03 - Deriving the Essential Laplace Transforms, Part 2

hello I'm Jason welcome to this lesson in the Laplace transform tutor here we're going to learn how to take the Laplace transform of the cosine function and also the sine function they're gonna look kind of similar what we'll do is we'll apply the full-blown integral to the cosine and we'll do that solution and then along the way we'll kind of pick up the sine function as well I don't think I have to explain to you how important sine and cosine are they're gonna pop up and almost on lots of lots of problems that you do but what we're gonna do is derive it straight through the integral gives you practice with doing that also sharpens or calculus skills now I am gonna say that it's unlikely that you would probably see exactly how to do this unless you see it once sometimes you know derivations you do them and you're like oh that makes sense but if you had just stared at this yourself it you it may occur to you what to do and it may not so what I'm gonna do is go through the integral assuming that you don't know anything and just explain everything step by step all right so let's say you wanted to find the Laplace transform of cosine beta times T now this in your book could be anything could C cosine a T cosine BT cosine whatever but there's a constant in front of the T sometimes you might see it as Omega if you're dealing with electrical engineering there's usually a cosine of a number in front of the T it's gonna end up being the frequency of the cosine that you're talking about but it's it's a function of time and this beta is just a constant that's the point to realize so what we're going to do is we're going to apply the definition 0 to infinity e to the minus s T cosine beta T DT so again if you know how to integrate this then you're done that's bill applause transform that's what we're gonna do what we need to do though is do integration by parts and it actually turns out if you remember back to calculus sometimes you have to do integration by parts more than one time you've done those kinds of problems back in calculus and it's gonna happen in this problem I'm gonna warn you so we're gonna fill up the board doing integration by parts once then we're gonna fill up a board doing it again I know it's a little bit to get through but these are kinds of questions that if I were professor I might ask you just as a bonus question or something because it really shows me that you know your cow Kilis now doing these kinds of integrals in such a detail or not something you have to do for every Laplace problem but you do have to do it here so let's go ahead and set up our integration by parts so we'll say U is equal to whatever the exponent is e to minus s T DV we'll just say that's the rest of the interval cosine beta TD T and from this you need to derive other pieces of information that will plug in so for this let me write it in blue for this do you what is the derivative of this with respect to time it's minus s e to the minus s T the T right think of it D u DT this is what you would take the derivative the derivative of the exponent comes out and then here we have an indestructible guy right there and then whenever you're calculating what V is you're integrating this the integral of cosine is going to be a sine but the 1 over beta is going to come out and you'll have sine beta T I'm not showing a lot of these steps because what's really happening is basically to get this to get V you're integrating both sides you're integrating this to give you B and you're integrating this to give you this I think a lot of you guys have had enough calculus now you can integrate a real simple cosine function to give you a sign and what comes in here comes out as 1 over beta so that's the set up for the integration by parts and now we have to actually pull it off so we say u v minus the integral of V D u u is going to be this so we'll have e to the minus s T V is gonna be this so we'll have 1 over beta sine beta T and then we go from 0 to infinity because when we do the first part we have to evaluate at those limits of integration right there and then we have minus the integral again from 0 to infinity V D u so V is 1 over beta sine beta T that's V D U is this guy here so I'm going to put a parenthesis so we don't get confused it's minus s e to the minus s T DT if I didn't put the parentheses you might think these were subtracted but these are really multiplied together so it's you V minus integral mean to you all right it looks really ugly and in truth it's it's really not that hard but you do have to be cognizant of getting the signs and everything right so what I'm going to do here is evaluate this at infinity and zero so I open a bracket and then I say all right at infinity I'm gonna have the 1 over beta and we just put out in front and then I'll have e to the negative infinity plugging it in here and then sine of infinity when I put infinity inside of that right and then I subtract off from that putting in zero so I'll have again 1 over beta from this and then this would be e to the 0 and this would be sine of 0 don't try to do too many things at once don't try to evaluate all this stuff as you go in one step write it down and then in the next step evaluate and you eliminate your sign errors that way now what's going to come of this here we have a minus sign but we have a minus sign here so that can be pulled out and may to be a plus sign what's a constant here beta is a constant s it acts as a constant because we're integrating over time we can't pull out the sign because it has a time and we can't pull out this because it also has a time but we can pull out s over beta integral 0 to infinity and just to rewrite things a little bit I'll put the exponent first e to the minus s T and then I'll put the sign sine beta T D T so what we have done is we have come down and done our initial pass through the integration by parts so let's go in here what am I going to have here well everything is trumped by this because the sine infinity is kind of impossible to really evaluate but here everything is divided by e to the Infinity so this whole thing goes to 0 here sine of 0 is 0 this actually goes to 1 e to the 0 goes to 1 but sine of 0 goes to 0 so this whole thing goes to 0 so again everything in front is nothing so this first pass through the integration by parts gives us this now this is not the answer but I want to point out to you something and I'm gonna do it right here it's not really a great place to do it this is the Laplace transform of sine beta T just keep that in mind because we're gonna use that fact later in other words the Laplace transform of a cosine gives you some junk out in front times the Laplace transform of the sine function notice beta still in there too so basically what happens is when you take a little applies transform of a cosine you get some stuff times Laplace transform of a sine we're gonna use that fact later to derive what this is I'm just pointing it out to you now so keep that in mind so what we want to do let me kind of continue on the next board let me rewrite this so far what I will do is I'll say that the Laplace transform of cosine beta times T is equal to s over beta and then I have an integral 0 to infinity e to the minus s T times the sine beta T DT now here's the part that you may or may not see it's not something that you were just born with and you would just know what to do it's just that sometimes when you do these kinds of problems you realize that here I started with a cosine and then I ended up with a Laplace of a sine all right and then you're stuck again because you're like well this is gonna lead to nowhere because I can do this again but it doesn't seem to converge on something like an answer because this integral looks to be just as complicated as the first one I started with but then you start thinking back to your intuition when you learned about integration by parts and since you have this cosine that ended up transforming into a sine there's a pretty good chance that if you do integration by parts again on this that this sine will flip flop back into a cosine and since this cosine beta T changed into a sine beta T there's a good chance that if I do it again this sine beta T will change back into a cosine beta T and then I might be able to solve the problem you may not be able to see the exact path to the answer but you can see that sometimes when you do integration by parts the the complexity of the integral remains the same but with these sinusoidal functions they just flip back and forth back and forth back and forth if you do it over and over again so that's the intuition we're going to run with and we're going to do integration by parts on this integral so we'll have to again define au which would be e to the minus st and a DV which is sine of beta T DT basically exactly the same or very very similar to what we have done the first time so then I'm gonna switch colors again and I'm gonna say alright then D U is negative s e to the minus s T DT same derivative as before really and then I have to solve for V so I'm gonna integrate both sides and then I'm going to basically say well what's the integral of sine it's a negative cosine but I have this beta here so it's a negative one over beta cosine theta T so when you integrate sine you get negative cosine that's where the negative comes from that's what the cosine comes from but one of her beta happens because if you were to do this in a calculus one kind of way you would do a substitution here and you would do a derivative and you would find that the beta ends up coming out as one over that so that's basically what you get for V and that's what you get for you and that's what you get for all of these other things so let's go and apply the integration by parts method so I'll draw on line there UV minus integral V D u so U is e to the minus s T V is 1 over beta times cosine beta T u times V I have to evaluate it from 0 to infinity minus the integral 0 to infinity VD u negative 1 over beta cosine beta T V and then I have D u which is this and I'll open up a parenthesis to make it clear s e to the minus s T DT u times V minus integral V D u that's what you're basically doing here and so I'll switch colors back to black let's evaluate this at the limits of integration so over here and infinity the 1 over beta let's carry that long in front then I'll have e to the negative infinity from plugging infinity in here and then cosign of infinity from putting cosign into the T right here and then I will subtract off plugging in the zero so again I'll have negative 1 over beta here and then it will be e to the 0 and they'll be cosine of zero okay cosine of zero like this so that's this and we'll evaluate what that really boils down to here in just a minute let's turn our attention to the second part so here I have a negative a negative and a negative and don't forget this is actually multiplied together inside of here so what's gonna happen is I'm still gonna have a negative sign here because these will make a positive this one will remain negative I'm integrating over time so this s is acting as a constant and this beta is acting as a constant just like before so I'll pull out s over beta and then the integral sign remains from 0 to infinity and then e to the minus st comes in and then cosine beta t d t and look exactly what we suspected happen to happen we did the first one and it changed into a sine we did the thing again and it changed into a cosine and believe it or not this really helps the situation because let's evaluate the limits of integration right here e to the negative infinity is going to drive this whole thing to zero no matter what else is there all right when you see e to the negative infinity it's going to drive to zero when we have this e to the zero gives you one cosine of zero also gives you one so actually this whole thing including the negative sign out in front is going to give you one over beta like that one over beta so we have negative times negative gives me a positive one over beta this becomes a 1 so that's what this whole thing is gonna be including the negative sign up front is gonna give you that guy right there and then we see that this integral transforms back again so at first glance you might look at that and say the integral doesn't get any simpler how am i possibly ever supposed to solve this until you realize that you can do the following a little thing and I know that you did learn this in calculus but you probably don't use it all that much let's go ahead and do that right here let's take it from the top all right let's go back up to here the Laplace transform of the cosine all right applies transform of the cosine let's write it down here Laplace transform of the cosine of beta times T was equal notice there's an S over beta times the result of this integral which I just did so I have to carry the S over beta down times the result of what I have in here now the result of everything down below is what I've gotten right here so let me open a bracket and say that what I get here is 1 over beta minus s over beta now instead of writing this as an integral just realize that this is the Laplace transform of cosine beta T so I'll just write this as Laplace transform of cosine beta times T and then you can kind of see how it might be okay because what's happening here is I've got Laplace transform of cosine beta T on the left and I've got Laplace transform of cosine beta T on the right so I can treat it like any equation I can solve the Laplace transform of cosine beta T and that's gonna give us the answer all right so what you're gonna end up having is on the right-hand side you're gonna have when you just distribute this guy in like this what you're gonna have is s over beta squared minus s squared over beta times the Laplace transform of cosine beta T like this and on the left you'll have Laplace transform cosine beta T just like that and there's one square right here because when I multiply this then I get s over beta squared when I multiply this into here I get s squared over beta squared the minus stays here everything stays here so let me move over to the next board and I'm not gonna try to continue yet until I just recopy everything that I have so Laplace transform cosine beta T is equal to s over beta squared s s squared over beta squared applause transform cosine beta T like that all right so now what we want to do is we want to move this whole term over to the right-hand side so what we'll have is Laplace transform cosine beta t plus s squared over beta squared plus transform cosine beta t equals s over beta squared so we just move that term over let me switch colors now you can treat this as a variable it's like X and X right so we can factor it out so let's pull it out and we'll say Laplace transform cosine beta T and then on the inside here we have 1 plus s squared over beta squared and on the right hand side we have s over beta squared so now you can kind of see what we're going because here I just divided by this term here and so what I'm gonna get is Laplace transform of cosine beta T is equal to s over beta squared over 1 plus s squared over beta square like that so basically you take this thing and you divide the whole thing if it comes on the denominator and then you realize that because of the fact that this is a fraction you can further write this as s over the beta squared can go in the bottom and then this becomes s squared over beta squared like that so it just basically becomes on the bottom and so what you have then is you can multiply through and you can say s over beta squared plus the beta squared will cancel with this one when I distribute in giving me s squared so the way you usually see this thing written the way you usually see this thing written is Laplace transform of cosine some number times time is equal to s over usually write the s squared plus beta squared and then you just say s greater than zero this is the Laplace transform of the cosine function notice that notice that it's a function of s only beta is just a number so let's say it's cosine 3t well then it will be s over s squared plus 3 squared so it would be 9 over there so that's what we get there that's how we declare the Laplace transform of the cosine function now how do we find the Laplace transform of the sine function how do we find a little boss transform of the sine function I could just give it to you but it's too tempting just to show you and remind you that way back is we were starting to do this whole process way back in the first Laplace transform we did we got the Laplace transform of the cosine was basically equal to this stuff times this integral which we said was the Laplace transform of the sine so what I will do then is I will kind of do a little aside let me see what color I should probably pick black so let me kind of draw double squiggly lines here because this is kind of a separate thing I'll do triple squiggly lines right so what we learned is the Laplace transform of cosine of beta times T was equal after all of this first integration by parts s over beta times the Laplace transform of the sine of beta times T but see now we know a lot more we know what this Laplace transform is we just calculated it the Laplace transform of cosine is s over s squared plus beta squared so all we do is we say s over s squared plus beta squared is equal to s over beta times the Laplace transform of the sine of beta times T and we just want to solve for this that's what we want to solve for and we know we can see right away that since we have s on the top in both cases these are going to cancel and then - in order to get this by itself we can just multiply by beta on both sides and so I can directly write then with the Laplace transform of sine of beta times T once we multiplied by beta to get it over there is gonna be beta okay over s squared plus beta squared and we can just say s greater than zero just like always like that this is the Laplace transform of the sine function beta over s squared plus beta squared it's a lot of work I'm not gonna lie to you and you know you could be asked this on a test or an exam mostly I did this because when you try to learn how this is derived in a book is very difficult to follow unless you jump through all the hoops because you know they leave out lots of critical steps like what is he doing how's he solving for it but basically what it boils down to is it's a little bit of a trick you have to just kind of have some intuition you have to realize that integration by parts will be useful here you set your setup self up for the first integration by parts and when you do it all everything in front drops to zero and you end up with this integral here and you notice that the cosine flipped into a sign in your next integral so then you decide to do it again and you say alright I'm gonna do that again I'm gonna do integration by parts again setting this almost the same substitutions and I'm hoping that that integral will flip back so I get this stuff in the front I evaluate it here and then it turns out that this interval does indeed flip back into a cosine which is what I get right here all right so because of that then what's basically happening here this is basically drops to zero this resolves to a number so then I can write kind of a summary of what I've done the Laplace transform is equal to s over B times whatever I got here which is what I've written down here I've got the the 1 over beta here and I've got minus this and then this guy turns into the Laplace transform of beta T so basically what it happens is whenever you do integration by parts and then do it a second time and the integral on the right hand side flips back to be equal to the integral that you were trying to solve to begin with then bingo you've won when you get an integral on the right which is the same as the integral you were looking for on the left and you can just solve the whole darn thing for that integral almost like algebra right so that's what we're doing here this integrals will we're trying to find and we find out that it's related to itself so then what we do on the next basically the next couple of steps here is we're just doing that stuff we first multiply it through we add this whole term on the left hand side factor out the Laplace transform giving us this junk over here and then we divide this whole term it gets a little bit ugly we bring the beta squared down we distribute it in and then we find out that this is the Laplace transform of cosine beta T so that's basically the answer and then we go and say well remember from before when we got down to this part this is the Laplace transform of the side so we simply summarize by saying Laplace transform of cosine is s over B times Laplace transform of sine we plug in what we know for cosine and then basically solve everything for the Laplace transform of sine so again a lot of books in a lot of classes your professor or your book would probably just say hey guys Laplace transform of a sine is equal to this if it's sine of 3t it's just three over s squared plus three squared right if you're doing sine and the Laplace transform of cosine looks sort of kind of similar but it's a little bit different and you basically apply at the same sort of way you're getting a function of s out in both cases they look kind of similar but they're not quite the same a lot of your professors a lot of your classes a lot of your books are just gonna say use me and that's fine that's totally fine but sometimes I want to show you where things come from because it builds your calculus skills get you comfortable with that Laplace integral and then you can kind of see that the stuff doesn't come out of thin air there is some some rigor behind it and it's not something you can do you know in five minutes but it's something that's understandable all right so we're almost to the end of our road here as far as doing the derivations I wanted to do that because I wanted to show you where this stuff comes from but we have enough essential Laplace transforms derived so that what we're gonna do in the next lesson is we're gonna write them all down the board almost like a table of integrals you know we use table of integrals back in calculus we'll have the table of Laplace transforms for you to understand and I don't want to say memorize but become really comfortable with and what's gonna happen is we're gonna use those Laplace transforms a lot over and over again to solve real problems and you'll find out that now you know how to take the Laplace transform of the sine the cosine and exponential a constant which is Laplace transform of 1 and also the Laplace transform of a power like T to the enth or T to the fifth or teeth of the fourth then you have almost all the essential bases covered to do a lot of practical problems so that's the path that we're going down I know the this course was front loaded with a lot of heavy calculus but I wanted to show you how to do it because I think that's what's missing from in a lot of books so follow me on to the next lesson we'll summarize our results in a table and then we'll begin using these Laplace transforms to do real problems in solving real differential equations and learning how Laplace transform is powerful and useful

Channel: Math and Science

Views: 66,181

Keywords: laplace, laplace transform, laplacian, differential equations, laplace equation, calculus, engineering math, engineering, laplace differential equation, differential, transform, deriving, transforms, math, equation, laplace transform from first principle, laplace from first principle, laplace transform using first principle, laplace first principles, education, math laplace transform, calculus laplace transform, laplace transform rules, laplace transform problems

Id: mwWjCNITvc8

Length: 25min 1sec (1501 seconds)

Published: Sat Aug 12 2017