Mathematica or any CAS for that matter, gives a general solution, and does not customize for specific conditions. There are three cases to consider here: Underdamped, critically damped, and overdamped. Each has to be solved separately and gives a different solution since the roots are different.
$$
mx^{\prime\prime}\left( t\right) +cx^{\prime}\left( t\right) +kx\left(
t\right) =F-F\cos\left( t\right)
$$
Writing in standard form for vibration analysis (to make it easy to follow and this is the standard also)
\begin{align*}
x^{\prime\prime}\left( t\right) +\frac{c}{m}x^{\prime}\left( t\right)
+\frac{k}{m}x\left( t\right) & =\frac{F}{m}-\frac{F}{m}\cos\left(
t\right) \\
x^{\prime\prime}\left( t\right) +2\xi\omega x^{\prime}\left( t\right)
+\omega^{2}x\left( t\right) & =\frac{F}{m}-\frac{F}{m}\cos\left( \varpi
t\right)
\end{align*}
Where $\varpi$ is the forcing frequency (1 rad per second in this case),
$\omega$ is the natural frequency $\omega=\sqrt{\frac{k}{m}}$ and $\xi$ is
damping ratio, which is the most important quantity here as it decides which solution will come out
$$
\xi=\frac{c}{c_{r}}
$$
Where $c$ is the damping coefficient and $c_{r}$ is critical damping
coefficient given by $c_{r}=2\sqrt{km}$. There are three solutions for the
above ODE depending on $\xi<1$ (underdamped)$,\xi=1$ (critically damped) and
$\xi>1$ (over damped)
For underdamped, which is when $\xi<1$ or $c<c_{r}$ or $c<2\sqrt{km}$. The
roots of the characteristic equation\ $\lambda^{2}+2\xi\omega+\omega^{2}=0$
for the homogeneous part of the ODE are
$$
-\xi\omega\pm i\omega\sqrt{1-\xi}
$$
Hence
$$
x_{h}\left( t\right) =e^{-\xi\omega t}\left( A\cos\omega_{d}t+B\sin
\omega_{d}t\right)
$$
Where $\omega_{d}$ is the damped natural frequency, defined only for
underdamped case, and given by $\omega_{d}=\omega\sqrt{1-\xi^{2}}$. Adding the
particular solution, which for $\frac{F}{m}$ is $x_{p}=\frac{1}{m}\frac{F}{k}$
and for $\frac{F}{m}\cos\left( \varpi t\right) $ the particular solution is
$x_{p}=\frac{1}{m}\frac{F}{k}\frac{1}{\sqrt{\left( 1-r^{2}\right)
^{2}+\left( 2\xi r\right) ^{2}}}\sin\left( \varpi-\theta\right) $ where
$r=\frac{\varpi}{\omega}$ and $\theta=\arctan\left( \frac{c\varpi}
{k-m\varpi^{2}}\right) =\arctan\left( \frac{2\xi r}{1-r^{2}}\right) $.
Hence the complete solution for underdamped is
$$
x\left( t\right) =e^{-\xi\omega t}\left( A\cos\omega_{d}t+B\sin\omega
_{d}t\right) +\frac{1}{m}\frac{F}{k}-\frac{1}{m}\frac{F}{k}\frac{1}
{\sqrt{\left( 1-r^{2}\right) ^{2}+\left( 2\xi r\right) ^{2}}}\sin\left(
\varpi-\theta\right)
$$
Now $A,B$ are found from initial conditions. For the case of critically
damped, $\xi=1$ or $c=2\sqrt{km}$ the roots become $-\xi\omega$ only and hence
$x_{h}\left( t\right) =\left( A+Bt\right) e^{-\xi\omega t}$ and therefore,
the full solution is
$$
x\left( t\right) =\left( A+Bt\right) e^{-\xi\omega t}+\frac{1}{m}\frac
{F}{k}-\frac{1}{m}\frac{F}{k}\frac{1}{\sqrt{\left( 1-r^{2}\right)
^{2}+\left( 2\xi r\right) ^{2}}}\sin\left( \varpi-\theta\right)
$$
For the case of overdamped, when $\xi>1$ or $c>2\sqrt{km}$ then roots become
$$
\left\{ \frac{-c}{2m}+\sqrt{\left( \frac{c}{2m}\right) ^{2}-\frac{k}{m}%
},\frac{-c}{2m}-\sqrt{\left( \frac{c}{2m}\right) ^{2}-\frac{k}{m}}\right\}
$$
or
$$
\left\{ -\omega\xi+\omega\sqrt{\xi^{2}-1},-\omega\xi-\omega\sqrt{\xi^{2}%
-1}\right\}
$$
And since $\xi>1$ then both roots are real, and there is no oscillation. Let
the above roots be $p_{1},p_{2}$, hence the solution is
$$
x_{h}\left( t\right) =Ae^{p_{1}t}+Be^{p_{2}t}
$$
And the complete solution is
$$
x\left( t\right) =Ae^{p_{1}t}+Be^{p_{2}t}+\frac{1}{m}\frac{F}{k}-\frac{1}
{m}\frac{F}{k}\frac{1}{\sqrt{\left( 1-r^{2}\right) ^{2}+\left( 2\xi
r\right) ^{2}}}\sin\left( \varpi-\theta\right)
$$
Where again, the constants $A,B$ are found from initial conditions. Hence in summary
For $c<2\sqrt{km}$
$$x\left( t\right) =e^{-\xi\omega t}\left( A\cos\omega
_{d}t+B\sin\omega_{d}t\right) +\frac{1}{m}\frac{F}{k}-\frac{1}{m}\frac{F}
{k}\frac{1}{\sqrt{\left( 1-r^{2}\right) ^{2}+\left( 2\xi r\right) ^{2}}
}\sin\left( \varpi-\theta\right)
$$
For $c=2\sqrt{km}$
$$x\left( t\right) =\left( A+Bt\right) e^{-\xi\omega
t}+\frac{1}{m}\frac{F}{k}-\frac{1}{m}\frac{F}{k}\frac{1}{\sqrt{\left(
1-r^{2}\right) ^{2}+\left( 2\xi r\right) ^{2}}}\sin\left( \varpi
-\theta\right)
$$
For $c>2\sqrt{km}$
$$x\left( t\right) =Ae^{p_{1}t}+Be^{p_{2}t}+\frac{1}{m}
\frac{F}{k}-\frac{1}{m}\frac{F}{k}\frac{1}{\sqrt{\left( 1-r^{2}\right)
^{2}+\left( 2\xi r\right) ^{2}}}\sin\left( \varpi-\theta\right)
$$
Where $r=\frac{\varpi}{\omega}$ and $\theta=\arctan\left( \frac{c\varpi
}{k-m\varpi^{2}}\right) =\arctan\left( \frac{2\xi r}{1-r^{2}}\right)
,\omega=\sqrt{\frac{k}{m}}$ and $\xi$ is damping ratio $\xi=\frac{c}{c_{r}}$
And $c_{r}=2\sqrt{km}$
You can use Mathematica now to obtain the solution given the initial conditions, by using the above three cases each time. i.e. knowing if the system is underdamped, critically damped, or overdamped, you use the correct solution for that case.
