Exterior differential system on $SO(3;\mathbb R) \times \mathbb R$ I have the following exterior differential system for one forms $\alpha, \beta, \gamma$, where the $\theta^i$ are a cotangent basis on $SO(3)$, i.e. they satisfy $d \theta^i = \epsilon_{ijk} \theta^j \wedge \theta^k$ for the antisymmetric tensor $\epsilon$:
$\theta^2 \wedge \alpha = 0$
$\theta^2 \wedge \gamma = 0$
$d \alpha = - \theta^3 \wedge \beta$
$d \beta = \theta^3 \wedge \alpha - \theta^1 \wedge \gamma$
$d \gamma = \theta^1 \wedge \beta$
In addition there is a coordinate $\rho$, i.e. the one forms can be expanded as $\alpha = \alpha_a \theta^a + \alpha_\rho d\rho$ etc.
I guess it doesn't look that imposing (I don't know much of EDSses), but I'm having a seminar talk in a few days, and I've still got plenty to do... It would be awesome if I had a solution to the system above to show the audience! I'd be happy with a proof of existence (a la Frobenius or something I guess) and SOME info on the solution as well or pointers to maybe similar systems in the literature. I'm aware of the text books by Bryant et al.(Google books link) and Ivey&Landsberg (Google books link), but these are thick books and I'm almost outta time!!
I'll think about this myself too, of course, and I'll post here about it too, but now I need to get some sleep...
 A: The only solutions are:
$ \alpha = a \theta^2, \qquad  \beta = -a \theta^1 - c \theta^3, \qquad \gamma = c \theta^2, $
where $a, c$ are constants.  (No $d\rho$ terms appear.)  Here's how the computation goes:
From your first two equations, we have
$ \alpha = a \theta^2, \qquad \gamma = c \theta^2$
for some functions $a, c$.  We can also assume that
$ \beta = b_1 \theta^1 + b_2 \theta^2 + b_3 \theta^3 + b_\rho d\rho$
for some functions $b_1, b_2, b_3, b_\rho$.  
Now, substitute these expressions into your equation for $d\alpha$.  Reducing modulo $\theta^2$ implies that $b_\rho = 0$ and $b_1 = -a$.  Then the remaining terms imply that
$ da = a_2 \theta^2 - b_2 \theta^3$
for some function $a_2$. Similarly, reducing the equation for $d\gamma$ modulo $\theta^2$ implies that $b_3 = -c$, and then the remaining terms in $d\gamma$ imply that 
$dc = b_2 \theta^1 + c_2 \theta^2$
for some function $c_2$.  Now, substitute all this into your equation for $d\beta$.  Reducing modulo $\theta^2$ implies that $b_2=0$, and then the remaining terms imply that $a_2 = c_2 = 0$ as well.  Therefore, $da = dc = 0$, and so $a$ and $c$ are constants, which leads to the solutions above.  (It is straightforward to check that these are, in fact, solutions for any constants $a, c$.)
