# 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.

• You already said you showed that fraction satisfies the PDE. That means it is a solution to that PDE; you don't have to do anything else. Or do you mean, rather than the question of "how can we show this formula is a solution", to ask the hypothetical question about how you would go about solving the PDE if you hadn't encountered this formula previously?
– user14972
Oct 30 '17 at 4:29
• @Hurkyl: Yes, but I am trying to explain where the fraction that defines $\mu(x,y)$ comes from. You can just take the formula and replace into the PDE and verify that the it satisfies the PDE, but if you want to explain more the euristics I think you can do the second answer. Oct 30 '17 at 4:33
• @Hurkyl: Exactly, yes. I meant to ask the hypothetical question about how you would go about solving the PDE if you hadn't encountered this formula previously :) Oct 30 '17 at 4:35
• @Hurkyl: Before reading the Serre book I didn't know where the formula for $\mu$ comes from. This was suggested by user Francois Ziegler. Oct 30 '17 at 4:37

(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 $\displaystyle{\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}}.$

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