00:01
Okay, so here we have that one -fourth times the second derivative of q with 3 to t plus 20 times dqdt plus 300 q is equal to zero.
00:10
Or we have the second derivative of q with respect to t plus 80 times the qdt plus 1 ,200 q is equal to zero.
00:18
And our initial conditions, q of 0 is equal to 4 and q prime of 0 is equal to 0.
00:24
So from m squared plus 80m plus 1 ,200 is equal to 0, we get here m.
00:30
Plus 20 times m plus 60 is equal to zero, giving us that m is equal to negative 20, and m is equal to negative 60.
00:42
So therefore we have that q of t then is going to be equal to c1 times e to the negative 20t plus c2 times e to the negative 60t.
00:56
And then our derivative, well, q prime of t, is going to be equal to negative 20t, 20, c1 times e to negative 20 t, minus 60 times c2, times e to the negative 60t.
01:13
Okay, and then q of zero is equal to four implies that c1 plus c2 is equal to four, and c prime of zero is equal to zero implies that negative 20 c1, minus 60 c2 is equal to zero.
01:27
So we get c1 plus 3c2 is equal to zero...