1)=Ce^(arctant)
==>√(x²+y²)=Ce^(arctant) (C是积分常数).