Here's another in my long series of study questions on ODEs (and more to come). I want to prove the following:
The (non-exact) equation: $$P(x,y)\,dx + Q(x,y)\,dy = 0$$ Admits an integrating factor $\mu = f(\Phi)\in\mathcal C^r$ iff :
$\mathcal C^r \ni \Phi, P, Q : W\subset\Bbb R^2\to\Bbb R$ are such that $$\frac{P_y-Q_x}{Q\Phi_x - P\Phi_y} = g(\Phi(x,y)) \in\mathcal C^{r-1}$$ ($Q\Phi_x - P\Phi_y\neq 0$ and the quotient is exclusively a function of $\Phi$)
The forwards implication is easy enough, we just write $\frac{\partial}{\partial y}(f(\Phi)P) = \frac{\partial}{\partial x}(f(\Phi)Q)$, and find that $$\frac{P_y-Q_x}{Q\Phi_x - P\Phi_y} = \frac{f'(\Phi)}{f(\Phi)}$$ as desired. We of course assume that the denominator is non-zero, and if it is we find $\mu = 0$ (I'm not sure what this says about the equation?).
The converse is the part I have trouble with. It's easy to see that $g\circ\Phi$ and $1/\mu$ satisfy the same a similar (?) PDE, and I'm sure there's some uniqueness theorem for this, but I haven't studied PDEs at all so I can't really take that route. Can somebody help?
P.S. The part I'm least concerned about is the $\mathcal C^r$-related conditions. I'm sure these follow from any proof.
To prove the converse statement, fix a point $(a,b) \in W$ and set $$\mu(x,y) = \exp\left\{\int_{\Phi(a,b)}^{\Phi(x,y)} g(u)\, du\right\}.$$ Then $\mu \in C^r$ and $\mu = f(\Phi)$, where $f(t) = e^{\int_{\Phi(a,b)}^t g(u)\, du}$. Furthermore,
\begin{align}(\mu P)_y - (\mu Q)_x &= (\mu_y P + \mu P_y) - (\mu_x Q + \mu Q_x)\\ &= (\mu\, g(\Phi)\Phi_y P + \mu P_y) - (\mu\, g(\Phi)\Phi_x Q + \mu Q_x)\\ &= \mu[g(\Phi)(\Phi_y P - \Phi_x Q) + P_y - Q_x]\\ &= \mu(Q_x - P_y + P_y - Q_x)\\ &= 0. \end{align}
Therefore, $\mu = f(\Phi)$ is an integrating factor of $P(x,y)\, dx + Q(x,y)\, dy = 0$.
Your work for the forward implication looks correct. Keep in mind though that an integrating factor of $P\, dx + Q\, dy = 0$ is not identically zero. This condition is implicit in the definition of an integrating factor; note that the equation $(0P)_y = (0Q)_x$ holds for all functions $P(x,y)$ and $Q(x,y)$.