Euler's identity to find integrating factor for an homogeneous 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 $N$: 
$$ x\cdot M_{x}(x,y)+y\cdot M_{y}(x,y)=d\cdot M(x,y)$$
$$ x\cdot N_{x}(x,y)+y\cdot N_{y}(x,y)=d\cdot N(x,y)$$
Euler's identity are true for each because $M$ and $N$ are homogenoeus functions of the same degree $d$.
I showed that the function
$$\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)}$$
will satisfy:
$$ \frac{\partial}{\partial y}\left(\,\mu\cdot M\right)=\frac{\partial}{\partial x}\left(\mu\cdot N\right).$$
My question: If we suppose that the PDE above is true how we can show that one solution for $\mu(x,y)$ is the fraction given above. 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 DiPrima.
 A: (From Serret J.A. Cours de Calcul Differentiel Et Integral... Volume 1 book)  Consider the homogeneous 1-form:
$$ M(x,y)dx+N(x,y)dy=0 $$
Since $M(x,y)$ and $N(x,y)$ are both homogeneous functions of the same degree we can find a one variable function $f$ such that:
$$ \frac{M(x,y)}{N(x,y)}=f\left(\frac{y}{x}\right), $$
then our original form becomes:
$$ N(x,y)\bigg(\frac{M(x,y)}{N(x,y)}dx+dy\bigg)=0 $$
$$ N(x,y)\bigg(f\left(\frac{y}{x}\right)dx+dy\bigg)=0 $$
We know that an homogeneous differential 1-form can be turned into a separable differential equation using the change of variables 
$$z=\frac{y}{x}$$
and this give us:
$$ N(x,zx)\bigg(f(z)dx+d(zx)\bigg)=0 $$
then
$$ N(x,zx)\bigg(f(z)dx+xdz+zdx\bigg)=0 $$
$$ N(x,zx)\bigg(\left(f(z)+z\right)dx+xdz\bigg)=0 $$
$$ xN(x,zx)\big(f(z)+z\big)\left(\frac{dx}{x}+\frac{dz}{f(z)+z}\right)=0$$
From here we see that multiplying the last equation by $\displaystyle{\frac{1}{xN(x,zx)\big(f(z)+z\big)}=\frac{1}{xM(x,y)+yN(x,y)}}$ give us the following equation:
$$ \frac{dx}{x}+\frac{dz}{f(z)+z}=0 $$
which is exact because:
$$ \frac{\partial}{\partial z}\left(\frac{1}{x}\right)=0=\frac{\partial}{\partial x}\left(\frac{1}{f(z)+z}\right)$$
Remark: $f(z)+z\neq 0$ because $\displaystyle{\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}}.$
A: (From Serret J.A. Cours de Calcul Differentiel Et Integral... Volume 1 book) 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:
$$\mu(x,y)=\frac{1}{xM+yN}.$$
