# 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.

• As I read the questoin, you have shown that your particular $\mu$ satisfies the PDE, and you want to show that it is one solution of the PDE. This doesn't make sense to me. Perhaps you could rephrase the question. Oct 29, 2017 at 16:02
• @BenMcKay: I woluld like to prove that $\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)}$. I want to know where the formula for $\mu$ comes from. Oct 29, 2017 at 16:03
• @BenMcKay: I showed if $\mu$ has the formula I wrote above, then the equation is satisfied but starting from the PDE how can I get $\mu$ ? Oct 29, 2017 at 16:05
• @FrancoisZiegler: Yes, exactly. I have to correct the statement. Oct 29, 2017 at 16:05
• @FrancoisZiegler: Thank you, exactly integrating factors are never unique. The thing is where the formula for $\mu$ comes from? I will correct the statement. Oct 29, 2017 at 16:16

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 $−d−1$).

• No, nothing. It was a misktake. I completely acdept the answer ! :) There is No problem. Nov 29, 2017 at 0:43

(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: $$\mu(x,y)=\frac{1}{xM+yN}.$$