This equation comes from Physics but here I need your mathematical skills to verify it 2 ways:
First of consider the holy Differential equation of forced oscillation: $$\boxed{m\,\ddot{x}+\beta\,\dot{x}+m\,{\omega_0}^2\,x = F_A\,\cos(\omega_t\,t)}$$
It's an equation that describes a fundamental equilibrium of forces.
The mathematical solution already has been provided here, but in my eyes incomplete.
Here I want to write out the complete solution of that ODE for $x(t)$.
$\color{gray}{\textbf{First Method}}$
$\textbf{Homogenous Part:}$
$\textsf{setting} \quad$ $m\,\ddot{x}+\beta\,\dot{x}+m\,{\omega_0}^2\,x = 0$ $\quad \textsf{and using an exponential ansatz yields:}$
$\underline{\text{Homogenous Part:}} \quad$
$$\boxed{x_h(t) = C_1\,e^{\textstyle-\frac{\beta}{2\,m}\,t}\,e^{\textstyle{i\sqrt{{w_0}^2-\left(\frac{\beta}{2\,m}\right)^2}}\,t}+ C_2\,e^{\textstyle-\frac{\beta}{2\,m}\,t}\,e^{\textstyle{-i\sqrt{{w_0}^2-\left(\frac{\beta}{2\,m}\right)^2}}\,t}}$$
Provided that
$\bullet \quad$$C_1$ and $C_1$ are real constants $c_1$ and $c_2$ $\bullet\quad{\omega_0}^2 > \left(\frac{\beta}{2\,m}\right)^2$ $\bullet \quad \omega_e= \sqrt{{\omega_0}^2-\left(\frac{\beta}{2\,m}\right)^2}$ $\bullet\quad \frac{\beta}{2\,m} = \gamma$
$\textsf{this can be rewritten:}$
$$\boxed{x_h(t) = e^{\textstyle -\gamma\,t}\,\left[c_1\cos(\omega_e\,t)+c_2\sin(\omega_e\,t)\right]}$$
Actually there exist starting conditions like $x(0) = A_0$, $\dot{x}(0) = 0$, but we can spare that.
$\textbf{Inhomogeneous Part}:$
$\textsf{looking at $\quad m\,\ddot{x}+\beta\,\dot{x}+m\,{\omega_0}^2\,x = F_A\,\cos(\omega_t\,t)\quad$ using the ansatz $x_p(t) = a\,e^{\textstyle i\varphi}\,e^{\textstyle i\,\omega_t\,t}$ brings:}$
$\underline{\text{Inhomogeneous Part:}}$
$$\boxed{x_p(t) = \frac{F_A}{\sqrt{m^2\,\left(w_0^2-{w_t}^2\right)^2+(\beta\,\omega_t)^2}}\,\cos\left(\omega_t\,t-\arctan\left(\frac{\beta\,\omega_t}{m\,\left({\omega_0}^2-\omega_t\right)}\right)\right)}$$
$\textbf{Total Solution:}$
The complete solution is given by $x(t) = x_h(t)+x_p(t)$. I won't write that out additionally. It's there.
Note that $\omega_t$ is a fixed excitation frequency for all times $t$. The Only changing variable is $t$ itself
I just mentioned all of that because I didn't see it appearing here but it is building to my question.
$\textbf{Question 1:}$
First of, what's the shape of the function? This might be physical, but just looking at the solution:
Is it really a decaying wave going over to a stabilised wave?
$\color{gray}{\textbf{Second Method}}$
$\textbf{Fourier Transform:}$
$\textsf{Transforming $\quad m\,\ddot{x}+\beta\,\dot{x}+m\,{\omega_0}^2\,x = F_A\,\cos(\omega_t\,t)\quad$ to frequency domain gives:}$
$\underline{\text{Fourier Transform:}}$
$$\begin{align}\left(-m\,\omega^2-i\,\omega\,\beta+m\,{\omega_0}^2\right)\,X(\omega) = \\\\ F_A\,\pi\,[\delta(\omega-\omega_t)+\delta(\omega+\omega_t)]\end{align}$$
$\textsf{so after rearranging:}$
$$X(\omega) = \dfrac{F_A\,\pi\,[\delta(\omega-\omega_t)+\delta(\omega+\omega_t)]}{\left(-m\,\omega^2-i\,\omega\,\beta+m\,{\omega_0}^2\right)}$$
Here I'm totally not sure if this is the right one. It's for sure, $x(\omega)$ depends no longer on $t$, but $\omega$
There I ask myself $\cdots$
$\textbf{Question 2}$
Where the frequency $\omega$ belongs to? Is it from the oscillation source? I assumed that's fixed.
$\textbf{Question 3}$
Does the Inverse FT bring back the full solution derived for $x(t)$? I know I had to solve:
$$x(t) = \dfrac{1}{2\,\pi}\,\int_{-\infty}^{\infty}\left(\dfrac{F_A\,\pi\,[\delta(\omega-\omega_t)+\delta(\omega+\omega_t)]}{\left(-m\,\omega^2-i\,\beta+m\,{\omega_0}^2\right)}\right)\,e^{\textstyle i\,\omega\,t}\,\mathrm{d\omega}$$
$\textbf{Conclusion:}$
At this point I lack the experience to solve for $x(t)$. On top I haven't seen the complete solution.
Consequently I beg you to bring this to an end with me.
Your solutions to the two parts give the shape of the solution. As long as $\beta$ is small the homogeneous part is a decaying sine wave at a slightly lower frequency than the frequency with no resistance. When $\beta$ is too large $\omega_e$ becomes imaginary, the oscillation disappears, and the system slowly approaches equilibrium. This part depends on the initial conditions. The inhomogeneous part is a sine wave at the driving frequency. It does not depend on the initial conditions and will persist as long as the driving term continues.
$x(\omega)$ (it would be better to use a different variable, maybe call it $X(\omega)$) is the Fourier transform of the position. $\omega$ ranges over all frequencies, positive and negative. You get back to $x_p(t)$ by doing an inverse Fourier transform. Your expression only has the effect of the driven oscillation in it.