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求全微分函数Z(X,Y),已知方程 Z^2*ln(X+Z)=XY
人气:258 ℃ 时间:2020-06-05 05:02:38
解答
Z^2*ln(X+Z)=XY
两边对X求微分 2Zln(X+Z)dZ/dX+(1+dZ/dX)Z^2/(X+Z)=Y
dZ/dX=(Y-Z^2/(X+Z))/(2Zln(X+Z)+Z^2/(X+Z))
两边对Y求微分
2Zln(X+Z)dZ/dY+(1+dZ/dY)Z^2/(X+Z)=X
dZ/dY=(X-Z^2/(X+Z))/(2Zln(X+Z)+Z^2/(X+Z))
所以dZ=(Y-Z^2/(X+Z))/(2Zln(X+Z)+Z^2/(X+Z))dX+(X-Z^2/(X+Z))/(2Zln(X+Z)+Z^2/(X+Z))dY
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