I've been trying to reduce and/or eliminate the exponential terms in this system of differential equations to solve for x/y (both functions of t), but I am having trouble finding the relation between x and y:
$$ (ih)(dx/dt)=-u(B0)x-u(B1)e^{-2iwt}y\\ (ih)(dy/dt)=-u(B1)e^{2iwt}x+u(B0)y $$ I think there is some relation I can make between x and y to make this easier to manage, but I haven't been able to find it yet.
Copy of your equations : $$ (ih)(dx/dt)=-u(B0)x-u(B1)e^{-2iwt}y\\ (ih)(dy/dt)=-u(B1)e^{2iwt}x+u(B0)y $$ $h,b_0,b_1,w$ are given constants. $x(t)$ and $y(t)$ are unknown functions that you are looking for.
HINT :
Let $a=\frac{b_0u}{ih}$ which is a given constant.
Let $b=\frac{b_1u}{ih}$ which is a given constant. $$\frac{dx}{dt}=-ax-be^{-2iwt}y$$ $$\frac{dy}{dt}=-be^{2iwt}x+ay$$
Change of functions : $$x(t)=e^{-iwt}X(t)\quad\implies\quad \frac{dx}{dt}=e^{-iwt}(-iwX+\frac{dX}{dt})$$ $$y(t)=e^{iwt}Y(t)\quad\implies\quad \frac{dy}{dt}=e^{iwt}(iwY+\frac{dY}{dt})$$
$\frac{dx}{dt}=-ae^{-iwt}X-be^{-2iwt}e^{iwt}Y=e^{-iwt}(-iwX+\frac{dX}{dt})$
$\frac{dy}{dt}=-be^{2iwt}e^{-iwt}X+ae^{iwt}Y=e^{iwt}(iwY+\frac{dY}{dt})$
After simplification :
$$\begin{cases} \frac{dX}{dt}=(-a+iw)X-bY\\ \frac{dY}{dt}=(a-iw)Y-bX \end{cases}$$ This is a classical system of two first order linear ODEs. I suppose that yo can take it from here.