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

Question

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 $g=\frac{1}{Mx+Ny}$

Answer 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 $\frac{\partial}{\partial x}N(Mx+ Ny)^{-1}= -\frac{MN}{(Mx+ Ny)^2}$

Yes, assuming M and N are constants, then that is an integrating factor.

Answer 2

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 $$ \frac{\partial Q}{\partial x}=-\frac{f\!\left(\frac{y}{x}\right)-\frac{y}{x}f'\!\left(\frac{y}{x}\right)}{\left(xf\!\left(\frac{y}{x}\right)+y\right)^2} =\frac{\partial P}{\partial y}. \quad{\square}\tag{5} $$

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$^{(*)}$ 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 $(1)$.