Solving the equation of damped oscillator

Question

I'm asked to prove that any solution of the equation $$\ddot\Phi+\Gamma\dot\Phi+\omega_0^2\Phi=0;\qquad \omega_0>\frac\Gamma 2$$ is $$\Phi=A_0e^{-\frac{\Gamma}{2} t}e^{i(\omega t-\beta)};\qquad \omega^2\equiv\omega_0 ^2-\left(\frac{\Gamma}2\right)^2$$ maybe show is simple, but I would get the process of obtaining the solution, many thanks in advance.

Answer 1

Let $x=(\Phi, \dot{\Phi})^T$. Then the equation becomes $\dot{x}(t) = A x(t)$, where $A= \begin{bmatrix} 0 & 1 \\ -\omega_0^2 & -\Gamma \end{bmatrix} $ which has the solution $x(t) = e^{At} x_0$.

The eigenvalues of $A$ are given by solutions to $\lambda(\lambda+\Gamma) + \omega_0^2 = 0$, which are $\lambda = {1 \over 2} (-\Gamma \pm \sqrt{\Gamma^2-4 \omega_0^2}) ={1 \over 2} (-\Gamma \pm i\sqrt{4 \omega_0^2-\Gamma^2}) =-{ 1\over 2} \Gamma \pm i \sqrt{\omega_0^2-({\Gamma \over 2})^2} = -{ 1\over 2} \Gamma \pm i \omega$.

Since the eigenvalues are distinct, $A$ is diagonalisable and can be written as $A= V \begin{bmatrix} -{ 1\over 2} \Gamma + i \omega & 0 \\ 0 & -{ 1\over 2} \Gamma - i \omega \end{bmatrix} V^{-1}$, and so $e^{At} = V \begin{bmatrix} e^{(-{ 1\over 2} \Gamma + i \omega)t} & 0 \\ 0 & e^{(-{ 1\over 2} \Gamma - i \omega)t} \end{bmatrix} V^{-1}$.

Hence the general solution is given by $\Phi(t) = c_+ e^{(-{ 1\over 2} \Gamma + i \omega)t} + c_- e^{(-{ 1\over 2} \Gamma - i \omega)t}$ for some constants $c_+,c_-$.