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My attempt at the question

What I don't know is do we integrate partially w.r.t. z to find the value of Φ (z) here?

$$ \require{begingroup} \begingroup \newcommand{\dd}{\;\mathrm{d}}$$ $$yz\dd x-2xz \dd y + (xy-y^3z) \dd z=0.\tag{1}$$ $P=yz$, $Q=-2xz$ and $R=xy-y^3z$
Now \begin{align*} &P\left(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y}\right)+ Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+ R\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial z}\right)\\[.8em] &=P(-2x-x+3y^2z)+Q(y-y)+R(z+2z)\\ &=yz(-3x+3y^2z)+(-2xz)(0)+(xy-y^3z)(z+2z)\\ &=-3xzy+3y^3z^2+0+3xyz-3y^3z^2\\ &=0 \end{align*} Hence equation $(1)$ is integrable.
Let $z$ be a constant in $(1)$, then $\dd z=0$. Integrating both sides \begin{align*} \int\frac1{2xz}\dd x-\int\frac1{2yz}\dd y&=0\\ \frac1{2z}\ln x - \frac1z \ln y &= const\\ \frac{x^{1/2}}y &= \Phi(z) \tag{2} \end{align*} Differentiating w.r.t $x$, $y$, $z$ $$\frac{x^{-1/2}}{2y} \dd x + \left(-\frac{x^{1/2}}{y^2}\right)\dd y+(-\Phi'(z)) \dd z = 0 \tag{3}$$ From equations $(1)$ and $(3)$ $$\frac{yz}{1/(2x^{1/2}y)}=\frac{-2xz}{-x^{1/2}/y^2}=\frac{xy-y^3z}{-\Phi'(z)}.$$ $$\endgroup$$

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You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find $$ y\,d(xz)-2(xz)\,dy=y^3zdz $$ and from that that you get an integrable expression by dividing by $y^3$, that is, the integrating factor should turn out to be $y^{-3}$.


In your solution, I'd have avoided the square root to get $$ xy^{-2}=\Phi(z) $$ as the equation for the solution surface. Then the derivative of that is $$ y^{-2}\,dx-2xy^{-3}\,dy =\Phi'(z)\,dz\\ yz\,dx - 2xz\,dy = y^3z\Phi'(z)\,dz\\ \implies y^3z\Phi'(z)\,dz=(y^3z-xy)\,dz\implies \Phi'(z)=1-z^{-1}\Phi(z) $$ which can now be easily solved.


In your version with $Φ(z)=\frac{x^{1/2}}y$, pick one of the relations and eliminate $y$ against $Φ$. The first two fractions are equal after simplification, the equality to the last term gives $$ 2x^{1/2}y^2z=y^3\frac{Φ(z)^2-z}{-Φ'(z)}\iff 2zΦ'(z)Φ(z)=z-Φ(z)^2\implies zΦ(z)^2=\frac12z^2+C $$

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  • $\begingroup$ Thanks a lot ! :) $\endgroup$
    – M.M.
    Commented Dec 4, 2018 at 9:26

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