Consider a harmonic oscillator subject to a frictional force proportional to velocity: $$\ddot{x}+2\gamma\dot{x}+\omega^2x=0.$$ Here $\dot{x}$ and $\ddot{x}$ are $\frac{dx}{dt}$ and $\frac{d^2x}{dt^2}.$ Assume $\omega>\gamma.$ Interpreting this equation as a system of equations for two functions $x\left ( t\right ) $ and $\nu\left ( t\right ) = \dot{x} \left ( t\right ) $, solve this system of equations and show that the solutions are:
$$\begin{gathered} x\left(t\right)=\mathrm{e}^{-\gamma t}\left[x_0\cos\left(\Omega t\right)+x_0\frac\gamma\Omega\sin\left(\Omega t\right)+\nu_0\frac1\Omega\sin\left(\Omega t\right)\right]; \\ \nu\left(t\right)=\mathrm{e}^{-\gamma t}\left[-x_{0}\frac{\omega^{2}}\Omega\sin\left(\Omega t\right)+\nu_{0}\cos\left(\Omega t\right)-\nu_{0}\frac\gamma\Omega\sin\left(\Omega t\right)\right]. \end{gathered}$$ Here, $x_0= x\left ( 0\right ) $ and $\nu_0= \nu\left ( 0\right ) $ are initial conditions. What is the frequency $\Omega$ of these oscillations?
My attempt:
Let's introduce a new variable $v(t) = \dot{x}(t)$ so that we have $$\begin{cases}\dot{x}&=v\\\dot{v}&=-2\gamma v-\omega^2x&\end{cases}$$ Now, we have a system of first-order ordinary differential equations. We can rewrite this in matrix form as: $$\frac d{dt}\begin{bmatrix}x\\v\end{bmatrix}=\begin{bmatrix}0\times x+1\times v\\-\omega^2\times x+(-2\gamma)\times v\end{bmatrix}=\begin{bmatrix}0&1\\-\omega^2&-2\gamma\end{bmatrix}\begin{bmatrix}x\\v\end{bmatrix}$$ The solution to this system can be found by diagonalizing the coefficient matrix. Let's denote the coefficient matrix by $A$: $$A=\begin{bmatrix}0&1\\-\omega^2&-2\gamma\end{bmatrix}$$ The eigenvalues $\lambda$ and eigenvectors $\mathbf{v}$ of $A$ satisfy the equation: $\det(A-\lambda I)=0$, so we have $$\det\begin{pmatrix}-\lambda&1\\-\omega^2&-2\gamma-\lambda\end{pmatrix}=0 \implies\lambda=-\gamma\pm\sqrt{\gamma^2-\omega^2}$$ Since $\omega > \gamma$, the term under the square root is negative, resulting in complex conjugate eigenvalues. Let's denote $\Omega = \sqrt{\omega^2 - \gamma^2}$, so the eigenvalues become: $$\lambda=-\gamma\pm i\Omega.$$
Now, we can write down the general solution using these eigenvalues. The general solution has the form: $$\begin{bmatrix}x(t)\\v(t)\end{bmatrix}=c_1\begin{bmatrix}1\\-\gamma+i\Omega\end{bmatrix}e^{(-\gamma+i\Omega)t}+c_2\begin{bmatrix}1\\-\gamma-i\Omega\end{bmatrix}e^{(-\gamma-i\Omega)t}$$ $$\begin{aligned} &x\left(t\right) =c_1e^{(-\gamma+i\Omega)t}+c_2e^{(-\gamma-i\Omega)t} \\ &v(t) =(-\gamma+i\Omega)c_1e^{(-\gamma+i\Omega)t}+(-\gamma-i\Omega)c_2e^{(-\gamma-i\Omega)t} \end{aligned}$$ Then, finding the constants $c_1$ and $c_2$ in terms of $x_0$ and $v_0$ I get $$\frac{x_0 i \Omega+v_0+\gamma x_0}{2i \Omega}=c_1$$ and $$-\frac{v_0+\gamma x_0-i \Omega x_0}{2i \Omega}=c_2,$$ if I haven't made any errors.
Thus plugging in and applying Euler's formula, $$\begin{gathered} x(t)=\left(\frac{x_0i\Omega+v_0+\gamma x_0}{2i\Omega}\right)e^{-\gamma t}(\cos(\Omega t)+i\sin(\Omega t)) \\ +\left(-\frac{v_0+\gamma x_0-i\Omega x_0}{2i\Omega}\right)e^{-\gamma t}(\cos(\Omega t)-i\sin(\Omega t)) \end{gathered}$$ and $$\begin{gathered} \begin{aligned}v(t)=(-\gamma+i\Omega)\left(\frac{x_0i\Omega+v_0+\gamma x_0}{2i\Omega}\right)e^{-\gamma t}(\cos(\Omega t)+i\sin(\Omega t))\end{aligned} \\ +(-\gamma-i\Omega)\left(-\frac{v_0+\gamma x_0-i\Omega x_0}{2i\Omega}\right)e^{-\gamma t}(\cos(\Omega t)-i\sin(\Omega t)) \end{gathered}$$ But I am having trouble simplifying it to get to the solutions as written.
Btw, am I correct in assuming $\Omega$ as frequency in this physical context? Furthermore, how do we calculate the frequency $Ω$ of these oscillations?