I have reduced an Ito diffusion process to the following: $$ dX^i_s = \sigma_j^i(X_s) \,dB_s^j $$ where $i,j$ are the indices of the components and $\sigma$ is derived from the inverse metric tensor (which was only a function of the $z$ coordinate). Note there is no drift term.
Expanded out, my system is given by: \begin{aligned} dX^1_s &= \frac{1}{2}\left[ (a+b\cos(c X^3_s))dB^1_s + b\sin(cX^3_s)dB^2_s \right]\\ dX^2_s &= \frac{1}{2}\left[ (a-b\cos(c X^3_s))dB^2_s + b\sin(cX^3_s)dB^1_s \right]\\ dX^3_s &= \frac{k}{2}\, dB^3_s \end{aligned} where $a,b,c,k$ are real, positive constants. (Also, the superscripts are of course indices, not powers.)
My naive solution attempt is to first integrate equation 3, giving: $$ X^3_s = {k} B^3_s $$ Now I'll let $X_0^i=0\;\forall\; i$. Then, by (blind) substitution and integration, the first two equations become: $$ 2X_t^1 = aB^1_t + b\int\limits_0^t\cos(ck B^3_s/2)\,dB^1_s + b\int\limits_0^t\sin(ck B^3_s/2)\,dB^2_s $$ $$ 2X_t^2 = aB^2_t + b\int\limits_0^t\sin(ck B^3_s/2)\,dB^1_s - b\int\limits_0^t\cos(ck B^3_s/2)\,dB^2_s $$ I do not know how to proceed from here (or whether the prior steps are correct). I would like to compute $X_t=(X_t^1,X_t^2,X_t^3)$, but $\mathbb{E}[X_t]$ and $\mathbb{E}[X_t^2]$ would be great too. Ito's isometry and formula are not helping me, because I need to integrate a Brownian motion term with respect to an independent Brownian motion term.
If there is an obvious mistake or solution, please forgive me, as I have only just starting reading in this area. Any literature or tutorials about this would be appreciated too!