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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 a little joke the Laplace transform of f of t let me just get some notation down we can write that is 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 had to deal with a a the unit step function so if I said to the Laplace transform of the unit step function that 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 C s times the Laplace transform of just this function right there so times the F of s times 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 in 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 the Laplace transform of e to the T times F of T that this is equal to your instead of and I want to make this distinction very clear here we shifted the f of T and we got just a kind of a regular f of s in this situation when we multiply it times the e to the positive 80 we end up shifting 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 gonna have to figure out which one of these to apply let's write all the other stuff that we learn as well the very first thing we learned was that the Laplace transform of 1 was equal to 1 over 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 and as a positive integer it equaled n factorial over 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 let me do it I'll do it right here the Laplace transform of sine of a T is equal to a over s squared plus a squared and the Laplace transform of the cosine of a T 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 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 transform so let's try to do a few so let's say I were to give you the Laplace transform and this is just a 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 transform 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 the Laplace transform our so it looks like n is equal to 3 so if you write the plus transform of T to the 3 this rule that we showed right here this means that it would be equal to three factorial over s to the fourth now this thing isn't exactly this thing they're not quite the same thing and just so 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 in in when I first learned this I would actually go through this step because I was you know you definitely want to make a careless mistake and you definitely want to kind of make sure you have a good handle on what you're doing just like okay 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 right if we call if we call this if we call this expression right here F of s we call this expression right here F of s then what's this expression this expression right here is f of s minus 2 right that expression is f of s minus 2 so what are we dealing with here so you see here you have a shifted F of S right so in this case a a would be equal to 2 so this is the Laplace transform of e to the 80 times our f of T so this is the Laplace transform so let me write this down this is little applause transform of e to the and what's a a is what we shifted by right it's what we shifted by minus a so you have a positive there so e to the 2t times R times the actual function if this was just an F of s what would F of t be well we figured that it was T to the 3 T to the third power so this 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 if that seemed confusing to you you can kind of go forward you can kind of let's apply this let's go to the other direction and maybe this will make it a little bit clearer for you so let's go from this direction if I had to take the Laplace transform of this thing I'd say okay well the Laplace transform of T to the third is easy the Laplace transform of T to the third I'll do it here I think it's but 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 to the third this is an easy one it's equal to three factorial over s to the fourth that's three plus one then it the Laplace transform of e to the 2t times T to 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 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 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 a careless mistake and I think doing a lot of examples probably helps a lot so let's let's do a couple of more to kind of make sure things things really get get hammered home in your brain so let's try this one right here this looks like 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 2 s all of that over s squared minus 2 s plus 2 now this looks very daunting how do you do this have an e here have something shifted here I have this polynomial on the in the in the denominator here what can I do with this so the first thing I when I look at these polynomials in the denominator say can i factor it somehow and I can i factor it some out fairly simply and in actually in the exams that you'll you'll find a differential equation class they'll never give you something that's factorable into you know these weird numbers allow it tends to be integers but you see okay what two numbers they have to be positive when you give their product you get two and then when you add them you get negative two what you can't have or they could both be negative but there's no two easy numbers not one and two none of those work so if you can't factor this out right the next idea is maybe you can complete the square and maybe this will match one of the cosine or the sine formula so how can we complete the square on this denominator well this can be re-written as this could be re-written as let me rewrite this as s squared minus 2 s and I'm going to put a plus two out here and this hopefully if and you can watch my I have a bunch of videos on completing of 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 s the C so something when I add to itself twice becomes minus two and so that when I square it well and what when I add it to itself twice it becomes minus two it's minus one and when I square it'll become plus 1 plus one I can't just add plus one arbitrarily to some expression I have to make it neutral so let me subtract one I haven't changed this I added one and I subtracted one 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 2 s make sure I'm not clipping off at the top e to the minus 2 s all of that over s minus 1 squared plus 1 so a couple of interesting things are starting to seem to be seemed to be happening here so if I just had let's just do a couple of tests laplace transform' so if I took the 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 we actually from the last video we saw well what if we took the Laplace transform the Laplace transform of e to the I'll call it 1 T but let's say e to the Alice right is the what to the 1 T times cosine of T well then this will just shift this Laplace transform by 1 it'll shift it by one to the right so this will just used 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'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 the if you have the Laplace transform of the unit the unit step function of T times some F of T shifted by some value of C then that this is equal to this is equal to e to the minus C s times F of s right ok so if we know and this can get very confusing this can get very confusing so I want to be very careful here let's let's ignore all of this I call this f of s before but now I'm going to backtrack a little bit and let's say just ignore this because I want to I'm going to redefine our F of s 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 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 is this and then our F of s is that right now we have a situation here this expression let's ignore the two here the two we is just kind of a scaling factor we can this expression right here we can rewrite as that expression is equal to if this is our F of s this expression right here is equal to two times our F of s times e to the minus 2's or let me just write let me switch the order just so we make it make it look right 2 times e to the minus 2 s times F of s well that looks just like this if our 2 was equal to rc right if r2 is equal to rc 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 two well this is what we could just write let's 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 the unit step function what's our C if you just you can just pattern match you have a 2 here you have a C minus C minus 2 so C is 2 the unit step function that is 0 until it gets to 2 times T or or of T so you know then it becomes 1 after T is equal to 2 times our function shifted by 2 50 by 2 so this is our inverse Laplace transform now what was our function our function was this thing right here so 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 kind of our big result we find we were established that the inverse Laplace transform of that big thing that I had written before two times s minus one times e to the minus two sorry e to the minus 2 s wouldn't have a T there if we over s squared minus 2 s plus 2 is equal to this thing where F of T is this or we could just rewrite this as 2 2 times the unit step function starting at 2 or that's when it becomes nonzero 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 it was actually a surprisingly good problem I didn't realize it when I first decided to do it we saw this thing we we wanted to get this denominator into some form that is vaguely vaguely useful to us so I completed the square there and then we rewrote our our Laplace transform this our F of s like this and then we use 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 is an 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 we're multiplying our original time domain function we're multiplying our f of T times e to the 1t e to the 1 T and that's what we got there right 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 2 s this entire time and that's where we said hey if we have E minus e to the minus 2's in our Laplace transform then 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 two hanging out the whole time and I could have used at any time but the just simple constants just scale so you can just you know if two times a function is equal to two times the Laplace transform of that function and vice versa so the twos 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 I have if f of s is this and if I'm multiplying it times e to the minus two s 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 two out there the two times the unit step function times my F of T shifted by C and now we established this was our F of T so we just shifted it by C we shifted it by two 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