Good evening. I'm having some trouble with this exercice:
Determine f$\in C^1(R) $such that $f(0)=1$ and the differential form $$ \omega = \frac{2xy(f(x))^2}{1+(f(x))^2} dx - \arctan (f(x)) dy$$ be exact.
I reasoned like this: For the differential form to be exact it must be closed. So it must hold true
$$f_y \left(\frac{2xy(f(x))^2}{ 1+(f(x))^2}\right) = f_x (- \arctan (f(x))$$
doing the calculations I found
$$\frac{2x (f(x))^2}{(1+(f(x))^2)^2} =-\frac {f'(x)}{1+(f(x))^2}$$
that is
$$f'(x) = - \frac{2xf(x)}{1+f(x)^2}$$
At this point I don't know how to continue to find $ f (x)$.