00:01
All right, let's take a look at the difference equation in red for the spring mass system with the initial conditions given in blue.
00:09
So basically, to solve this, we're going to first look at the homogeneous case, and that's when basically we'll set the right side equal to zero.
00:19
We're going to let x be e to the mt so that if we take the first derivative, we get e, m, e to the mt, and the second derivative, we get m, e to the mt, and the second derivative, we get m, squared, e to the mt.
00:34
Then i'm going to go ahead and substitute these values into my homogeneous equation.
00:39
That will give me m squared, e to the mt plus nine, e to the mt.
00:47
And we set that equal to zero.
00:50
Notice i can factor out e to the mt, leaving behind m squared, m squared plus nine equal to 0.
00:57
E of the mt never goes to 0, but m squared plus 9.
01:01
That then i'll set equal to 0.
01:03
And this is known as my characteristic equation.
01:14
And there we go, characteristic equation.
01:18
Okay, so if i subtract 9 from both sides and square root, you can see that m will be plus or minus 3i.
01:26
Now just as a reminder, when i have complex roots like this, the general form of my homogeneous solution is equal to e.
01:35
If i have complex roots that are like a plus or minus b i, then the general form is e to the a t, c1, cosine, b, t plus c2 sine of b t.
01:51
My a is zero.
01:53
So for us, the homogeneous equation is equal to e to the zero, which is just one.
01:59
So we just get c1 cosine of 3t plus c2 sign of 3t.
02:08
After we solve for a particular solution, then we can actually solve also for these coefficients.
02:15
But anyway, we're not done yet.
02:16
We still need to get the particular solution with this exponential driving force.
02:21
So that's our next steps.
02:23
Let me clear enough of the board so that we have room to finish off.
02:28
And well, actually, you know what? i'm well, yeah, i'll clear up just enough.
02:35
All right.
02:36
So now for our particular solution, we're going to use a method of undetermined coefficient.
02:42
So method of undetermined coefficients.
02:49
And basically how this works is we're going to guess a solution very similar to what we see here.
02:56
And we check for repeat routes, which we don't.
02:59
If we look our homogeneous solution, they don't look the same at all.
03:03
So therefore, i'm going to guess that my particular solution is just a, e to the minus 4t.
03:11
Then i take the derivatives.
03:13
First derivative is minus 4ae to the minus 4t.
03:19
And the second derivative, oh, that's kind of getting, for some reason, my pen is acting funny.
03:27
Okay, yp double prime is equal to then, 16ae to the minus 4t with a bit of chain roll there.
03:35
All right.
03:35
So now we're going to substitute into our differential equation.
03:38
So yp double prime goes in and i get 16a, e to the minus 4t.
03:43
And then i do nine times y particular.
03:46
So 9a, e to the minus 4t.
03:50
And this has to equal 75e to the minus 4t.
03:54
So you can see that if i add the left side together, i get 25a, e to the minus 4t.
04:00
Minus 4t equal 75e to the minus 4t.
04:05
So if i match up terms, then 25a equals 75.
04:11
If i divide both sides by 25, a then is 3.
04:15
So now i have my particular solution.
04:18
And so why particular is 3e to the minus 4t.
04:24
Okay, excellent.
04:26
So let me make a little room and we'll solve for our coefficient...