求方程x(1+y^2)dx+y(1+x^2)dy=0的通解
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求方程x(1+y^2)dx+y(1+x^2)dy=0的通解
求方程x(1+y^2)dx+y(1+x^2)dy=0的通解
求方程x(1+y^2)dx+y(1+x^2)dy=0的通解
x(1+y^2)dx+y(1+x^2)dy=0
x(1+y^2)dx=-y(1+x^2)dy
xdx/(1+x^2)=-ydy/(1+y^2)
dx^2/2(1+x^2)=-dy^2/(1+y^2)
d(1+x^2)/2(1+x^2)=-d(1+y^2)/(1+y^2)
ln(1+x^2) / 2 =-ln(1+y^2) /2 +C
ln(1+x^2) =-ln(1+y^2) +2C
所以通解是
ln(1+x^2) +ln(1+y^2) +C=0