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What is the general solution of the differential equation \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+2k\frac{\mathrm{d}x}{\mathrm{d}t}+x=0 \] for each of the cases: (i) \(k>1;\) (ii) \(k=1\); (iii) \(0 < x < 1\)? In case (iii) the equation represents damped simple harmonic motion with damping factor \(k\). Let \(x(0)=0\) and let \(x_{1},x_{2},\ldots,x_{n},\ldots\) be the sequence of successive maxima and minima, so that if \(x_{n}\) is a maximum then \(x_{n+1}\) is the next minimum. Show that \(\left|x_{n+1}/x_{n}\right|\) takes a value \(\alpha\) which is independent of \(n\), and that \[ k^{2}=\frac{(\ln\alpha)^{2}}{\pi^{2}+(\ln\alpha)^{2}}. \]
Solution: The auxiliary equation is \(\lambda^2 + 2k\lambda + 1 = (\lambda + k)^2+1-k^2 = 0\) (i) If \(k > 1\) then the solution is \(A\exp \left ({(-k + \sqrt{k^2-1})t} \right)+B\exp\left((-k-\sqrt{k^2-1})t \right)\). (ii) If \(k = 1\) then the solution is \(x = (A+Bt)e^{-kt}\) (iii) If \(k < 1\) then the solution is \(x = Ae^{-kt} \sin \left ( \sqrt{1-k^2} t \right)+Be^{-kt} \cos \left ( \sqrt{1-k^2} t \right)\) If \(x(0) = 0\) then \begin{align*} && x &= Ae^{-kt} \sin(\sqrt{1-k^2}t)\\ && \dot{x} &= Ae^{-kt} \left (-k \sin(\sqrt{1-k^2}t)+\sqrt{1-k^2} \cos(\sqrt{1-k^2}t) \right) \\ (\dot{x} =0): && \tan (\sqrt{1-k^2}t) &= \frac{\sqrt{1-k^2}}{k}\\ \end{align*} Therefore maxima and minima occur every \(\frac{\pi}{\sqrt{1-k^2}}\), so \begin{align*} && \frac{x_{n+1}}{x_n} &= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \frac{\sin\left (\sqrt{1-k^2}\left(t+\frac{\pi}{\sqrt{1-k^2}}\right)\right)}{\sin(\sqrt{1-k^2}t)} \\ &&&= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \left (-1+0 \right)\\ &&&= -\exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \\ \Rightarrow && \ln \alpha &= - \frac{k\pi}{\sqrt{1-k^2}} \\ \Rightarrow && (\ln \alpha)^2 &= \frac{k^2\pi^2}{1-k^2} \\ \Rightarrow && (1-k^2)(\ln \alpha)^2 &= k^2 \pi^2 \\ \Rightarrow && k^2(\pi^2+(\ln \alpha)^2) &= (\ln \alpha)^2 \\ \Rightarrow && k^2 &= \frac{(\ln \alpha)^2}{\pi^2 + (\ln \alpha)^2} \end{align*}