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Mar 23, 2017 at 7:58 vote accept AJMansfield
Mar 15, 2017 at 18:59 comment added AJMansfield Ok, thanks! That worked perfectly to get me to what I needed: $$y(t) = \left(1 + a\frac{e^{\epsilon t} - 1}{\epsilon}\right)e^{at}y_0 + \frac{e^{\epsilon t} - 1}{\epsilon}e^{at}y_0' = (1+at)e^{at}y_0+te^{at}y_0' = \left(\cos \omega t - a \frac{\sin \omega t}{\omega}\right)e^{at}y_0+\frac{\sin \omega t}{\omega}e^{at}y_0'$$
Mar 15, 2017 at 18:23 comment added Chappers For the other form, it should work if you write $b = a+\epsilon$ and take $\epsilon \to 0$: then $$ c_1 e^{at}+c_2 e^{bt} = e^{at}\left( \frac{(a+\epsilon)y_0-y_0'}{\epsilon} + (y'_0-ay_0)\frac{e^{\epsilon t}}{\epsilon} \right) = e^{at} \left( y_0 + (y_0'-ay_0)\frac{e^{\epsilon t}-1}{\epsilon} \right), $$ and the second term tends to $t$ as $\epsilon \to 0$. (Note that your $c_2$ has $y_0$ and $y_0'$ the wrong way round: should be $(y'_0-ay_0)/(b-a)$, if only on dimensional grounds.)
Mar 15, 2017 at 18:12 comment added Chappers You can write the overdamped solution in the same way, as $$ Ae^{-bt}\cosh{\xi t} + Be^{-bt} \frac{\sinh{\xi t}}{\xi}, $$ where $\xi = \sqrt{b^2-c} = \frac{\alpha-\beta}{2}$, and the limit works in the same way (indeed, this is what I meant when I said that it held for real and imaginary $\omega$: the two forms are the same with $i\omega \leftrightarrow \xi$).
Mar 15, 2017 at 17:41 comment added AJMansfield Or for example taking $\lim_{a\to b} c_1 e^{at} + c_2 e^{bt}$, with $c_1=\frac{by_0-y_0'}{b-a}$ and $c_2=\frac{y_0-ay_0'}{b-a}$.
Mar 15, 2017 at 17:34 comment added AJMansfield The part with the limit of the underdamped equation as $\omega\to 0$ makes sense, but is there a way to take the limit of the overdamped form to get the critical form? I tried writing it out in terms of the fundamental solution set, but all the terms appear to diverge when taking the limit.
Mar 15, 2017 at 16:21 history answered Chappers CC BY-SA 3.0