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$$ \begin{align} i x'(t) &= c A(t)(e^{-i q(t)}) y(t)\\ i y'(t) &= c A(t)(e^{i q(t)}) x(t)\\ \\ A(t)&=a cos(wt-r)\\ q(t)&=b+dt+k sin(wt+r) \end{align} $$ where all of $a,b,c,d,k,w,r$ are constants ... and the boundary condition is as follows: $$ X(t=0)=1\\ Y(t=0)=0 $$ It might be possible to slightly simplify the solutions by considering special cases. For example assume that when one of the parameters of the theory, say $d$ has some special values say $d_0=0$ or $1$ or something. Then if in this case an analytical solution $x_0(t)$ and $y_0(t)$ can be found, in the next step you may assume that the solutions $x(t)=x_0(t)\cdot X(t)$ and $y(t)=y_0(t)\cdot Y(t)$. Hopefully the equations for X(t) and Y(t) can be much simpler??

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Welcome to MSE! Pleae make sure I edited your question properly with MathJax. It greatly helps readability and is better to post them as such. Regards – Amzoti Jan 30 '13 at 17:31
Thanks a lot robjohn . You edited properly but what is $ in the second equation? – illia Jan 30 '13 at 18:59
The parameter "rho" is in d. For example d=m*rho. – illia Jan 30 '13 at 19:18
Can you just update the question to include that tidbit? $d=m*rho$. Regards – Amzoti Jan 30 '13 at 19:46

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