<s>[INST] Solving $y' = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}$ without tricks 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!
 [/INST] $$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy  - x^2 - 2} $$
$$\dfrac {dy}{y^2 - 2xy - x^2 + 2} = \dfrac{dx}{y^2 + 2xy  - x^2 - 2}$$
$$\dfrac {d(x+y)}{-y^2 +x^2} = \dfrac{d(x-y)}{- 2xy  +2}$$
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.
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