x(1+y^2)+y(1+x^2)dy/dx=0 尤其是积分的步骤请详细讲解

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x(1+y^2)+y(1+x^2)dy/dx=0
x(1+y^2)=-y(1+x^2)dy/dx
y/(1+y²)dy=-x/(1+x²)dx
2y/(1+y²)dy=-2x/(1+x²)dx
1/(1+y²)dy²=-1/(1+x²)dx²
两边同时积分,得
ln(1+y²)=-ln(1+x²)+lnc
所以
1+y²=-c(1+x²)

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