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Question Number 110262 by Ar Brandon last updated on 28/Aug/20

Solve for X(x,y,z), Y(x,y,z), Z(x,y,z) { (((∂Z/∂y)−(∂Y/∂z)=1−x^2 )),(((∂Z/∂x)−(∂X/∂z)=−(y^2 /2))),(((∂Y/∂x)−(∂X/∂y)=z(2x−y))) :} where { ((X(x,y,0)=0)),((Y(x,y,0)=0)),((Z(x,y,0)=0)) :}

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{X}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right),\:\mathrm{Y}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right),\:\mathrm{Z}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$$$\begin{cases}{\frac{\partial\mathrm{Z}}{\partial\mathrm{y}}−\frac{\partial\mathrm{Y}}{\partial\mathrm{z}}=\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\\{\frac{\partial\mathrm{Z}}{\partial\mathrm{x}}−\frac{\partial\mathrm{X}}{\partial\mathrm{z}}=−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\\{\frac{\partial\mathrm{Y}}{\partial\mathrm{x}}−\frac{\partial\mathrm{X}}{\partial\mathrm{y}}=\mathrm{z}\left(\mathrm{2x}−\mathrm{y}\right)}\end{cases}\:\mathrm{where}\:\begin{cases}{\mathrm{X}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\\{\mathrm{Y}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\\{\mathrm{Z}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\end{cases} \\ $$

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