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Current time:0:00Total duration:10:46

Video transcript

welcome back we're finally using the Laplace transform to do something useful in the first part of this problem we just had this fairly straightforward differential equation and I know it's a little bit frustrating right now because like this is such an easy one to solve using the characteristic equation why are we doing Laplace transform so I just want to show you that they can solve even these problems but later on they're going to be classes of problems that frankly are our traditional methods aren't as good as the Laplace transform but anyway how did we solve this we just took the Laplace transform of both sides of this equation we got all of this hairy mess we use the property of the derivatives of functions when you take the Laplace transform and we ended up after doing a lot of algebra essentially we got this we've got the Laplace transform of Y is equal to this thing we just took the Laplace transform of both sides and manipulated algebraically so now our task in this video is to figure out what Y is Laplace transform is this thing and essentially what we're trying to do is we're trying to take the inverse Laplace transform of both sides of this equation so another way to say it if we could say that Y we could take the inverse Laplace transform of both sides we could say that Y is equal to the inverse Laplace transform of this thing 2's plus 13 over s squared plus 5 s plus 6 now we'll eventually actually learn the formal definition of the inverse Laplace transform how do you go from the S domain to the T domain or how do you go from the frequency domain to the time domain we're not going to worry that about that right now what we're going to do is we're trying to we're going to get this into a form that we recognize and say oh I know those functions though that's the Laplace transform of whatever and whatever and then we will know what Y is so let's try to do that so what we're going to use is something that you probably haven't used since algebra 2 which as I think we're when it's taught and you know 8th 9th or 10th grade depending and you finally see it now in differential equations it actually has some use let me write we're going to use partial fraction expansion and I'll do a little primer on that in case you don't remember it so anyway let's just factor the bottom part right here and you'll see where I'm going with this so if i factor the bottom I get s plus 2 times s plus 3 and what we want to do is we want to rewrite this this fraction as the sum of two I guess you could also call it partial fractions I think that's why it's called partial fraction expansion so we want to write this as the sum of a over s plus two plus B over s plus three and if we can do this then and you might already bells my dog to be ringing in your in your head we could we know we know the Laplace tween oh that these things that look like this are the Laplace transform of functions that we've already solved for and I'll do a review on that in a second but anyway how do we figure out a and B well if we were to actually add a and B if we were to let's do a little side right here so if we said that a so if we were to give them a common denominator which is this s plus 2 times s plus 3 then what would a become we'd have to multiply a times s plus 3 right so we'd get a s plus 3 a write this as I've written it right now is the same thing as a over s plus 2 you could you could you could cancel out an S Plus 3 on the top and the bottom and now we're going to add the B to it so plus other than in a different color plus well if we have this as the denominator we could to multiply the numerator in the denominator by s plus 2 right to get B times s plus 2 B and that's going to equal this thing that's going to equal it all I did is I added these two fractions nothing fancier than there that was algebra to actually I think I should do actual video on that as well but that's going to equal this thing 2's plus 13 all of that over s plus 2 times s plus three notice in all differential equations the hairiest parts always the algebra so now what we do is we match up we say well well let's add the S terms here and we could just say that the numerators have to equal each other because the denominators are equal so we have a plus B s a plus B s plus three a plus 2 B is equal to 2's plus B so the coefficient on s on the right hand side is 2 the coefficient on the left hand side is a plus B so we know that a plus B is equal to 2 a plus B is equal to 2 and then on the right hand side we see 3 a plus 2 B must be equal to what oh this is a 13 did I say B this is a 13 it's a 13 they look looks just like a B right that was 2's plus 13 anyway so on the right-hand side I get what is it 3 3 a plus 2 B is equal to 13 now we have two equations with two unknowns and what do we get I know this is very tiresome but it'll be satisfying in the end because you'll actually solve something with a Laplace transform so let's multiply the top equation by two let's let's say minus two so we get minus 2a minus 2b equals minus four and then we get some add the two equations you get a is equal to these cancel out a is equal to nine great if a is equal to nine what is B equal to B is equal to nine plus what is equal to 2 or 2 minus 9 is minus seven and we have done some serious simplification because now we can rewrite this whole expression as the Laplace transform of Y the plus transform of Y is equal to a over s plus 2 is equal to 9 over s plus 2 - seven over s plus three or another way another way of writing it we could write it is equal to nine times one over s plus two minus seven times one over s plus three why did I take the trouble to do this well hopefully you'll recognize this was actually the second Laplace transform we figured out this was the second Laplace transform we figured out what was that I'll write it down here just so you remember it it was the Laplace transform of e to the eighty was equal to one over s minus a that was the second Laplace transform we figured out so this is interesting so if we were to take this is the Laplace transform of what so if we were take the inverse Laplace transformation let me just stay consistent so that means that this is the Laplace transform of Y is equal to nine times the Laplace transform of what if we just do pattern matching if this is s minus a then a is minus two so nine times the Laplace transform of e to the minus 2t does that make sense take this put it in this book which we figured out and you get one over s plus two and let me clear it clean this up a little bit because I'm going to need that real estate let me I'll write this I'll leave that there cuz we will still use that and then we have minus seven times this is the Laplace transform of what this is the Laplace transform of e to the minus 3t e to the minus 3t right it's just pattern matching you're like wow you know this this if you if you saw this you would go to your little plans form table if you didn't remember it you would see this you're like wow that looks a lot like that it's just I just have to figure out what a is I have s plus three I have s minus a so in this case a is equal to minus three so if a is equal to minus three this is the Laplace transform of e to - 3 T so now we can take the inverse Laplace transform II do that we know that this because the Laplace transform is a linear operator and actually now I can delete this down here we know that the Laplace transform is a linear operator so we can write this and you normally wouldn't go through all of these steps I just really want to make you understand that we're what we're doing so we could say that this is the same thing as the Laplace transform of 9 e to the minus 2t - 7 e to the minus 3t now we have something interesting the Laplace transform of Y is equal to the Laplace transform of this well if that's the case then y must be equal to 9 e to the minus 2t minus 7 e to the minus 3t and I never proved to you but the Laplace transform is actually one-to-one transformation that if a function is Laplace transform if I take a function against Laplace transform and then if I were to take the inverse Laplace transform the only function whose Laplace transform it that is is that original function it's not like two different functions can have the same Laplace transform anyway a couple of things to think about here notice we had that thing that kind of looked like a characteristic equation pop up here and there and we still have to fall solve a system of two equations with two unknowns those are both things that we had to do when we when we solve an initial value problem when we use just just traditional the characteristic equation but here it happened all at once and frankly it was a little bit hairier because we have to do all this partial fraction expansion but it's pretty neat the Laplace transform got us something useful in the next video I'll actually do a non-homogeneous equation and show you that the Laplace transform applies equally well there so it's kind of a more consistent theory of solving differential equations instead of kind of guessing solutions and solving for coefficients and all of that see you in the next video