急!!在线等!!下面这道微分题该怎样解答啊??
设z=z(x,y)是由方程e^(-xy)+2z-e^z=2确定求dz|(x=2,y=-1/2)求详细解答过程!!谢谢!!!...
设z=z(x,y)是由方程e^(-xy)+2z-e^z=2确定 求dz|(x=2,y=-1/2)
求详细解答过程!!谢谢!!! 展开
求详细解答过程!!谢谢!!! 展开
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对方程e^(-xy)+2z-e^z=2两边微分,
有:e^(-xy)*d(-xy) + 2*dz - e^z*dz = 0
-e^(-xy)*(x*dy + y*dx) + 2*dz - e^z*dz = 0
移项,得:(e^z - 2)*dz = -y*e^(-xy)*dx - x*e^(-xy)*dy
当x=2,y=-1/2时,代入e^(-xy)+2z-e^z=2,得:z = 1
所以dz|(x=2,y=-1/2) = [-y*e^(-xy)*dx - x*e^(-xy)*dy]/(e^z - 2)(其中x=2,y=-1/2,z=1)
所以dz|(x=2,y=-1/2) = e*(1/2 * dx - 2*dy)/(e - 2)
有:e^(-xy)*d(-xy) + 2*dz - e^z*dz = 0
-e^(-xy)*(x*dy + y*dx) + 2*dz - e^z*dz = 0
移项,得:(e^z - 2)*dz = -y*e^(-xy)*dx - x*e^(-xy)*dy
当x=2,y=-1/2时,代入e^(-xy)+2z-e^z=2,得:z = 1
所以dz|(x=2,y=-1/2) = [-y*e^(-xy)*dx - x*e^(-xy)*dy]/(e^z - 2)(其中x=2,y=-1/2,z=1)
所以dz|(x=2,y=-1/2) = e*(1/2 * dx - 2*dy)/(e - 2)
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