# Does nudging an exact differential equation nudge or destroy the identity integrating factor?

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?

• 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. – Evgeny Mar 24 at 18:20
• Hmm, I think you're right. Is this any better? – user10478 Mar 24 at 20:38
• Seems good to me :) Also, don't forget that user gets a notification about a comment is when you mention them with @ . – Evgeny Mar 26 at 18:20