设函数z=(x,y)由方程x^2+z^2=2ye^z所确定,求dz

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设函数z=(x,y)由方程x^2+z^2=2ye^z所确定,求dz

设函数z=(x,y)由方程x^2+z^2=2ye^z所确定,求dz
设函数z=(x,y)由方程x^2+z^2=2ye^z所确定,求dz

设函数z=(x,y)由方程x^2+z^2=2ye^z所确定,求dz
两边求微分的 2xdx+2zdz=2e^zdy+2ye^zdz
解得 dz=(2e^zdy-2xdx)/(2z-2ye^z)=(e^zdy-xdx)/(z-ye^z)

直接用一介微分不变性有
2xdx + 2zdz = 2e^zdy+2ye^z dz
整理有 dz = (xdx-e^zdy)/(ye^z-z)