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(I kind-of skipped ODEs in undergrad, so my knowledge is a little sketchy; please excuse me if this question is elementary.)

(In addition, this problem is a bit of a sketch; I'll flesh it out with calculations in a day or so.)

I've been looking over the theory of integrating factors, and they appear to focus on making the resulting differential form $\mu\omega$ closed, not necessarily exact.

This appears to work fine most of the time. For instance, if I take the ODE

$$-y\ dx + x\ dy = g(t) dt$$

the left-hand side is neither closed nor exact. I recall reading in $\textit{Advanced Calculus}$ by Buck https://smile.amazon.com/Advanced-Calculus-SECOND-Creighton-Buck/dp/B000OG0SRK/ that the differential equation for an integrating factor is called a "Pfaffian"; if that is the case and my back-of-the-envelope calculations from last night are correct [I'll flesh this out with calculations in a day or so], the Pfaffian is

$$x\frac{\partial \mu}{\partial x} + y\frac{\partial \mu}{\partial y} = -2\mu$$

This appears to admit the solution integrating factor

$$\mu(x,y) = \frac{1}{xy}$$

and so the ODE has the implicit solution

$$\left(\frac{y}{x}\right)^{xy} = A\exp\left(xy\int \frac{g}{xy}\ dt\right)$$

(Maybe one would want to take $g(t) \equiv 0$ ?)

The situation is different with the closed but not exact differential form (left-hand side of the equation)/ODE

$$\frac{-y}{x^2+y^2}\ dx + \frac{x}{x^2+y^2}\ dy = g(t)\ dt$$

The Pfaffian is [again, I'll flesh this out with calculations in a day or so]

$$\frac{\partial \mu}{\partial x}x + \frac{\partial \mu}{\partial y}y = 0$$

which doesn't appear to admit a solution, as nearly as I can figure.

Does this second ODE admit an integrating factor and I'm doing something wrong? More generally, if one has a closed but not exact differential form, is it possible to find an integrating factor that makes the differential form exact?

Thanks in advance.

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  • $\begingroup$ These differential equations fall under the umbra of "implicit ODEs" (at least, they generally have implicit solutions). The variables x and y are themselves functions of t in the paradigm. $\endgroup$ – Jeffrey Rolland Mar 19 at 2:14
  • $\begingroup$ Note that every closed form is locally exact ... Solving the differential equation is usually a local question; topology will dictate whether you can get a global integrating factor. $\endgroup$ – Ted Shifrin Mar 19 at 17:21
  • $\begingroup$ @TedShifrin Mmmm, so, "in a flow box" (to use terminology from Abraham and Marsden smile.amazon.com/Foundations-Mechanics-Ams-Chelsea-Publishing/…), that is to say, in a rectangle with respect to t, x, and y in which all functions and their partial derivatives are continuous, the second ODE actually is exact, and we have an implicit solution; very nice, thank you. $\endgroup$ – Jeffrey Rolland Mar 19 at 19:19

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