# Integrating Factor for Closed but not Exact Differential Form

(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?