Consider the diferential equation $$M(x,y)dx+N(x,y)dy=0$$ if $f(x)$ and $g(y)$ are integrating factors then the diferential equation can be solved by separation of variables.
I don’t know how to prove it, for now, I only have used the hypothesis, that is, we know that $f(x)$ and $g(y)$ are integrating factors so the equations $$f(x)M(x,y)dx+f(x)N(x,y)dy=0 \qquad\text{and}\qquad g(y)M(x,y)dx+g(y)N(x,y)dy=0 $$ are exact but then I don’t know how to continue. I don’t know if multiplying the diferential equation by both functions at the same time can be useful.
I also know that if a diferential equation can be solved by separation of variables then the equation is exact but this isn’t useful.
Any idea?
With some careful application of definition and a lot of computation, we can prove the separability of an equation with integrating factors.
Consider an integrating factor depending only on $x$ (you gave it as $f(x)$). Then we can do as you said:
$$f(x)M(x,y)dx + f(x)N(x,y)dy = 0$$
Now by definition of exactness of the above equation (which is an assumption you gave), we have that:
$$\frac{\partial \left( f(x)M(x,y)\right)}{\partial y} = \frac{\partial \left( f(x)N(x,y)\right)}{\partial x}$$
Just differentiate this:
$$\frac{\partial f}{\partial y}M(x,y) + f(x)\frac{\partial M}{\partial y} = \frac{\partial f}{\partial x}N(x,y) + f(x)\frac{\partial N}{\partial x}$$
However, since the integrating factor $f$ is only a variable of $x$, the term with $\frac{\partial f}{\partial y} = 0$, leading us to this equation:
$$f(x)\frac{\partial M}{\partial y} = \frac{\partial f}{\partial x}N(x,y) + f(x)\frac{\partial N}{\partial x}$$
Therefore,
$$\frac{\partial f}{\partial x} = -\frac{f(x)}{N(x,y)}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)$$
Now a bit of logic here, if $f$ is an integrating factor depending only on $x$, then $\frac{\partial f}{\partial x} := \frac{df}{dx}$, and actually, $\frac{df}{dx}$ itself is only a function of $x$. This means that the RHS of the above equation is also only a function of $x$, i.e.
$$q(x) := -\frac{1}{N(x,y)}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)$$
Therefore, we have transformed the original equation:
$$f(x)M(x,y)dx + f(x)N(x,y)dy = 0$$
into:
$$\frac{df}{dx} = -f(x)q(x)$$
Using an entirely similar set of steps and reasoning for the case of $g(y)$, (i.e. $g(y)M(x,y)dx + g(y)N(x,y) = 0$), we obtain:
$$\frac{dg}{dy} = -g(y)p(y)$$
Where $p(y)$ is defined as:
$$p(y) := -\frac{1}{M(x,y)}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)$$
This allows us to solve for integrating factors $f$ and $g$ directly through separation of variables, the solution of which are:
$$f(x) = \exp\left(- \int q(x)dx\right) \qquad g(x) = \exp\left(- \int p(y)dy\right)$$
To complete this...
We can now easily find a general solution for the equation of the form:
$$k(x,y) = C_{0}$$
where $C_{0}$ is a constant and $k(x,y)$ satisfies:
$$\frac{\partial k}{\partial x}dx + \frac{\partial k}{\partial y}dy = 0$$
P.S. Feel free to correct me if there's a mistake in any step of my work!