Preamble

Consider a damped harmonic oscillator, with his well know differential equation \begin{equation*} m \ddot{x} + c \dot{x} + kx=0 \end{equation*} and let's find the solution that satisfies $x(0)=x_0$ and $\dot{x} (0) = v_0$. The problem admit a lovely exact solution (exploiting the considerations I found in books, I derived the complete explicit solution I write below). We have three cases. In the particular case in which $c=2\sqrt{mk}$, we have \begin{equation*} x(t) = \left[ x_0 + \left( v_0 + x_0 \sqrt{\frac{k}{m}} \right) t \right] e^{- \sqrt{\frac{k}{m}} t} \end{equation*} Otherwise we have to consider separately the two cases $c \gtrless 2\sqrt{mk}$, and to don't make heavier equations we have to introduce this parameters (we can observe that $\alpha$ and $\beta$ are length and depend to initial conditions too, while $\gamma$ e $\xi$ are adimensional and depend only by physical characteristics of the system) \begin{equation*} \alpha \equiv \frac{x_0}{2} \qquad \beta \equiv \frac{m v_0}{c} \qquad \gamma \equiv \frac{4mk}{c^2} \qquad \xi \equiv \frac{ct}{2m} \end{equation*} In the case $c>2\sqrt{mk}$ (i.e. $\gamma<1$) we have \begin{equation*} \begin{split} \displaystyle x(t) = & \frac{ \alpha \left(\sqrt{1 - \gamma } + 1 \right) + \beta}{\sqrt{1 - \gamma }} \cdot \exp \left[ {\left({-1 + \sqrt{1 - \gamma}} \right) \cdot \xi} \right] + \\ \displaystyle &\frac{ \alpha \left(\sqrt{1 - \gamma } - 1 \right) - \beta}{\sqrt{1 - \gamma }} \cdot \exp \left[ {\left({-1 - \sqrt{1 - \gamma}} \right) \cdot \xi} \right] \end{split} \end{equation*} While if $c<2\sqrt{mk}$ (i.e. $\gamma>1$) the solution is \begin{equation*} x(t) = 2 \alpha \cdot \frac{{\sin \left[ \tan^{-1} \left( \frac{\sqrt{\gamma - 1}}{1+\frac{\beta}{\alpha}} \right) + \sqrt{\gamma - 1} \cdot \xi \right]}} {\sin \left[ \tan^{-1} \left( \frac{\sqrt{\gamma - 1}}{1+\frac{\beta}{\alpha}} \right) \right]} \cdot {\exp(-\xi)} \end{equation*} Note that in this expressions, time dependence is included in $\xi$.

Question

Is it possible to do the same with the sinusoidal forced case? We have \begin{equation*} m \ddot{x} + c \dot{x} + kx= F \cos(\omega t + \phi) \end{equation*} but what is the solution of this differential equation that satisfies the condition $x(0)=x_0$ and $\dot{x} (0) = v_0$? I can't write it, even in the simpler case with $\phi = 0$.

Preamble

Consider a damped harmonic oscillator, with his well know differential equation \begin{equation*} m \ddot{x} + c \dot{x} + kx=0 \end{equation*} and let's find the solution that satisfies $x(0)=x_0$ and $\dot{x} (0) = v_0$. The problem admit a lovely exact solution (exploiting the considerations I found in books, I derived the complete explicit solution I write below). We have three cases. In the particular case in which $c=2\sqrt{mk}$, we have \begin{equation*} x(t) = \left[ x_0 + \left( v_0 + x_0 \sqrt{\frac{k}{m}} \right) t \right] e^{- \sqrt{\frac{k}{m}} t} \end{equation*} Otherwise we have to consider separately the two cases $c \gtrless 2\sqrt{mk}$, and to don't make heavier equations we have to introduce this parameters (we can observe that $\alpha$ and $\beta$ are length and depend to initial conditions too, while $\gamma$ e $\xi$ are adimensional and depend only by physical characteristics of the system) \begin{equation*} \alpha \equiv \frac{x_0}{2} \qquad \beta \equiv \frac{m v_0}{c} \qquad \gamma \equiv \frac{4mk}{c^2} \qquad \xi \equiv \frac{ct}{2m} \end{equation*} In the case $c>2\sqrt{mk}$ (i.e. $\gamma<1$) we have \begin{equation*} \begin{split} \displaystyle x(t) = & \frac{ \alpha \left(\sqrt{1 - \gamma } + 1 \right) + \beta}{\sqrt{1 - \gamma }} \cdot \exp \left[ {\left({-1 + \sqrt{1 - \gamma}} \right) \cdot \xi} \right] + \\ \displaystyle &\frac{ \alpha \left(\sqrt{1 - \gamma } - 1 \right) - \beta}{\sqrt{1 - \gamma }} \cdot \exp \left[ {\left({-1 - \sqrt{1 - \gamma}} \right) \cdot \xi} \right] \end{split} \end{equation*} While if $c<2\sqrt{mk}$ (i.e. $\gamma>1$) the solution is \begin{equation*} x(t) = 2 \alpha \cdot \frac{{\sin \left[ \tan^{-1} \left( \frac{\sqrt{\gamma - 1}}{1+\frac{\beta}{\alpha}} \right) + \sqrt{\gamma - 1} \cdot \xi \right]}} {\sin \left[ \tan^{-1} \left( \frac{\sqrt{\gamma - 1}}{1+\frac{\beta}{\alpha}} \right) \right]} \cdot {\exp(-\xi)} \end{equation*} Note that in this expressions, time dependence is included in $\xi$.

Question

Is it possible to do the same with the sinusoidal forced case? We have \begin{equation*} m \ddot{x} + c \dot{x} + kx= F \cos(\omega t + \phi) \end{equation*} but what is the solution of this differential equation that satisfies the condition $x(0)=x_0$ and $\dot{x} (0) = v_0$? I can't write it, even in the simpler case with $\phi = 0$.