# A formula for homogeneous differential form integrating factor

(From Boyce and Di Prima book) Suppose $M$ and $N$ are differentiable function such that $$M(x,y)\,dx+N(x,y)\,dy=0$$ is an homogeneous differential form. I can show that the 2 variables functions $\mu$ defined as: $$\mu(x,y)=\frac{1}{x\,M(x,y)+y\,N(x,y)}$$ is an integrating factor that transform any homogeneous equation $M(x,y)\,dx+N(x,y)\,dy=0$ into an exact form, that is: $$\frac{\partial}{\partial y}\big(\mu(x,y)\cdot M(x,y)\big)=\frac{\partial}{\partial x}\big(\mu(x,y)\cdot N(x,y)\big).$$

My questions:

1) Where this integrating factor comes from ?

• I'm not sure if this is true. Are there other conditions? – Dylan Oct 25 '17 at 20:07
• As far as I know there are no others conditions. But I put in Maple using symbolic calculations and it works. Please, if you find one counter example let me know :) – Hector Blandin Oct 25 '17 at 20:09
• @Dylan: I would like to really understan this exercise. I tried several times to deduce the formula they give but I couldn't until now. – Hector Blandin Oct 25 '17 at 20:10
• @Dylan: I looked again and the only hypothesis is that the equation is homogeneous. – Hector Blandin Oct 25 '17 at 20:14
• Well I tried taking partials and plugging it in but the condition is not satisfied – Dylan Oct 25 '17 at 20:16

I will show you how to obtain an integrating factor with an example. Suppose you want to find an integrating factor of the form $\mu=\mu(xy)$ for the ODE $$yf(xy)\cdot dx+xg(xy)\cdot dy=0.$$

First, we will find a formula for integrating factors of the form $\mu(u)$, where $u=xy$ for the equation $$M(x,y)dx+N(x,y)dy=0.$$ By definition, $\mu$ must satisfy the condition $$\dfrac{\partial}{\partial y}[\mu(xy)M(x,y)]=\dfrac{\partial}{\partial x}[\mu(xy)N(x,y)] \Leftrightarrow \mu \dfrac{\partial M}{\partial y}+M\dfrac{\partial \mu}{\partial y}=\mu \dfrac{\partial N}{\partial x}+N\dfrac{\partial \mu}{\partial x}.$$ For $u:=xy$ we have $$\dfrac{\partial \mu}{\partial x}=\dfrac{d\mu}{du}\cdot \dfrac{\partial u}{\partial x}=\dfrac{d\mu}{du}y \: \: \: \mathrm{y} \: \: \: \dfrac{\partial \mu}{\partial y}=\dfrac{d\mu}{du} \cdot \dfrac{\partial u}{\partial x}=\dfrac{d\mu}{du}x,$$ and hence $$\mu \dfrac{\partial M}{\partial y}+M\dfrac{d \mu}{d u}x=\mu \dfrac{\partial N}{\partial x}+N\dfrac{d \mu}{d u}y.$$ And so $$\dfrac{d \mu}{du}\left( Mx-Ny \right)=\mu \left( \dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y} \right)\ \Leftrightarrow \dfrac{\partial N/\partial x-\partial M/\partial y}{-(Ny-Mx)}du=\dfrac{d\mu}{\mu}.$$ Integrating both members of the equaity and solving for $\mu$ we get $$\mu=\exp\left(-\int \dfrac{\partial N/\partial x-\partial M/\partial y}{Ny-Mx}du \right).$$

In this particular case we have $$N(x,y)=xg(xy) \Rightarrow \dfrac{\partial N}{\partial x}=g(xy)+xyg'(xy) \: \: \: \mathrm{y} \: \: \: M(x,y)=yf(xy)$$ $$\Rightarrow \dfrac{\partial M}{\partial y}=f(xy)+xyf'(xy).$$ Finally, by substitution, we get $$\mu(u)=\exp\left(-\int \dfrac{g(xy)+xyg'(xy) -f(xy)-x y f'(xy)}{g(xy)xy-f(xy)xy}du \right)$$ $$=\exp\left(-\int \dfrac{g(u)+ug'(u) -f(u)-u f'(u)}{u[g(u)-f(u)]}du \right), \: \: \: \mathrm{where} \: \: \: u:=xy$$ $$=\exp\left(-\int \dfrac{1}{u}du-\int \dfrac{f'(u)-g'(u)}{f(u)-g(u)}du \right)$$ $$= \exp\left(-\left[\ln(u)+\ln[f(u)-g(u)]\right]\right)$$ $$=\exp\left[ -(\ln\left(u(f(u)-g(u)) \right) \right]=\dfrac{1}{xy(f(xy)-g(xy))}.$$

The process you have to use to find any integrating factor of any ODE is analogous to this one.

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