Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Cheers in advance!

share|improve this question
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? –  B. 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
add comment

2 Answers

up vote 3 down vote accepted

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

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


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

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 $x$.

share|improve this answer
add comment

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 $$\frac{M_y-N_x}{N}=\frac{-1}x$$

share|improve this answer
I love simplicity! +1 –  Amzoti Feb 27 '13 at 15:04
Nice observation! +1 –  amWhy Feb 28 '13 at 0:16
@amWhy: Thanks. –  B. S. Feb 28 '13 at 5:27
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.