00:01
Here, we want to get the solution after a long time to the differential equation, x double dot, plus 2x dot, plus n squared, plus 1, x equals sine of n of t.
00:16
Now, in this problem, we are told that we'll have some oscillation.
00:24
However, we want to know the long -term behavior.
00:27
To get the long -term behavior, we can entirely ignore the fact that there's a decaying oscillation in the first place.
00:35
And we can instead simply solve for the particular solution of this differential equation, ignoring homogenous, decaying solutions.
00:44
If we assume that the particular solution x is equal to a, e to the i, omega, t, and we make sine of nt go to plus phi, we make sine of nt become e.
01:07
To the i n t plus pi over two so we said omega equal to n and pi over two equal to phi so basically we're taking the real so we're basically making this using complex exponentials and taking the real part thus if x is this then that means that then x then if x is a e to the i and t plus 5 over 2, we can then have our equation become x.
01:55
Dot equals to i .n .x and x double dot be minus n squared x now therefore we end up with and we have that sign of nt goes to x over a where a is the amplitude that we're solving for that is the capital x amplitude that we're solving for plugging all this into our equation we have minus n squared plus 2 i n plus n squared plus n squared plus n squared plus 1, all times x, is equal to x over a.
02:52
X is cancel.
02:54
We notice that this n squared cancels as well, and if we bring the a to the other side, we have a times 2 .i .n...