Inverse Laplace examples

A lot of what we do with Laplace transforms, taking them and taking their inverse, it's a lot of pattern matching. And it shouldn't just be a mechanical thing, and that's why I've gone through the exercise of showing you why they work. But in order to just kind of make sure we don't get confused, I think it might be useful to review a little bit of everything that we've learned so far. So in the last video, we saw that the Laplace transform-- well, let me just write something. The Laplace transform of f of t, let me just get some notation down, and we can write that as big capital F of s, and I've told you that before. And so given that, in the last video I showed you that if we have to deal with the unit step function, so if I said, look, the Laplace transform of the unit step function, it becomes 1 at some value c times some shifted function f of t minus c, in the last video, we saw that this is just equal to e to the minus cs times the Laplace transform of just this function right there, so times the F of s. And it's really important not to get this confused with another Laplace transform property or rule, or whatever you want to call it, that we figured out. I think it was one of the videos that I made last year, but if you're just following these in order I think it's three or four videos ago. And that one told us that the Laplace transform of e to the at, times f of t, that this is equal to-- and I want to make this distinction very clear. Here we shifted the f of t and we got just kind of a regular F of s. In this situation, when we multiply it times a e to the positive at, we end up shifting the actual transform. So this becomes F of s minus a. And these two rules, or properties, or whatever you want to call them, they're very easy to confuse with each other. So we're going to do a couple of examples that we're going to have to figure out which one of these two apply. Let's write all the other stuff that we learned as well. The very first thing we learned was that the Laplace transform of 1 was equal to 1/s. We know that that's a pretty straightforward one, easy to prove to yourself. And more generally, we learned that the Laplace transform of t to the n, where n is a positive integer, it equaled n factorial over s to the n plus 1. And then we had our trig functions that we've gone over. Let me do this in a different color. I'll do it right here. The Laplace transform of sine of at is equal to a over s squared, plus a squared. And the Laplace transform of the cosine of at is equal to s over s squared plus a squared. And you'll be amazed by how far we can go with just what I've written here. In future videos, we're going to broaden our toolkit even further, but just these right here, you can already do a whole set of Laplace transforms and inverse Laplace transforms. So let's try to do a few. So let's say I were to give you the Laplace transform. And you know, this is just the hard part. I think you know how to solve a differential equation, if you know how to take the Laplace transforms and go back and forth. The hard part is just recognizing or inverting your Laplace transforms. So let's say we had the Laplace transform of some function F of s. Let's say it's 3 factorial over s minus 2 to the fourth. Now, your pattern matching, or your pattern recognition part of your brain, should immediately say, look, I have a Laplace transform of something that has a factorial in it, and it's over an exponent. This must be something related to this thing right here, right? If I just had the Laplace transform-- let me write that down-- the Laplace transform of-- you see a 3 factorial and a fourth power, so it looks like n is equal to 3. So if you write the Laplace transform of t to the 3, this rule that we showed right here, this means that it would be equal to 3 factorial over s to the fourth. Now, this thing isn't exactly this thing. They're not quite the same thing. You know, I'm doing this to instruct you, but I find these, when I'm actually doing them on an exam-- I remember when I did them when I first learned this, I would actually go through this step because you definitely don't want to make a careless mistake and you definitely want to make sure you have a good handle on what you're doing. So you're like, OK, it's something related to this, but what's the difference between this expression right here and the expression that we're trying to take the inverse Laplace transform of, and this one here? Well, we've shifted our s. If we call this expression right here F of s, then what's this expression? This expression right here is F of s minus 2. So what are we dealing with here? So you see here, you have a shifted F of s. So in this case, a would be equal to 2. So this is the Laplace transform of e to the at times our f of t. So let me write this down. This is the Laplace transform of e to the-- and what's a? a is what we shifted by. It's what we shifted by minus a, so you have a positive a, so e to the 2t times the actual function. If this was just an F of s, what would f of t be? Well, we figured out, it's t the 3, t to the third power. So the Laplace transform of this is equal to that. Or we could write that the inverse Laplace transform of 3 factorial over s minus 2 to the fourth is equal to e to the 2t times t to the third. Now, if that seemed confusing to you, you can kind of go forward. Let's go the other direction, and maybe this will make it a little bit clearer for you. So let's go from this direction. If I have to take the Laplace transform of this thing, I'd say, OK, well, the Laplace transform of t to the third is easy. I think the tool isn't working right there properly. Let me scroll up a little bit. So I could write it right here. So if I wanted to figure out the Laplace transform of e to the 2t times t to the third, I'll say, well, you know, this e to the 2t, I remember that it shifts something. So if I know that the Laplace transform of t the third, this is an easy one. It's equal to 3 factorial over s to the fourth. That's 3 plus 1. Then the Laplace transform of e to the 2t times t the third is going to be this shifted. This is equal to F of s. Then this is going to be f of s minus 2. So what's F of s minus 2? It's going to be equal to 3 factorial over s minus 2 to the fourth. I think you're already getting an appreciation that the hardest thing about these Laplace transform problems are really kind of all of these shifts and kind of recognizing the patterns and recognizing what's your a, and what's your c, and being very careful about it so you don't make a careless mistake. And I think doing a lot of examples probably helps a lot, so let's do a couple of more to kind of make sure things really get hammered home in your brain. So let's try this one right here. This looks a little bit more complicated. They give us that the Laplace transform of some function is equal to 2 times s minus 1 times e to the minus 2s, all of that over s squared minus 2s plus 2. Now this looks very daunting. How do you do this? I have an e here. I have something shifted here. I have this polynomial in the denominator here. What can I do with this? So the first thing, when I look at these polynomials in the denominator, I say can I factor it somehow? Can I factor it fairly simply? And actually, in the exams that you'll find in differential equation class, they'll never give you something that's factorable into these weird numbers. It tends to be integers. So you see, OK, what two numbers? They have to be positive. When you give their product, you get 2. And then when you add them, you get negative 2, or they could both be negative. But there's no two easy numbers, not 1 and 2. None of those work. So if you can't factor this outright, the next idea is maybe we could complete the square and maybe this will match one of the cosine or the sine formulas. So how can we complete the square in this denominator? Well, this can be rewritten as s squared minus 2s. And I'm going to put a plus 2 out here. And you can watch, I have a bunch of videos on the completing of the square, if all of this looks foreign to you. And to complete the square, we just want to turn this into a perfect square. So to turn this into a perfect square-- so something when I add it to itself twice becomes minus 2, and so that when I square it, when I add it to itself twice, it becomes minus 2, it's minus 1. And when I square it, it'll become plus 1. I can't just add plus 1 arbitrarily to some expression, I have to make it neutral. So let me subtract 1. I haven't changed this. I added 1 and I subtracted 1. A little bit of a primer on completing the square. But by doing this, I now can call this expression right here, I can now say that this thing is s minus 1 squared. And then this stuff out here, this out here is 2 minus 1. This is just plus 1. So I can rewrite my entire expression now as 2 times s minus 1 times e to the minus 2s-- make sure I'm not clipping off at the top-- e to the minus 2s, all of that over s minus 1 squared plus 1. So a couple of interesting things seem to be happening here. Let's just do a couple of test Laplace transforms. So if a Laplace transform of cosine of t, we know that this is equal to s over s squared plus 1, which this kind of looks like if this was an s and this was an s squared plus 1. If this was F of s, then what is this? Well, let's ignore this guy right here for a little bit. So what is it? We know, actually, from the last video. We saw, well, what if we took the Laplace transform of e to the-- I'll call it 1t. But let's say e to the-- yeah I'll just write it e to the 1t times cosine of t? Well, then this will just shift this Laplace transform by 1. It will shift it by 1 to the right. Wherever you see an s, you would put an s minus a 1. So this will be equal to s minus 1 over s minus 1 squared plus 1. We're getting close. We now figured out this part right here. Now, in the previous video, I think it was two videos ago, or maybe it was the last video, I forget. Memory fails me. I showed you that if you have the Laplace transform of the unit step function of t times some f of t shifted by some value of c, then that this is equal to e to the minus cs times F of s. OK, And this can get very confusing. This can get very confusing, so I want to be very careful here. Let's ignore all of this. I called this F of s before, but now I'm going to backtrack a little bit. And let's just ignore this, because I'm going to redefine our F of s. So let's just ignore that for a second. Let's define our new f of t to be this. Let's say that that is f of t. Let's say f of t is equal to e to the t cosine of t. Then if you take the Laplace transform of that, that means that F of s is equal to s minus 1 over s minus 1 squared plus 1. Nothing fancy there. I just defined our f of t as this, and then our F of s is that. Now, we have a situation here. Let's ignore the 2 here. The 2 is just kind of a scaling factor. This expression right here, we can rewrite as that expression is equal to-- this is our F of s. This expression right here is equal to 2 times our F of s times e to the minus 2s. Or let me just write it. Let me switch the order, just so we make it look right. 2 times e to the minus 2s times F of s. Well, that looks just like this if our 2 was equal to our c. So what does that tell us? That tells us that the inverse Laplace transform, if we take the inverse Laplace transform-- and let's ignore the 2. Let's do the inverse Laplace transform of the whole thing. The inverse Laplace transform of this thing is going to be equal to-- we can just write the 2 there as a scaling factor, 2 there times this thing times the unit step function. What's our c? You can just pattern match. You have a 2 here. You have a c, a minus c, a minus 2, so c is 2. The unit step function is zero until it gets to 2 times t, or of t, so, then it becomes 1 after t is equal to 2, times our function shifted by 2. So this is our inverse Laplace transform. Now, what was our function? Our function was this thing right here. So if our inverse Laplace transform of that thing that I had written is this thing, an f of t, f of t is equal to e to the t cosine of t. Then our inverse-- let me write all of this down. Let me write our big result. We established that the inverse Laplace transform of that big thing that I had written before, 2 times s minus 1 times e to the minus 2-- sorry, e to the minus 2s over s squared minus 2s plus 2 is equal to this thing where f of t is this. 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. I'll do it in another color, just to ease the monotony. So it would be e to the t minus 2 cosine of t minus 2. Now, you might be thinking, Sal, you know, he must have taken all these baby steps with this problem, because he's trying to explain it to me. But I'm taking baby steps with this problem so that I myself don't get confused. And I think it's essential that you do take these baby steps. And let's just think about what baby steps we took. And I really want to review this. This is actually a surprisingly good problem. I didn't realize it when I first decided to do it. We solved this thing. We wanted to get this denominator into some form that is vaguely useful to us, so I completed the square there and then we rewrote our Laplace transform, our f of s like this. And then we used a little pattern recognition. We said, look, if I take the Laplace transform of cosine of t, I'd get s over s squared plus 1. But this isn't s over s squared plus 1. It's s minus 1 over s minus 1 squared plus 1. So we said, oh, well, that means that we're multiplying our original time domain function. We're multiplying our f of t times e to the 1t. And that's what we got there. So the Laplace transform of e to the t cosine of t became s minus 1 over s minus 1 squared plus 1. And then we had this e to the minus 2s this entire time. And that's where we said, hey, if we have e to the minus 2s in our Laplace transform, when you take the inverse Laplace transform, it must be the step function times the shifted version of that function. And that's why I was very careful. And you had this 2 hanging out the whole time, and I could have used that any time. But the simple constants just scale. A function is equal to two times the Laplace transform of that function and vice versa. So the 2's are very easy to deal with, so I kind of ignored that most of the time. But that's why I was very careful. I redefined f of t to be this, F of s to be this, and said, gee, if F of s is this, and if I'm multiplying it times e to the minus 2s, then what I'm essentially doing, I'm fitting this pattern right here. And so the answer to my problem is going to be the unit step function-- I just throw the 2 out there-- the 2 times the unit step function times my f of t shifted by c. And we established this was our f of t, so we just shifted it by c. We shifted it by 2, and we got our final answer. So this is about as hard up to this point as you'll see an inverse Laplace transform problem. So, hopefully, you found that pretty interesting.