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This question will be related to this one, if for no other reason because a positive answer to the latter would likely help to solve the former.

Consider the differential equation $(y)\ dx + (x)\ dy = 0$. It is already exact, so we can think of the multiplicative identity, $u(x) = 1$, as an integrating factor. We can also observe by separating variables that $u(x) = \frac{1}{xy}$ is an integrating factor, yielding $(\frac{1}{x})\ dx + (\frac{1}{y})\ dy = 0$. From here, $u(x) = xy$ allows us to return to the original form, so we can toggle freely between the two, and both produce the same solution, $y = \frac{C}{x}$.

Now instead consider the inexact equation $(y)\ dx + (x + \epsilon\ x)\ dy = 0$, for some small value of $\epsilon$. By separating variables, we find the integrating factor $u(x) = \frac{1}{(x + \epsilon x)y}$, and the solution $y = \frac{C}{x^{\frac{1}{1 + \epsilon}}}$, which respectively are very close to $u(x) = \frac{1}{xy}$ and $ = \frac{C}{x}$ from the previous problem. However, the original inexact form has $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = \epsilon$, which is very close to $0$. Does this imply (not only for this problem, but in general) that an integrating factor very close to $u(x) = 1$ exists, or is the nearby exact form destroyed entirely when we nudge $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}$ even by a tiny amount?

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  • $\begingroup$ Consider slightly changing the title of the question. If I were to answer the question in title, I would say that exactness is a fragile property (non structurally stable, using terminology from dynamical systems): you can always perturb the exact equation with non-degenerate extremum of potential and get an equation which can't be exact at all. However, the question in your post is a bit different. $\endgroup$ – Evgeny Mar 24 at 18:20
  • $\begingroup$ Hmm, I think you're right. Is this any better? $\endgroup$ – user10478 Mar 24 at 20:38
  • $\begingroup$ Seems good to me :) Also, don't forget that user gets a notification about a comment is when you mention them with @ . $\endgroup$ – Evgeny Mar 26 at 18:20

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