Time evolution of state in rotating $B$-field spin system

I’m given a state $|psi(t)rangle$ that solves the time dependent Schroedinger equation with the Hamilton operator $hat{H}(t)$. $hat{H}(t)$ describes a Spin-$frac{1}{2}$ in a rotating magnetic field ($vec{B}(t) = B_1[cos(omega t)vec{e_1} + sin(omega t) vec{e_2}] + B_3 vec{e_3}$) rotating around the z-axis with frequency $omega$. Furthermore I have a transformed state $|widetilde{psi}(t)rangle = hat{D}(t) |psi(t)rangle$ with $hat{D}(t)$ being the rotation operator $hat{D}(t) = e^{iomega t hat{sigma_3} /2}$. I have calculated that therefore the Hamiltonian solving the SG in this case becomes:
begin{align}
hat{widetilde{H}}(t) = frac{hbar Omega_1}{2} hat{sigma_1} + frac{hbar(Omega_3-omega)}{2}hat{sigma_3}, quad Omega_1 = -2frac{mu_0 B_1}{hbar}, quad Omega_3 = -2frac{mu_0 B_3}{hbar}
end{align}
The task is to calculate the two time evolution operators $hat{widetilde{U}}(t)$ and $hat{{U}}(t)$ given by $|widetilde{psi}(t)rangle = hat{widetilde{U}}(t) |widetilde{psi}(0)rangle$ and $|psi(t)rangle = hat{{U}}(t) |psi(0)rangle$. I plugged in the standard formula for time evolution and tried to get a result by simply puttung in the Hamiltonians, but I don’t get any further. I seem to have a blackout how to deal with this problem