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If we have to valid integrating factors to a differential equation we must get the same general solution using each of them although we get two new equations by multiplying the original for each of the integral factors?

So if the original equation is not exact and we verify that the two new functions we get by multiplying it for each integrating factors are exact then you will get the same solution in both cases. Is that right? I´m referring as "new equations" because I read it in the bibliography. First you call P(x,y) and Q(x,y) when the equation is not exact and when you multiply it by the integrating factor you call M(x,y) and N(x,y) to the functions of the new exact equations. The new exact solutions will have the same solution that will also be solution to the original equation.

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The two integrating factors differ by a factor that is an (sometimes trivial) integral of the differential equation. That is, if $\mu(x,y)$ is an integrating factor giving the integral or implicit solution $F(x,y)=C$, then any product $g(F(x,y))\mu(x,y)$ with a differentiable (and bijective) function $g$ will also be an integrating factor leading to the integral $G(F(x,y))$, $G'=g$.

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  • $\begingroup$ Thanks @LutzLehmann! So that´s why both of them will end up giving us the same solution to the differential equation, right? $\endgroup$
    – Ignacio
    Jun 17 '20 at 10:53
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    $\begingroup$ Yes, in making the solution explicit in $y$ the transformation by $G$ is undone, so that you get, in general, the same solutions. There might be critical points in $G$ that lead to missing or "hidden" solutions. $\endgroup$ Jun 17 '20 at 10:57

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