Verify that the total differential equation(1+yz)dx+z(z-x)dy-(1+xy)dz=0 is integrable and hence find its integral.
P(∂Q∂z−∂R∂y)+Q(∂R∂x−∂P∂z)+R(∂P∂y−∂Q∂x)=P(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y})+Q(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z})+R(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})=P(∂z∂Q−∂y∂R)+Q(∂x∂R−∂z∂P)+R(∂y∂P−∂x∂Q)=
=(1+yz)(2z−x+x)+z(z−x)(−y−2z+x)−(1+xy)(z+z)==(1+yz)(2z-x+x)+z(z-x)(-y-2z+x)-(1+xy)(z+z)==(1+yz)(2z−x+x)+z(z−x)(−y−2z+x)−(1+xy)(z+z)=
=2yz2−2xyz−z2y−2z3+z2x+xyz+2z2x−zx2≠0=2yz^2-2xyz-z^2y-2z^3+z^2x+xyz+2z^2x-zx^2\neq0=2yz2−2xyz−z2y−2z3+z2x+xyz+2z2x−zx2=0
The equation is not integrable.
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Dear Sema, if we solve this question, then a solution will be published.
Please solve the problem