 I'm going to show you something quite neat. It's called operational properties. So I might know the Laplace transform of the f of t. But what if I multiply that f of t by e to the power at? So I can have, well, we'll look at an example after this. Instead of doing, I mean, it can become very onerous to do these Laplace transforms. There's a simple way, though. And if you multiply f of t by e to the power at, that becomes the f of not s, but s minus a. It's as simple as that. It's as simple as that. So if I were to ask the Laplace transform of e to the power 3t times, for instance, t to the power 3. t times t to the power 3, p cubed. What would that be? Well, first of all, I note that a equals 3 here, a equals 3. And I know what the Laplace transform is of the f of t here, which is the Laplace transform of t to the power 3. And here I have n equals 3. So for me, this is going to be 3 factorial divided by s to the power n plus 1. So that's 4. So I'm going to have 6 over s to the power 4. OK, so I have the Laplace transform of the f of t being 6 over s to the power 4. And I have this a equal to 3. So everywhere where I have the s, everywhere where I have s, I just have to replace that with s minus 4. I just have to replace every s f of s that was there for s. Now s minus a s minus 4. So the Laplace transform of this whole thing is now going to be 6 over. And in place of s, I'm going to have s minus 4. So a was 3. So you'll be there, drifting, a was 3, s minus 3 to the power 4. So just instead of that s, I now have s minus a, which is s minus 3 to the power 4. If I had to do this the long way, remember that's going to be this improper integral of e to the power negative s t times e to the power 3 t times t to the power 3 dt. Now at least with these, it's quite easy because I can realize this is the improper integral of e to the power s minus 3 t times t to the power 3 dt. I can have that, which I'll now have to run through various, sitting that equal to my u and this part to my v prime with the product rule. Then I'll have a t squared. Then I'll have a t to the power 1. Eventually t to the power 0. So I'll have this long pages and pages full of doing this improper integral. Instead, I can just use this property. If I multiply f of t by e to the power 80, I can just do my normal Laplace transform of the f of t. And wherever I see s, I just substitute it with s minus a, which is here was s minus 3.