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.
Best Answer
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 $−d−1$).