1The general solution changes depending on if the forcing function is \sin or \cos . But the particular solution is the same.
| ODE | solution |
| m\ddot{x}+kx=F_{0}\cos \omega t | x\left ( t\right ) =\left ( x_{0}-\frac{x_{st}}{1-r^{2}}\right ) \cos \omega _{n}t+\frac{\dot{x}_{0}}{\omega _{n}}\sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\cos \omega t |
| m\ddot{x}+kx=F_{0}\sin \omega t | x\left ( t\right ) =x_{0}\cos \omega _{n}t+\left ( \frac{\dot{x}_{0}}{\omega _{n}}-\frac{r}{1-r^{2}}x_{st}\right ) \sin \omega _{n}t+\frac{x_{st}}{1-r^{2}}\sin \omega t |