00:01
For this question, we're asked to solve this initial value problem.
00:04
And right off the bat, we can take this and divide through by 3.
00:08
So we're going to solve this differential equation, which is the same as solving this one.
00:13
Right? and then if you look at the auxiliary equation, we have this.
00:19
And we want to solve for r.
00:21
So since this is a polynomial of degree 3, we can't use the quadratic formula.
00:27
So if it has any rational roots, then by the rational roots theorem, they are plus or minus 1, plus or minus 2, just the divisors of 270.
00:39
So plus or minus 3, plus or minus 5, 6, and you know, dot, dot, dot, all the way up to plus or minus 135, plus or minus 270.
00:55
All right? and then looking at this, we see that a negative number cannot be a root, since you're just going to be adding three negative numbers, which will never equal zero.
01:03
And then, so 1 is too small, same with 2, 3, 5 is almost there, but not there yet.
01:13
And then 6 is actually a root.
01:16
So let's say note, it's a 6 cubed, plus 6 times 9, minus 270.
01:26
This is equal to 216 plus 54, minus 270.
01:34
And this is equal to zero.
01:36
Therefore, 6 is a root.
01:39
And then to find what's left over, we can use synthetic division, or just polynomial division.
01:53
And so i have this 1 times 6, this is 6, 6 times 6, this is 36, 45, this is 270, and then zero.
02:05
Okay.
02:07
So this tells us that this polynomial right here, it factors the following way, r minus 6 times r squared, plus 6r, and then plus 45.
02:26
This is all equal to zero.
02:29
And these coefficients, the 6 and the 45, they came from right here.
02:34
Okay.
02:34
And then what are the zeros of this quadratic polynomial? all right.
02:38
So we can use the quadratic formula.
02:40
And let me do that right here.
02:45
So i'll have r equals negative 6.
02:51
So this is for solving r squared plus 6r plus 45 is equal to zero.
03:00
So r is equal to negative 6 plus or minus, this would be 36 minus 180, and all over 2.
03:14
So this is equal to negative 6 plus or minus 12i, because i'll have the square root of minus 140, plus, minus 144, and then all over 2.
03:27
So this is equal to negative 3 plus or minus 6i.
03:36
Okay.
03:37
So therefore, r is equal to 6, negative 3 plus or minus 6i.
03:47
And therefore, the general solution is, so c1, let me call this y for right now, this is not the final answer, but the general solution.
04:09
So this is c1 e to the 6t, and then plus c2 e to the negative 3t cosine of 6t, and then plus c3 e to the negative 3t sine of 6t.
04:32
Let me bring this a little bit lower.
04:47
So this is the general solution.
04:49
And then now we have to find the constants c1, c2, and c3 using these initial conditions.
04:56
So first one, just y of zero, this is simple.
05:01
This is just c1 plus c2.
05:06
This is equal to 11.
05:10
We're told it's equal to 11...