Cosine of minus t and 2 sine of minus t.

So it would be e to the t minus 2 cosine of t minus 2.

We're T minus 20 seconds.

T minus 15 and counting.

So 1 times t. And v1 is minus 2 times t.

And that would be equal to the integral from 0 to t, of e to the minus t minus tau, sine of t, minus tau, times sine of alpha tau, dtau.

This is when t is equal to a, so minus t, or minus a, which is equal to what?

(t 1) times (z 2 minus z 1) is going to look like this. This is (t minus 1) times (z 2 minus z 1).

So Dirac delta of t minus c.

Delta t minus c and times dt.

This right here would be T times Z 2 minus Z 1. Now what would T minus 1 times Z 2 minus Z 1 look like?

I could say it's cosine of t minus the natural log of t.

Or we could just rewrite this as 2 times the unit step function starting at 2, where that's when it becomes non zero of t times f of t minus 2. f of t minus 2 is this with t being replaced by t minus 2.

So then our y, y, is equal to y of a plus b minus t. a plus b minus is t.

Well T minus 1 if you remember T is between zero and 1. So T is less than 1. So T minus one, this right here is going to be a negative number.

T he tax multiplier is around minus 1.1....

This is f of t minus 2 pi.

Oh sorry, not f of t minus c.

And then we have minus cosine of t.

Well, I would go to t minus 2.

So the Laplace transform of e to the t cosine of t became s minus 1 over s minus 1 squared plus 1.

But we could rewrite this thing right here as being equal to 1 minus T, I'm just factoring out the (1 minus t) times (z 1 minus z 2).

So I t is equal to minus 5 plus or minus the square root.

Or v1 is equal to minus 2v3 is equal to minus 2 times t.

And then finally I have this minus 1 2 sine squared t cosine of t.

This is f of t. e to the minus st times f of t dt.

But then if x is this, then y is going to be equal to I just substitute this back in cosine of t minus ln of t squared plus cosine of t minus ln of t.

So let me rewrite that sine of t minus tau, and we'll just use the trig identity, that the sine of t minus tau is just equal to the sine of t times the cosine of tau minus the sine of tau times the cosine of t.

T minus 31. You're go for auto sequence start.

The left hand side simplifies to t minus 5.

And then over here I have minus 1 2 sine squared t times cosine of t.

We could have written this expression as e to the minus t, sine of t, convoluted with sine of alpha t.

We could just take the integral from t is equal to c to t is equal to infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt.

So this is going to be minus 1 3 times the unit step function, where c is 2 pi of t times instead of sine of t sine of t minus 2pi.

So the Laplace transform of our shifted delta function t minus c times some function f of t, it equals e to the minus c.

I did that a little bit in the last video, but you can start at t equals minus a google or t equals minus infinity.

It's going to be zero everywhere until we get to minus t, and then at minus t, we're going to jump up to some level.

Countdown clock will hold at T minus 20 minutes and...

Let's say that x is equal to t minus c.

Wherever we had a t before, we're going to replace it with a t minus 2 pi.

It's the integral from 0 to t, of sine of the first function of t minus tau.

And then this one over here is minus 1 2 sine squared of t cosine of t.

This is t is equal to minus 3, minus 2, minus 1, 0, 1, 2, and so forth and so on.

That is equal to c1e to the minus st, f of t, plus c2e to the minus st, g of t, and all of that times dt.

And let's say that the transformation matrix for T, so let's say that T of x in standard coordinates, is equal to the matrix 3, 2, minus 2, minus 2, minus 2 times x.