Question #159068

Verify that the total differential equation(1+yz)dx+z(z-x)dy-(1+xy)dz=0 is integrable and hence find its integral.


1
Expert's answer
2021-01-29T04:20:24-0500

P(QzRy)+Q(RxPz)+R(PyQx)=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})=

=(1+yz)(2zx+x)+z(zx)(y2z+x)(1+xy)(z+z)==(1+yz)(2z-x+x)+z(z-x)(-y-2z+x)-(1+xy)(z+z)=

=2yz22xyzz2y2z3+z2x+xyz+2z2xzx20=2yz^2-2xyz-z^2y-2z^3+z^2x+xyz+2z^2x-zx^2\neq0

The equation is not integrable.


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Comments

Assignment Expert
29.01.21, 11:20

Dear Sema, if we solve this question, then a solution will be published.

Sema
29.01.21, 09:08

Please solve the problem

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