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Consider an inexact ODE $$M(x,y)dx + N(x,y)dy = 0\ . \quad (1) $$ Assume we can find an integrating factor $\mu(x)$.

This integrating factor $\mu$ fixes the integrating constants $C(x)$ and $C(y)$ of the integral of the second partial derivates of the potential function $\frac{\partial \tilde M}{\partial y}$ and $\frac{\partial \tilde N}{\partial x}$ respectively $(\tilde M = \mu M; \tilde N = \mu N)$.

Now, which degree of freedom is our choice of $\mu$ using, such that we can find the potential function $\psi$?

Since $\psi(x,y(x)) = C$ should give as a solution to the original ODE $(1)$, there must be, I suppose, some degree of freedom on which we impose the integrating constants $C(x)$ and $C(y)$.

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  • $\begingroup$ I'm not sure I understand your question, but I think the answer is: The degree of freedom isn't used in the choice of $\mu$, but when calculating $\psi$. The usual way is to integrate $\mu M$ with respect to $x$. The constant of integration in this operation is a function of $y$, and that's where the degree of freedom comes into play. $\endgroup$ – B. Goddard Sep 14 '16 at 13:17
  • $\begingroup$ My confusion lies in the manipulations needed to calculate $psi$, i.e. choosing integration constants in the process, whilst still solving the original ODE for a (possibly explicit) function $y(x)$. $\endgroup$ – Mussé Redi Sep 14 '16 at 13:46

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