I'm working through problem 6 in chapter 1 in Goldstein's classical mechanics book. I've reduced it to asking, if $x$ and $y$ are coordinates and function of time $t$, whether the differential constraint
$$ \frac { dy } { y } = \frac{ dx } { f ( t ) - x } $$
can be integrated for arbitrary $ f( t ) $, where $f ( t ) $ is differentiable, such that it could be written as a holomorphic constraint $$ F ( x , y , t ) = C $$ where $C$ is some constant. A constraint is integrable if and only if there exists an integrating factor. If we were dealing with just $x, y$, then there are some consistency conditions here that we could use to prove that such an integrating factor doesn't exist. But I'm not sure how to attack it in the case where $x, y $ are functions of $t$. I don't think this differential constraint is integrable, but I'm struggling to prove it. Help appreciated.
Edit: Thank you QMechanic for the hint. If we define the following form $$ \omega = \frac{ dy } { y } + \frac{ d x} { f( t ) - x } $$
Then, Frobenius' theorem states that this is integrable if and only if the wedge product of the form with its exterior derivative is zero.
$$ d \omega \wedge \omega = 0 $$
Doing the calculation, the exterior derivative is $$ d\omega = -\frac{ f ' ( t ) } { ( x + f( t ) ) ^ 2 } dt \wedge dx $$
Only the wedge of $d \omega $ and $dy $ is preserved since $ dx \wedge dx = 0 $. And so
$$ d \omega \wedge \omega = -\frac{ f ' ( t ) } { y ( x + f( t ) ) ^ 2 } dt \wedge dx \wedge dy $$
This is zero if and only if $ f' ( t ) = 0 $, and so $f( t ) $ can't be arbitrary.
