00:01
Okay, so to start solving this problem, let's find the auxiliary equation.
00:06
So that's going to be p of r.
00:08
It's going to be equal to r squared plus one -fifth r plus 1 over 100.
00:14
We set that equal to 0.
00:15
If you notice, if you take the square root of 1 over 100, that's going to be 1 over 10, and then twice of that is 1 over 5.
00:24
So this is actually going to factor into r plus 1 over 10 squared.
00:30
Is equal to 0.
00:34
Okay.
00:36
So then r is going to be equal to negative 1 over 10 with a multiplicity 2.
00:41
So our general solution is going to be a y of t is going to be equal to c1t e to the t divided by 10, or sorry, negative t divided by 10.
00:56
And then plus c2, actually i'll do that first.
01:01
And then plus c2, t, e to the negative 2.
01:05
T divided by 10.
01:09
Okay.
01:09
Now we need to use our initial conditions here.
01:14
So this is a critically damped system.
01:16
So y prime of t, y prime of t is equal to.
01:23
And then first we have negative 1 over 10, c1, e to the negative t divided by 10.
01:30
And then first times derivative of a second, first being this times derivative of this.
01:38
So that's going to be minus 1 over 10, and then c2, t, e to the negative t divided by 10, and then plus c2e to the negative t over 10.
01:57
So that's derivative first time second.
02:01
Okay.
02:01
Now we need to solve for c1 and c2 by plugging in 0.
02:06
So y of 0 equals 1.
02:08
So first we have y of 0.
02:11
That's going to be equal to.
02:13
So this becomes 0 because we have 0 here.
02:17
E to the 0 is equal to 1.
02:18
So then we just get c1 is equal to 1.
02:22
Next, y prime is 0.
02:25
So here we have this is all 1.
02:29
So we have negative 1 tenth and then minus, well this is 0.
02:35
So then next we just have c2.
02:38
So 2 is equal to 5.
02:41
So then c2 is going to be 5 plus 1 over 10.
02:48
5 plus 1 over 10.
02:50
So 5 is going to be 50 over 10 plus 1 over 10.
02:55
It's going to be 51 over 10.
02:59
So our final displacement, y of t, is going to be equal to e to the negative t divided by 10, and then plus 51 over 10, e to the negative t or sorry, t, e to negative t, e to negative t divided by 10.
03:19
Okay.
03:20
Now, for part b, in order to find the maximum displacement, we need to set the derivative equal to zero.
03:29
So let's take a look at the derivative.
03:32
So that's going to be this...