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

My try:

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

## 1 Answer

$\frac{1}{Mx+ Ny}$ 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.