# Integrating factor of a differential arising from thermodynamics

Let $\delta E = (xy^2 + xye^x)dx + (2x^2y + xe^x)dy$ I now need to find the integrating factor $\mu (x,y)$ s.t. $dS = \mu (x,y) \delta E$ is a exact differential.

Now as far as I know $\delta E$ is exact if $\int (xy^2 + xye^x)\mu (x,y)dx = \int (2x^2y + xe^x)\mu (x,y)dy$ i.e. $\frac{1}{2} x^2 y^2 + e^x(x-1)y + C_1(y) = x^2 y^2 + xe^x +C_2(x)$ which as given isn't exact.

Now $\delta S$ is exact if $\int (xy^2 + xye^x)\mu (x,y)dx = \int (2x^2y + xe^x)\mu (x,y)dy$ at least I think so. But how do I compute these integrals when I have no idea what $\mu (x,y)$ looks like? Can somebody explain this to me?

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Well, we can always let $\mu(x,y)=0$, but I doubt that's what you're looking for. – Cameron Buie Feb 27 '13 at 13:36
@CameronBuie that isn't even true, since with $\mu (x,y) = 0$ $\delta S$ isn't exact. – Howdy Ho Feb 27 '13 at 13:40
Oops. You're right. – Cameron Buie Feb 27 '13 at 13:46
@HowdyHo: Are you sure about the coefficients of the OE? I mean isn't there any typo when typing? – Babak S. Feb 27 '13 at 14:06
@BabakS. you mean the equation for $\delta E$? That one is correct 100%. Or which one do you mean? – Howdy Ho Feb 27 '13 at 14:16

The differential is exact if it's of the form

$$\mathrm dS=\frac{\partial S}{\partial x}\mathrm dx+\frac{\partial S}{\partial y}\mathrm dy\;,$$

and you can check this using the integrability condition

$$\frac{\partial}{\partial y}\frac{\partial S}{\partial x}=\frac{\partial}{\partial x}\frac{\partial S}{\partial y}\;.$$

In your case, this is

$$\frac{\partial}{\partial y}\left(\mu\left(xy^2+xy\mathrm e^y\right)\right)=\frac{\partial}{\partial x}\left(\mu\left(2x^2y+x\mathrm e^y\right)\right)\;,$$

that is,

$$\frac{\partial\mu}{\partial y}\left(xy^2+xy\mathrm e^x\right)+\mu\left(2xy+x\mathrm e^x\right)=\frac{\partial\mu}{\partial x}\left(2x^2y+x\mathrm e^x\right)+\mu\left(4xy+x\mathrm e^x+\mathrm e^x\right)\;.$$

The ansatz $\mu=x^\alpha y^\beta$ seems promising, and indeed substituting this, dividing through by $x^\alpha y^\beta$ and separately comparing the coefficients of the polynomial and exponential terms yields the two equations

$$\beta xy+2xy=2\alpha xy+4xy$$

and

$$\beta x\mathrm e^x+x\mathrm e^x=\alpha\mathrm e^x+x\mathrm e^x+\mathrm e^x\;,$$

which are simultaneously solved by $\alpha=-1$ and $\beta=0$. Thus the integrating factor is $\mu=1/x$; with hindsight, this might have been guessed from the fact that all coefficients contain a factor $x$.

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Besides to generalized method of @joriki, if you take $$M(x,y)=\left(xy^2+xy\mathrm e^x\right)$$ and $$N(x,y)=\left(2x^2y+x\mathrm e^x\right)$$ then $$\frac{M_y-N_x}{N}=\frac{-1}x$$

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I love simplicity! +1 – Amzoti Feb 27 '13 at 15:04
Nice observation! +1 – amWhy Feb 28 '13 at 0:16
@amWhy: Thanks. – Babak S. Feb 28 '13 at 5:27