which are time-independent.
Let us display some results for the case \(F=F_0\sin\left(\alpha t\right)\). Since the force term does not change the differential equations for \(\beta\) and\(\ \beta^{\ast}\), their solutions are the same as the ones proposed in this section. Now that we have a force term we need to calculate \(\mathcal{F}\) given by (\ref{eq:05}), resulting