00:01
To start this problem, first let's find the auxiliary equation.
00:03
P of r is equal to 4r squared plus 12r plus 5 is equal to 0.
00:10
Now, once we have this here, we see that we can, or we're going to attempt to factor this.
00:17
So the factors of 1 or 4 are 1 and 4 or 2 and 2.
00:25
So we can either have, when we have our r, we can either have 1 and 4, or 4 and 1, or 2 and 2, like so.
00:36
For 5, our only factors are going to be 1 and 5.
00:42
Okay, so we need to find the coefficients here of the r's.
00:46
So if we do 1 and 4, so this becomes 4 and then plus 5, which is 9.
00:53
So that's not equal to this here.
00:56
So let's try this instead.
00:58
So if we do that, we get 20 and then 1.
01:02
So 21 is not 12.
01:04
So let's try 2 and 2.
01:07
So if we do 2 and 2, so we get 10 and then 2, which does add up to 12.
01:14
So this is our factored form here.
01:17
So next, our roots are going to be negative 1ā2 and negative 5 halves.
01:22
So since we have two different roots, this is going to be.
01:26
To be an over -damped system, or we have two real roots.
01:30
So this is over -damped.
01:33
Next we need to find our solution.
01:37
So our general solution is going to take the form c1e to the negative 1 half t plus c2, e to the negative 5 -5 -2.
01:46
In order to use our initial condition here, we need to find y prime of t.
01:52
This is equal to negative 1 1 1ā2, e to the negative 1 half t.
01:57
And then minus 5 -half c2, e to the negative 5 -halfs t.
02:02
Right? so, y of 0, this is equal to c -1 plus c2, which is equal to 1.
02:09
So remember, this is equal to 1, 1.
02:12
So we just get c -1 plus c2.
02:16
So again, e to the 0 is going to be equal to 1.
02:21
So now we're going to get y prime of 0 is going to be negative 1 -half, c -1, then minus 5 -2 is equal 2, and then we get negative 3 over here.
02:39
So if you notice, if we multiply this by 2, then we can add these two together.
02:46
So i'm going to multiply all of this by 2...