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Given the differential equation $\ddot x + \gamma \dot x + \omega_0^2 x =0$, one could assume a solution of the form $e^{\lambda t}$.

The characteristic equation yields $$\lambda_{1,2} = \frac{-\gamma \pm \sqrt{\gamma ^2 - 4\omega_0^2}}{2}.$$

I'm concerned with the case in which $\gamma^2<4\omega_0^2$. Then, if $\lambda=\lambda_1=\lambda_2^*$, the general solution is:

$$x(t) = C_1e^{(\Re(\lambda)+i\Im(\lambda))t}+C_2e^{(\Re(\lambda)-i\Im(\lambda))t} =e^{\Re(\lambda)t}\left[(C_1+C_2)\cos\Im (\lambda) t + i(C_1-C_2)\sin\Im(\lambda)t\right].~\tag{1}$$

Typically in physics when one solves this equation, the assumption $Ae^{-\gamma t/2}\cos(\omega_1t-\phi)$ as a solution is taken, as demonstrated here. I don't see a clear way to go from $(1)$ to the solution mentioned. Can you help me out with this? Thanks a lot!

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    $\begingroup$ what, you don't know eulers formula $e^{i t} = \cos t + i\sin t?$ $\endgroup$ – abel May 2 '15 at 16:23
  • $\begingroup$ @abel, yes but how can I get rid of the terms $(C_1\pm C_2)$? $\endgroup$ – Vladimir Vargas May 2 '15 at 16:25
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    $\begingroup$ choose $C_2 = \bar{C_1}.$ $\endgroup$ – abel May 2 '15 at 16:26
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Set $C_1=A·e^{iϕ}$ with $A>0$ real. To obtain a real solution, you then necessarily get $$ C_2=\bar C_1=A·e^{-iϕ} $$ Combined with your first form of the solution and $λ=-γ/2+iω_1$ you get $$- x(t)=A·e^{iϕ}·e^{(-γ/2+iω_1)·t}+A·e^{-iϕ}·e^{(-γ/2-iω_1)·t} =2A·e^{-γt/2}·\cos(ω_1t+ϕ) $$ So with a little tweaking in the choice of representing the constants, you can get to the formula with amplitude and phase.

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  • $\begingroup$ I'm sorry if it's obvious but, why is it necessary to have $C_2=C_1^*$? $\endgroup$ – Vladimir Vargas May 2 '15 at 19:32
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    $\begingroup$ Because $\overline{x(t)}=x(t)$ for real solutions. The coefficient identity follows from comparing the coefficients of equal exponentials. $\endgroup$ – LutzL May 2 '15 at 19:45

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