∵(x+y)dx+(y-x)dy=0
==>(1+y/x)dx+(y/x-1)dy=0
设y=xt,则dy=tdx+xdt
∴(x+y)dy+(y-x)dx=0
==>(1+t)dx+(t-1)(tdx+xdt)=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|C1| (C是积分常数)
==>ln|x|+1/2∫d(t²+1)/(t²+1)-arctant=ln|C1|
==>ln|x|+1/2ln(t²+1)-arctant=ln|C1|
==>x√(t²+1)=Ce^(arctant)
==>√(x²+y²)=Ce^(arctany/x)
==>(x²+y²)=Ce^(2arctany/x)
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