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I would like to solve the equation:

$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy - x^2 - 2} \tag{1}$

From that, we have:

$y'(y^2 + 2xy - x^2 - 2) = y^2 - 2xy - x^2 + 2 \implies $

$\frac{(2x + 2yy')(x+y) - (x^2 + y^2 + 2)(1 + y')}{(x+y)^2} = 0 \implies $

$(\frac{x^2 + y^2 + 2}{x + y})' = 0 \implies $

$\frac{x^2 + y^2 + 2}{x + y} = C$

Where the last part agrees with the solution in the book.

The "trick" is that I first looked at the solution in the book, differentiated it, and then applied here the process in reverse, so it looks like a solution. I was obviously not happy with that approach, so I looked online for a better solution. Unfortunately, I could not see a generic pattern in neither of the two existing solutions [solution 1, solution 2] I found.

Does there exists a more standard solution to the equation?

Are the two existing solutions common, and are not considered as "tricks"? Why would they be less of a trick than the one I described above?

Thanks!

S11n
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    Does there exists a more standard solution to the equation? Maple gives this as type rational. If you do not want to use tricks, it says that you need to use Lie symmetry methods. "This type of ODE cannot be solved in a general manner. In some cases, the symmetry method succeeds in finding rational symmetries for the problem (see dsolve,Lie)" see maple help page – Nasser Aug 30 '22 at 10:53
  • this seems like a reverse implicit function theorem problem – Max0815 Aug 30 '22 at 11:01

1 Answers1

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$$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy - x^2 - 2} $$ $$\dfrac {dy}{y^2 - 2xy - x^2 + 2} = \dfrac{dx}{y^2 + 2xy - x^2 - 2}$$ $$\dfrac {d(x+y)}{-y^2 +x^2} = \dfrac{d(x-y)}{- 2xy +2}$$ Substitute $u=x+y; v=x-y$: $$\dfrac {du}{2uv} = \dfrac{dv}{v^2-u^2 +4}$$ $$ 2uv\dfrac{dv}{du}=v^2-u^2 +4$$ $$ u\dfrac{dv^2}{du}=v^2-u^2 +4$$ This is a first order linear DE. $$ us'-s=-u^2 +4$$ Where $s=v^2=(x-y)^2$.

I only used substitutions to transform the DE into a first order linear DE that can be easily solved with well-known methods.

user577215664
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  • Thank you for the answer! Can you please expand on how you got from the second line to the third one (i.e. from $dy$ to $d(x+y)$)? Also, it seems that when going from line 4 to line 5, you just divided by $dv$. Is there a more formal way to achieve the same thing? – S11n Aug 30 '22 at 12:10
  • Thanks for the explanation! Isn't that an abuse of the notation, because in the first line $y' = \frac{dy}{dx}$? Is there a more formal way to achieve the same thing you achieved when going from line $4$ to $5$, when you divided by $du$? I saw in other places already people treating $dx$ as a usual variable, while in the Tom Apostol's Calculus book I saw he avoids giving it any meaning, and letting it just be part of the notation (e.g. as in $\int x dx$), so I wonder why is it justified to treat it as a number as you did. – S11n Aug 31 '22 at 07:54
  • I looked at the top answer about $dy/dx$, and it basically says the same as I said, that in the modern calculus, $dx$ and $dy$ does not have the infinitesimal meaning (https://math.stackexchange.com/q/21199/942378). I don't see how the substitution for $u$ and $v$ leads me formally (without using infinitesimals) to a first order linear DE. Can you please edit your answer to add that, more formal, derivation? Otherwise I don't see much added value when compared to the previous answer I already mentioned in the question formulation (https://math.stackexchange.com/a/524990/942378)? – S11n Sep 02 '22 at 11:45