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