00:02
Okay, so if remember, the general solution for a critically damp system is y of t is equal to c1, or sorry, e to the negative ct divided by 2m times c1 plus c2t.
00:23
Now to solve for c1 and c2, so we're given that y0 is equal to y of 0 is equal to y not, and then dy, dt, or veloc, at 0 is equal to v0.
00:37
So we need to find y prime of t.
00:41
This is going to be so first times derivative of second.
00:45
So that's going to be c2, e to the negative c t divided by 2m, and then plus second times derivative of first.
00:53
So that's going to be negative c over 2m, and then e to the negative ct over 2m times c1 plus c2t like so.
01:06
Next, we need to plug in 0.
01:08
So, y of 0, right? this is going to be, so this becomes 1, and then we just get c1 plus c2t.
01:18
So that c1 is going to be equal to why not.
01:23
Then now for c2, so y prime of 0, so this becomes 1, and then this is 0, and this is 1...