Problem source

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 source

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*}