How to calculate this Differential Form

Question

Let $d\omega=(\frac{-3}{2}u^2D - Su + Tu - u^2 - Qu)(u^3D + Su^2 - Tu^2)^{-1}du \wedge \omega$ Where $D$, $S$, $T$ and $Q$ are constants.

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

Answer 1

My suggestion is to look for constants $A$ and $B$ such that $$ \mathrm{d}\bigl(u^A(Du+S-T)^B\,\omega\bigr) = 0, $$ then you can write $\omega = u^{-A}(Du+S-T)^{-B}\,\mathrm{d}v$ for some function $v$.

You will have to treat the special cases $S=T$ and/or $D=0$ separately, but you are still looking for an integrating factor in those cases as well.

Added remark: The point is that, when $D$ and $S{-}T$ are nonzero, we obviously have a partial fractions decomposition $$ \frac{-\tfrac32 u^2D-Su+Tu-u^2-Qu}{u^3D+Su^2-Tu^2} = -A\frac{1}{u} - B\frac{D}{Du+S-T} $$ for some constants $A$ and $B$. Calculating $A$ and $B$ is pure algebra. This means that $$ \frac{-\tfrac32 u^2D-Su+Tu-u^2-Qu}{u^3D+Su^2-Tu^2}\,\mathrm{d}u = \mathrm{d}\left(\log\bigl({u}^{-A} (Du+S{-}T)^{-B}\bigr)\right). $$ Now the rest should be easy.