General formula for integrating factor of an homogeneous differential 1 form This question is probably very elementary but I don't know how to tackle the conversely part of the following result. Let $M(x,y)$ and $N(x,y)$ be two differentiable and homogeneous functions of the same degree $d$ and such that $M(x,y)dx+N(x,y)dy$ is not exact that is:
$$ \frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x} $$. Using Euler's identity on $M$ and $N$: 
$$ xM_{x}(x,y)+yM_{y}(x,y)=d\cdot M(x,y)$$
$$ xN_{x}(x,y)+yN_{y}(x,y)=d\cdot N(x,y)$$
Euler's identity comes from Euler's homogeneous function theorem which is appicable in this case since $M$ and $N$ are both homogeneous functions.
I showed that the function
$$\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)}\tag1$$
will satisfy:
$$ \frac{\partial}{\partial y}\left(\,\mu\cdot M\right)=\frac{\partial}{\partial x}\left(\mu\cdot N\right).\tag2$$
My question: If we suppose that the PDE above is true,then how we can explain where the formula of $\mu(x,y)$ comes from?, that is:
$${If}\quad N(x,y)\mu_{x}-M(x,y)\mu_{y}=(N_{x}-M_{y})\mu,\quad \text{then where the formula} \quad\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)} \quad \text{comes from ?} $$
I tried to apply the method of characteristics but I don't see it. This result comes from an old edition of Boyce and Di prima.
 A: The last display asks whether (1) is the unique solution of (2). It isn’t: try $M=N=x$, $\mu=1/x$.
In fact an integrating factor is never unique: see e.g. Serret (1886, thm 681).
Now if you are asking for heuristics, then e.g. (ibid., §685) “derives” (1) under the Ansatz that $\mu$ is itself homogeneous (of degree $−d−1$).
A: (From Serret J.A. Cours de Calcul Differentiel Et Integral... Volume 1 book)
Following suggestions from Francois Ziegler, we will show that we can find an homogeneous function $\mu(x,y)$ of some degree $k\in\mathbb{Z}$ such that:
$$(\mu\cdot M)\,dx+(\mu\cdot N)\,dy=0$$
is an exact 1-form using the fact that $M$ and $N$ are both homogeneous functions of the same degree $d$. In effect, let $\mu(x,y)$ be such an homogeneous function of degree $k\in\mathbb{Z}$, then $\mu\cdot M$ will be an homogeneous function of degree $k+d$ and by Euler's homogeneous function theorem we have:
$$ x\frac{\partial}{\partial x}(\mu\cdot M)+y\frac{\partial}{\partial y}(\mu\cdot M)=(d+k)\,\mu\cdot M. $$
Since we want $\mu$ to be a factor such that the original equation is exact we must have:
$$ \frac{\partial}{\partial x}(\mu\cdot N)=\frac{\partial}{\partial y}(\mu\cdot M) $$
Then
$$ x\frac{\partial}{\partial x}(\mu\cdot M)+y\frac{\partial}{\partial x}(\mu\cdot N)=(d+k)\,\mu\cdot M $$
but
$$ y\frac{\partial}{\partial x}(\mu\cdot N)=\frac{\partial}{\partial x}(y\,\mu\cdot N) $$
and $$ x\frac{\partial}{\partial x}(\mu\cdot M)=\frac{\partial}{\partial x}(x\,\mu\cdot M)-\mu\cdot M $$
this implies:
$$ \frac{\partial}{\partial x}(x\,\mu\cdot M)-\mu\cdot M+\frac{\partial}{\partial x}(y\,\mu\cdot N)=(d+k)\,\mu\cdot M $$
so 
$$ \frac{\partial}{\partial x}(x\,\mu\cdot M)+\frac{\partial}{\partial x}(y\,\mu\cdot N)=(d+k+1)\,\mu\cdot M $$
then we get:
$$ \frac{\partial}{\partial x}(\mu\,(xM+yN))=(d+k+1)\,\mu\cdot M. $$
Let's choose $k=-d-1$, then 
$$ \frac{\partial}{\partial x}(\mu\,(xM+yN))=0.$$
Similarly, the function $\mu\cdot N$ is homogeneous of degree $k+d$, then again by Euler's homogeneous function theorem we get:
$$ x\frac{\partial}{\partial x}(\mu\cdot N)+y\frac{\partial}{\partial y}(\mu\cdot N)=(k+d)\,\mu\cdot N $$
as before we can write the last equation as follows:
$$ x\frac{\partial}{\partial y}(\mu\cdot M)+\frac{\partial}{\partial y}(\mu\cdot N\,y)-\mu N=(k+d)\,\mu\cdot N $$
then
$$ \frac{\partial}{\partial y}(x\,\mu\cdot M)+\frac{\partial}{\partial y}(\mu\cdot N\,y)=(k+d+1)\,\mu\cdot N=0 $$
so $$\frac{\partial}{\partial y}(\mu(xM+yN))=0$$
Then the expresion $\mu\cdot(xM+yN)$ satisfies both conditions:
$$ \frac{\partial}{\partial x}(\mu(xM+yN))=0$$
and
$$\frac{\partial}{\partial y}(\mu(xM+yN))=0$$
so $\mu\cdot(xM+yN)$ is any constant, in particular we can say 
$$ \mu\cdot(xM+yN) = 1 $$
and hence:
$$\mu(x,y)=\frac{1}{xM+yN}.$$
