Differential form

Question

Let $(u^3+A)^{-2}(-\frac{5}{2}u^2-Bu)du \wedge \omega= d\omega$

Where A and B are constant.

I don't know how to calculate $\omega$.

$\omega$ ?

Answer 1

To do this, you'd need to find the function $v(u)$ such that $$ \mathrm{d}(\ln(v)) = \frac{\mathrm{d}v}{v} = \frac{(5u^2+2Bu)\mathrm{d}u}{2(u^3+A)^2}\,, $$ which is a integral calculus problem. (It probably doesn't have any particularly nice answer for general constants $A$ and $B$, though. You could run it through Maple or Mathematica to see what happens.)

Then you'd have $\omega = \frac{\mathrm{d}x}{v}$ for some function $x$.