# What is the integrating factor for 0 = (3x^2 + 3y^2)dx + x(x^2 + 3y + 6y)dy?

Jan 20, 2017

I got:

$\mu \left(y\right) = {e}^{y}$

The point of an integrating factor is to turn an inexact differential into an exact one. One physical application of this is to turn a path function into a state function in chemistry (such as dividing by $T$ to turn ${q}_{\text{rev}}$, a path function, into $S$, a state function, entropy).

I assume that the second terms include $3 {y}^{2}$, not $3 y$ (it would be odd to not simply write $9 y$).

Two options to find the special integrating factor, as defined by Nagle, are:

$\boldsymbol{\mu \left(x\right) = \text{exp} \left[\int \frac{{\left(\frac{\partial M}{\partial y}\right)}_{x} - {\left(\frac{\partial N}{\partial x}\right)}_{y}}{N \left(x , y\right)} \mathrm{dx}\right]}$,

$\boldsymbol{\mu \left(y\right) = \text{exp} \left[\int \frac{{\left(\frac{\partial N}{\partial x}\right)}_{y} - {\left(\frac{\partial M}{\partial y}\right)}_{x}}{M \left(x , y\right)} \mathrm{dy}\right]}$,

for the differential

$\boldsymbol{\mathrm{dF} \left(x , y\right) = {\left(\frac{\partial F}{\partial x}\right)}_{y} \mathrm{dx} + {\left(\frac{\partial F}{\partial y}\right)}_{x} \mathrm{dy}}$,

where $M = {\left(\frac{\partial F}{\partial x}\right)}_{y}$ and $N = {\left(\frac{\partial F}{\partial y}\right)}_{x}$.

For now, let's find the partial derivatives. For your differential:

$\textcolor{g r e e n}{M \left(x , y\right) = 3 {x}^{2} + 3 {y}^{2}}$
$\textcolor{g r e e n}{N \left(x , y\right) = {x}^{3} + 3 x {y}^{2} + 6 x y}$

Therefore:

$\textcolor{g r e e n}{{\left(\frac{\partial M}{\partial y}\right)}_{x} = 6 y}$

$\textcolor{g r e e n}{{\left(\frac{\partial N}{\partial x}\right)}_{y} = 3 {x}^{2} + 3 {y}^{2} + 6 y}$

which are clearly not equal, so the current differential is inexact. So, let us divide by $N \left(x , y\right)$ and see if the integral with respect to $x$ is reasonable to do:

$\ln \mu \left(x\right) = \int \frac{6 y - 3 {x}^{2} - 3 {y}^{2} - 6 y}{{x}^{3} + 3 x {y}^{2} + 6 x y} \mathrm{dx}$

$= \int \frac{- 3 {x}^{2} - 3 {y}^{2}}{{x}^{3} + 3 x {y}^{2} + 6 x y} \mathrm{dx}$

This doesn't look all that nice (it cannot be readily factored to eliminate $y$ terms), so what if we try integrating with respect to $y$ instead, and dividing by $M \left(x , y\right)$ instead? Then:

$\mu \left(y\right) = \text{exp} \left[\int \frac{{\left(\frac{\partial N}{\partial x}\right)}_{y} - {\left(\frac{\partial M}{\partial y}\right)}_{x}}{M \left(x , y\right)} \mathrm{dy}\right]$

and:

$\ln \mu \left(y\right) = \int \frac{3 {x}^{2} + 3 {y}^{2} + 6 y - 6 y}{3 {x}^{2} + 3 {y}^{2}} \mathrm{dy}$

$= \int {\cancel{\frac{3 {x}^{2} + 3 {y}^{2}}{3 {x}^{2} + 3 {y}^{2}}}}^{1} \mathrm{dy}$

And this integral is easy. It's just $y$. Therefore, $\textcolor{b l u e}{\mu \left(y\right) = {e}^{y}}$ would be your integrating factor. Let's test it out:

$3 {e}^{y} \left({x}^{2} + {y}^{2}\right) \mathrm{dx} + x {e}^{y} \left({x}^{2} + 3 {y}^{2} + 6 y\right) \mathrm{dy} = 0$

$\left(3 {x}^{2} {e}^{y} + 3 {y}^{2} {e}^{y}\right) \mathrm{dx} + \left({x}^{3} {e}^{y} + 3 x {y}^{2} {e}^{y} + 6 x y {e}^{y}\right) \mathrm{dy} = 0$

Checking for exactness, we obtain:

((delM)/(dely))_x stackrel(?" ")(=) ((delN)/(delx))_y

3x^2e^y + 3(y^2e^y + 2ye^y) stackrel(?" ")(=) 3x^2e^y + 3y^2e^y + 6ye^y

$3 {x}^{2} {e}^{y} + 3 {y}^{2} {e}^{y} + 6 y {e}^{y} = 3 {x}^{2} {e}^{y} + 3 {y}^{2} {e}^{y} + 6 y {e}^{y}$ color(blue)(sqrt"")

so we know our integrating factor is correct!