# Why the Integrating Factor is $\frac{1}{Mx+Ny}$

If we have a Differential Equation of the form $$Mdx+Ndy=0$$ which is not Eaxct and $$Mx+Ny \ne 0$$

If the equation is Homogenous

Why is the Integrating Factor $$\frac{1}{Mx+Ny}$$ ?

My attempt:

Let $$g(x,y)$$ Be the Integrating Factor to make the equation exact.

Then we have

$$Mgdx+Ngdy=0$$ an Exact Differential Equation. So

$$\frac{\partial(Mg)}{\partial y}=\frac{\partial(Ng)}{\partial x}$$ $$\implies$$

$$M\frac{\partial(g)}{\partial y}+g \frac{\partial(M)}{\partial y}=N\frac{\partial(g)}{\partial x}+g\frac{\partial(N)}{\partial x}$$

Any clue here to prove $$g=\frac{1}{Mx+Ny}$$

• This question would need the characterization of "the equation is homogeneous" to get a useful answer. Commented Jul 11, 2021 at 8:12
• I’m voting to close this question because if an answer from @Goncalo is still considered "might be good or bad" by the community bot then it is totally hopeless. Commented Mar 7 at 10:10

A first order ODE of the form $$Mdx+Ndy=0$$ is said to be homogeneous if $$M(x,y)$$ and $$N(x,y)$$ are homogeneous functions of the same degree (see here). Thus, if $$N\neq 0$$, it follows that$$^{(*)}$$ $$\frac{M(x,y)}{N(x,y)}=f\!\left(\frac{y}{x}\right). \tag{1}$$ Therefore, multiplying the ODE by $$(Mx+Ny)^{-1}$$, we obtain $$Pdx+Qdy=0, \tag{2}$$ where $$P=\frac{f\!\left(\frac{y}{x}\right)}{xf\!\left(\frac{y}{x}\right)+y}, \qquad Q=\frac{1}{xf\!\left(\frac{y}{x}\right)+y}. \tag{3}$$ To prove that $$(2)$$ is exact, we have to show that $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$. Indeed, $$\frac{\partial P}{\partial y} =\frac{\frac{1}{x}f'\!\left(\frac{y}{x}\right)}{xf\!\left(\frac{y}{x}\right)+y} -\frac{f\!\left(\frac{y}{x}\right)\left(f'\!\left(\frac{y}{x}\right)+1\right)}{\left(xf\!\left(\frac{y}{x}\right)+y\right)^2} =\frac{\frac{y}{x}f'\!\left(\frac{y}{x}\right)-f\!\left(\frac{y}{x}\right)}{\left(xf\!\left(\frac{y}{x}\right)+y\right)^2} \tag{4}$$ and $$\frac{\partial Q}{\partial x}=-\frac{f\!\left(\frac{y}{x}\right)-\frac{y}{x}f'\!\left(\frac{y}{x}\right)}{\left(xf\!\left(\frac{y}{x}\right)+y\right)^2} =\frac{\partial P}{\partial y}. \quad{\square}\tag{5}$$
$$^{(*)}$$ Proof of $$(1)$$: Since $$M$$ and $$N$$ have the same degree of homogeneity, we have (for $$\lambda\neq 0$$) $$\frac{M(\lambda x, \lambda y)}{N(\lambda x,\lambda y)}=\frac{\lambda^nM(x,y)}{\lambda^nN(x,y)}=\frac{M(x,y)}{N(x,y)}. \tag{P1}$$ In particular, taking $$\lambda=x^{-1}$$ in $$(\text{P1})$$ yields $$\frac{M(x,y)}{N(x,y)}=\frac{M(1, x^{-1} y)}{N(1, x^{-1} y)}, \tag{P2}$$ from which follows $$(1)$$.
$\frac{1}{Mx+ Ny}$ is an integrating factor if and only if $\frac{1}{Mx+ Ny}\left(Mdx+ Ndy\right)$ is exact. In order for that to be true we must have $\frac{\partial}{\partial y}\frac{M}{Mx+ Ny}= \frac{\partial}{\partial x}\frac{M}{Mx+ Ny}$.
$\frac{\partial}{\partial y}M(Mx+ Ny)^{-1}= -\frac{MN}{(Mx+ Ny)^2}$ and $\frac{\partial}{\partial x}N(Mx+ Ny)^{-1}= -\frac{MN}{(Mx+ Ny)^2}$
• But if $M,N$ are constant, you do not need an integrating factor, $F(x,y)=Mx+Ny$ is already the integral. Commented Jul 11, 2021 at 8:11