...公式怎么求x(y²-1)dx+y(x²-1)dy=0的通解?,那个公式里面的x...
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热心网友
解：∵x(y²-1)dx+y(x²-1)dy=0
==>xy²dx+x²ydy-xdx-ydy=0
==>y²d(x²)/2+x²d(y²)/2-d(x²)/2-d(y²)/2=0
==>d(x²y²)/2-d(x²)/2-d(y²)/2=0
==>d(x²y²)-d(x²)-d(y²)=0
==>d(x²y²-x²-y²)=0
==>x²y²-x²-y²=C (C是积分常数)
∴原微分方程的通解是x²y²-x²-y²=C (C是积分常数)。
热心网友
x(y^2-1)dx+y(x^2-1)dy=0
xdx/(x^2-1)+ydy/(y^2-1)=0
d(x^2-1)/(x^2-1)+d(y^2-1)/(y^2-1)=0
ln(x^2-1)(y^2-1)=C
(x^2-1)(y^2-1)=C'
