We can write down the answer for \( x_{pn}(t) \), by substituting \( f_n/m \) or \( g_n/m \) for \( F_0/m \). By writing each factor \( 2n\pi t/\tau \) as \( n\omega t \), with \( \omega\equiv 2\pi/\tau \),
$$ \begin{equation} \tag{3} F(t)=\frac{f_0}{2}+\sum_{n>0}f_n\cos(n\omega t)+g_n\sin(n\omega t). \end{equation} $$The solutions for \( x(t) \) then come from replacing \( \omega \) with \( n\omega \) for each term in the particular solution,
$$ \begin{eqnarray} x_p(t)&=&\frac{f_0}{2k}+\sum_{n>0} \alpha_n\cos(n\omega t-\delta_n)+\beta_n\sin(n\omega t-\delta_n),\\ \nonumber \alpha_n&=&\frac{f_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\ \nonumber \beta_n&=&\frac{g_n/m}{\sqrt{((n\omega)^2-\omega_0^2)+4\beta^2n^2\omega^2}},\\ \nonumber \delta_n&=&\tan^{-1}\left(\frac{2\beta n\omega}{\omega_0^2-n^2\omega^2}\right). \end{eqnarray} $$