Help with first integral of PDE

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SUMMARY

The discussion focuses on solving a first integral of a partial differential equation (PDE) represented by the system: \(\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{xy(z^2 + 1)}\). The user successfully identifies a relationship between \(x\) and \(y\) leading to \(\frac{1}{x} - \frac{1}{y} = c\), but struggles to incorporate all three variables to find a second constant function. Key hints provided include isolating variables and manipulating the system as fractions, which are essential techniques for solving such PDEs.

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Hey guys, I'm having a little difficulty with a pde I'm trying to solve. It boils down to solving for a first integral. I don't want the answer, but I'd be glad to get a little help. We have the system:

\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{xy(z^2 + 1)}

We can use the first two and find that

\frac{1}{x} - \frac{1}{y} = c

I need to use all three to find a second function which is constant here. I've tried using the compendo and dividendo rule and I can't seem to get anywhere... I'm hoping just for a slight hint because I want to solve it myself but I'm really stuck at this point.

Thanks!
 
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Here are two hints you might find helpful:

(1) You have a relationship between x and y, so you can isolate for x or y, then substitute the result into the original system of equations.

(2) You can manipulate the original system as if the differentials (dx, dy, dz) were parts of a fraction. You can multiply both sides by the same factor, for example, or cancel out common factors, just like with fractions.
 
right... the problem is this solution isn't as easy as all that... there is a more trivial example solved with

\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}

Which can be solved by doing some proper addition and subtraction: so I know the idea. Its a matter of getting a form \frac{g*dx + h*dy}{f(x,y)} such that that solution is the differential of a higher form... as the dz part is quite easy, we can integrate it to arctan, we just need a better function of the xs I think...
 

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