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I have the following PDE,

$$(xz-y)p+(yz-x)q=xy-z$$where $p=z_x,\quad q=z_y$

Now having a hard time to get two solution from,

$$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}$$

I can't think of any multipliers trick which can help me to get one. In fact, $dx-dy$ or $dx+dy$ also couldn't help here. I guess the $xy-z$ term is the main culprit.

Any solution or hint will be appreciated. It will be a great help if you also describe how to guess the trick (if you used so).

Thanks in advance.

update

In fact, I got two more problems where I faced the similar issue:

$$\left(x z+y^{2}\right) p+\left(y z-2 x^{2}\right) z=-\left(2 x y+z^{2}\right)\qquad (1)$$ gives $$\frac{d x}{x z+y^2}=\frac{d y}{y z-2 x^{2}}=\frac{d z}{-\left(2 x y+z^{2}\right)}$$

$$(y+3 z)p+(z+5 x)q=x+7 y\qquad (2)$$

$$ \begin{align} &\frac{d x}{y+3 z}=\frac{d y}{z+5 x}=\frac{d z}{x+7 y}\\ \implies&\frac{d x-3d y}{y-15x}=\frac{d y-5d z}{z-7 y}=\frac{d z-7d x}{x-21 z} \end{align} $$

falamiw
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  • a PDE is supposed to contain partial derivatives... My hint is that you do not have a PDE. – PierreCarre May 19 '22 at 18:02
  • @PierreCarre $p = z_{x}, q = z_{y}$ – Matthew Cassell May 19 '22 at 18:03
  • Hint: $$\frac{x dx}{x^{2} z - xy} = \frac{y dy}{y^{2} z - yx} = \frac{x dx - y dy}{x^{2} z - y^{2} z} = \frac{d(x^{2} - y^{2})}{2(x^{2} - y^{2})z}$$ – Matthew Cassell May 19 '22 at 18:04
  • @mattos We should not have to guess from the solution... – PierreCarre May 19 '22 at 18:05
  • @PierreCarre I agree, it should be explicit, but in fairness using $p$ and $q$ is a widely used notation. – Matthew Cassell May 19 '22 at 18:06
  • I update the post with adding the meaning of those notations @PierreCarre. – falamiw May 19 '22 at 18:14
  • I also tried similar thing like $\frac{ydx-xdy}{x^2-y^2}$, but the $z$ variable still here in your hint. Then how to integral that? From my understanding, there shouldn't be any other variable @mattos – falamiw May 19 '22 at 18:17
  • You need to manipulate the other equalities (try $dx, dz$ and $dy, dz$ combinations) in a similar way and then set them equal to the result I gave above. – Matthew Cassell May 19 '22 at 18:21
  • $d y, d z$ combination with $y, z$ multipliers give me $\frac{y d y-z d z}{y^2{z}-x y-x y z-z^{2}}$ which I guess won't work with $\frac{d\left(x^{2}-y^{2}\right)}{2\left(x^{2}-y^{2}\right) z}$, similar with $d x, d z$ combination with $x, z$ multipliers @mattos – falamiw May 19 '22 at 18:51

2 Answers2

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This is not a complete answer rather expanding the @ShengtongZhang's one as the OP commented there,

characteristic equation of that PDE,

$$ \begin{align} &\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}=dt\\ &x'=xz-y,\quad y'=yz-x, \quad z'=xy-z \end{align} $$ Now, inspecting the $x'y-y'x$ term,

$$ \begin{align} x'y-y'x&=(xz-y)y-(yz-x)x\\ &=xyz-y^2-xyz+x^2\\ &=x^2-y^2 \text{(free of $z$ variable)} \end{align} $$

Which give some idea to guess one solution as $\frac{x}{y}$,

$$ \begin{align} \frac{d}{dt}\left(\frac{x}{y}\right)&=\frac{x'y-y'x}{y^2}\\ &=\frac{x^2-y^2}{y^2} \text{(solvable)} \end{align} $$

emonHR
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An idea: WLOG let $dt = \frac{dx}{xy - z}$. Then $$\frac{d\frac{x}{y}}{dt} = \frac{x'y - y'x}{y^2} = \frac{x^2 - y^2}{y^2} = (x/y)^2 - 1.$$ So you can now find $\frac{x}{y} = \coth(t + C)$. But I don't know how to go further.

abacaba
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