Question

Consider the overdamped harmonic oscillator
d^2y/dt^2 + 5dy/dt+ 2y = 0

Show that for every solution y(t) to this differential equation, there is at most one
value of t such that y(t) = 0 (i.e. none of the solutions oscillate).

Answer 1

@TuringTest

Answer 2

I'm not sure I understand
In this situation of over-damping we have that \(y(t)\to0\) as \(t\to\infty\), so it seems you are asking to prove something that is not quite true...
there is no finite value t at which y(t)=0

Answer 3

well that would explain why this isn't making sense.. is there a calculation that would prove that this is the casE?

Answer 4

check that the roots of the characteristic polynomial are both negative, so you have something of the form\[\large y(t)=c_1e^{r_1t}+c_2e^{r_2t}\]where\[r_1,r_2<0\]that means that \(y\to0\) as \(t\to\infty\)

Answer 5

ya they are both negative

Answer 6

you can prove, in general, that if the discriminant of the characteristic polynomial is >0, the roots will both be negative
a proof of that in general can be found here:
http://tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx
look at part 2 under "Free, Damped Vibrations"

Answer 7

thank you so so so much!!!

Answer 8

welcome :)