WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and today we are going to be looking at Laplace transforms and using them to solve initial value problems.
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Let me show you how that works out, the idea is that you will be given an initial value problem of this swarm (a)Y″ + (b)Y′ + (c)Y=g(t), then you will also have two initial values.
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y(0) is equal to y₀ and Y′ 0 is equal to y₀′ we have two initial values and you have the actual differential equation and the idea is that you take the Laplace transform of both sides of the differential equation.
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In particular we are just looking at this part right now, we are going to take the Laplace transform of both sides, we are going to take the Laplace transform of (a)Y″ + (b)Y′ +cy.
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Remember that the Laplace transform is linear, that splits up into a times the Laplace transform of Y″, b times Laplace transform of Y′ and c times Laplace transform of y.
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Over on the right hand side, we had g(t) in the original differential equation, we get the Laplace transform of g(t) on the right hand side.
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Let me show you how we use that, the idea is that we know what L(Y″) is, this is something that you can work out if you just fill around the Laplace transform a couple times.
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It turns out to be s² x L(y) - s x y(0) - Y′ (0) and then we have a similar expression for L(Y′), it turns into S x L (y) - y(0), remember the y(0) and Y′(0) those are all quantities that we know from the initial conditions.
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For each of these we can plug in numbers that we get from the initial conditions, we write that down numbers from initial conditions, those will all be given to us in the original problem.
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We are going to plug in L(Y″) L(Y′) in by using these two identities and will get an equation that we can solve for L(y), I will end up with L(y) is equal to blah- blah -- some kind of function that will be in terms of x.
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Then we are going to take the inverse Laplace transform and start with this function of s and go backwards to find y which will be a function of t and that will be our complete solution to the initial value problem.
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That is the way it is going to work, let me mention one step here that is very important taking the inverse Laplace transform, that is something that we learned about in the previous lecture here on www.educator.com.
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That is the lecture immediately preceding this one in the differential equations lectures series, if you do not remember that, if you did not watch that lecture recently.
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You really want to go back and work through that lecture before you do this lecture because you want to be very comfortable with the inverse Laplace transform before we start using it to solve the the initial value problems.
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In particular the problems that I'm going to solve today are closely tied to the problems that we used in that previous lecture, every example that we study today was a problem that we already used in the previous lecture to study the inverse Laplace transform.
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We are going to be using the answers from that previous lecture as part of our work today, if you have not looked at that previous lecture recently, you might want to go back and look at that.
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In particular check out the answers to each of those problems because we are going to be using each one of those problems to solve our problems today on initial value problems.
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Let us go ahead and see some examples of this, we are going to solve the following initial value problem Y″ + Y′ -2y=0 and then we have 2 initial conditions y(0) is equal to 7 and Y′(0) is equal to -2.
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Let me remind you of the two identities that we had at the beginning of this lecture, we had L(Y″) this is something we learned at the beginning of the lecture is equal to s² L(y) - S x y(0) - Y′(0).
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That was something from the beginning of lecture and we also had one that was relevant to L(y′) which was s x L(y) - y(0) what we are going to do with each of these is plug in the initial conditions that we are given.
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The y(0) and Y′(0), for this first one we have s² L(y) - s x y(0) but y(0) is given to us to be 7, this is -7s - Y′(0) , Y′(0) -2, this is +2 and then over down here with L(y′) we have s times Laplace transform of y-y(0).
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That is -7 and we are going to go to this differential equation and we are going to take the Laplace transform of both sides and what we get on the left is L(Y″).
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I'm going to plug in what I have for L(Y″) that is s² L(y) - 7s + 2 + L(Y′) that is s x L(y) -7, this part all came from L(y″), this part all came from L(Y′) and -2y.
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That is-2 times L(y) is equal to L(0), that is just zero, what we have here is an equation now and you want to think of the variable is being L(y), we are going to try to solve for L(y).
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I'm going to factor out all the terms I can from L(y) I see I have s² times, let me see what other terms I have before I start writing down the L(y).
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s², I see have s², I have a plus S and then I have a -2, that takes care of that term and that term and that term, all those multiplied by L(y).
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I'm going to collect my other terms, I see I have a -7s +2 - 7 is -5 is equal to 0 and the goal here is to solve for L(y), I'm going to move the 7s -5 over to the other side.
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I have s² + s -2 x L(y) is equal to now moving them over to the other side and I get +7s +5 and finally I can get L(y) is equal to 7s + 5/s² + s - 2.
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What I need to do now is to do the inverse Laplace transform to figure out what the original y was, now this is exactly what we did in the answer to example 1 in the previous lecture.
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I'm not going to go through the details of that because that was quite a bit of work back then we are going to take the inverse Laplace transform and this is exactly the one that we did for example 1 of the previous lecture.
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I'm just going to give you a reference to that example 1 in the previous lecture, you can go back and check that out and see how the arithmetic worked out there and what you will find is that we derive y=3e⁻2t + 4e^.
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That is our solution to the original initial value problem that we we are given, 3e⁻2t + 4e⁺t, let me go back and recap what we did there, what we started out by doing was taking the Laplace transform of both sides.
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But then that give us an L(y″) on the left and we had an identity for that at the beginning of this lecture which was s² L(y) - minus sY(0) - Y′(0) and y(0) zero Y′ (0) those are the initial conditions given to us .
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I plug in those numbers here to get -7s + 2 and we also had L(Y′) we had an identity for that at the beginning of lecture which is sL(y) - y(0).
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Again I plug in y(0) given to me is the initial condition is -7, I took each of these two expressions, this one and this one, I plugged them in for Y′ and Y″ here and so I get s² L(y) - 7s + 2 , sL(Y -7)
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This 2y gave us 2L(y) is equal to zero and what l want to do is think of L(y) as the variable and I want to solve for that variable.
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I found all my terms that were multiplied by L(y) and that was s² + s -2, found all my other terms separated those out, move them over to the other side and then I divided by the coefficient of L(y) and I got an expression for L(y).
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This point out have to take the inverse Laplace transform this is exactly what we did in the previous lecture, if we go back and look at example 1 from the previous lecture.
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You will see the arithmetic for that where we did partial fractions on this expression that we looked back at our Laplace transform chart.
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What we end up with from the previous lecture is that y=3e⁻2t + 4e⁺t, that is our answer for example 1, let us go ahead look at another one.
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We had the initial value problem Y″ +3Y′ equal zero and then some initial values Y(0) is equal to -2 and Y′(0) is equal to three, we are going to start out just like we did before.
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We are going to take the Laplace transform of both sides of the differential equation but we are going to need to know what L(Y′) and L(Y″) are.
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We have those identities from the beginning of the lecture L(Y″) is the same identity as before it would be the same every time.
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It is s² L(y) - s x y(0) - Y′(0), that is the same every time and now we fill in what the initial conditions tell us about y(0) and Y′(0), we got s² L(y) still nothing I can do about that yet, minus s x y(0) that is + 2 x y(0) term plus 2s - Y′(0) is - 3 there.
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Then L(Y′) again using the identity from the beginning of this lecture is s times L(y) - y(0), I will fill in what I know about y(0); well that is -2.
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Subtracting that will give me a + 2 and I'm going to plug both of those back into the differential equation, I will get s² L(y) + 2s -3, that came from my L(y)″.
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Now +3 Y′, I'm going to multiply this by 3, +3s L(y) + 6, that came from calculating 3L(y)′ is equal to zero and remember you want to think of L(y) being the variable and we are going to try to solve for L(y) .
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I'm going to final my terms of L(y), I see I have an s² + 3s x L(y) and now I'm just going to look at all the other terms, I see I have 2s - 3 and a +6, that is plus 2s + 3 is equal to zero.
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I will move that over to the other side, minus 2s -3 over on the other side and now if I solve for L(y) that means I have to divide both sides by s² + 3s.
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I get L(y) - 2s -3/s² + 3s, I need to take the inverse Laplace transform of this and we learn how to take inverse Laplace transforms in the previous lecture here on the differential equations lectures series on www.educator.com.
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I will take the inverse Laplace transform and we did this one as one of the examples in the previous lecture this was example 2 in the previous lecture.
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You can check back at the previous lecture and you will see the arithmetic that we went through to take the inverse Laplace transform for this example.
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You will see what we figured out is that y itself is equal to -1 - e⁻3t, that is our solution to the initial value problem.
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Let me go back over that and just remind you what the steps were, we took the Laplace transform of both sides L(y)″ + 3(Y′).
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We use the identity that we learned in the beginning of this lecture L(Y ″) is equal to s² L(y) - sY(0) - Y′(0). that identity always holds it very safe, you are going to use that same identity in every problem.
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What you do with that is you plug in the initial conditions that you are given for y(0) and Y′(0), that gave me s² L(y), y(0) is -2, we got a plus 2s here and then Y′(0) is 3, we got -3 there.
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L(y) we are going to use that because of this L(Y′) is equal to sL(y) - y(0), again that is the identity from the beginning of this lecture and what we can fill in is that y(0) is 2.
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We plug both those back into the differential equation here and here, the L(Y′) gets multiplied by 3, that is why we got that 3 and the 2 turn into a 6 and you want to think of L(y) being the variable.
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We segregate all the terms that have an L(y) in them and I see that I have an s² L(y) and 3sL(y) and then all the extra terms were 2s - 3, that - 3 in that 6 combined to give me a +3 here.
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I wanted to move the 2s +3 over to the other side because I was trying to solve for the L(y), I moved those over the other side, they turn into negatives and I got L(y) equal to minus -2s -3/s² + 3s.
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Now it is an inverse Laplace transform problem, in fact it is the exact one that we solved in example 2 on the previous lecture, go back and check that out if you have not seen that for a while.
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What we figured out is that, that corresponds to the original function y(-1) - e⁻3t, that is our answer to the initial value problem.
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In example 3 we got another initial value problem Y″ + 4Y′ + 5y=0 and a couple of initial values y(0)=1 and Y′(0)=0.
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We are going to work it out exactly the same way as the first two examples using the same identities that we learned back in the beginning of lecture.
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L(Y″) is equal to s² L(y) - S x y(0) - Y′(0) and again I'm going to fill in the initial conditions for y(0) and Y′(0), this is s² L(y), now s x y(0), y(0) is 1, this is just -s.
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Y′(0) is 0, there is just a zero for that term there, that will just drop right out, next we have L(Y′) and again our identity from the beginning of lecture this never changes, it is always the same s x L(y) - y(0).
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In this case, if that is just s x L(y) and y(0) we are given that as an initial condition that is 1.
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Now I'm going to take each of these and plug them into the differential equation, that means I have to multiply L(Y′) by 4, I get s² L(y) - s, because I already simplified it down to s² L(y) - s and now I have 4 Y′.
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I'm going to put + 4s L(y) - 4, I multiply everything here by 4 to get that term right there, +5y, +5 times the Laplace transform of y is equal to 0 and remember L(y) you want to think of that as your variable, you want to solve for that variable.
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Let us see what terms I have multiplied by L(y), I on the left I have s² L(y), I see a 4s L(y) and I see a 5 L(y) and let me see what other terms I have.
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I see just -s and -4, -s -4 is equal to zero and if I move that over to the other side, on the other side I will get an s + 4 and my L(y) if I divide both sides by the coefficient there would s + 4/s² + 4s + 5.
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That is my Laplace transform of the y that I'm looking for and to solve that out I need to take the inverse Laplace transform.
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I synched up all these examples to the examples in the previous lecture on inverse Laplace transforms, this was exactly the function that we used in example 3 of the previous lecture on inverse Laplace transforms.
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In example 3, in previous lecture which is where we solve this one out and what you will see is we figured out that the inverse Laplace transform of s + 4/s² + 4s + 5 is equal to e⁻2t x cos(t) + 2e⁻2t x sin(t).
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That is our answer to the initial value problem, let me go over the steps there.
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We started out with Y″ + 4Y′ + 5y=0, we want to take the Laplace transform of both sides there, that is why we needed to know the Laplace transform of Y″ and Y′.
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We have identities for those, s² L(y) - sY(0) - Y′(0) and then we plug in our initial conditions y(0) is 1, Y′(0) is 0, our L(Y′) simplifies down to s² L(y) minus s and L(Y ′).
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The identity from the beginning of the lecture is sL(y) - y(0), that is a universal rule and then we plug in y(0) is 1, that is where we got that term.
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We plug all those back into the differential equations, we get s² L(y) - s. that is our L(Y′), this term right here is L(Y″) and then we had 4 L(Y′), that 4 became those 4's right here.
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We had to multiply everything here by 4, such when we got the for SL of why -4 and then +5y that became + 5L(y) is equal to zero, you want to think of L(y) as your variable.
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You want to separate out all the terms multiplied by L(y) and then collect the other terms -S -4, move those other terms over to the other side that is what we did here and then divide by s² + 4s + 5.
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That became a denominator and now we get L(y) is a function of s and that is something that we need to apply the inverse Laplace transform to and this particular function is the one we studied in example 5 of the previous lecture.
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If you go back and look at example 5 of the previous lecture, we started out with this function of s and we did a lot of arithmetic on it, some partial fractions in completing a square.
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What we ended up with in the previous lecture was this function of ye⁻2t cos(t) + 2e⁻2t sin(t), what that tells us is that this is the solution to our initial value problem.
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For example 4, we got the following initial value problem y″ - 2Y′ + y=4e⁺t and then a couple of initial conditions, y(0)=4 and Y′(0)=1.
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This ones going to be very much similar to the other examples, we are going to take the Laplace transform of both sides, the one difference here is that we have an inhomogeneous differential equation.
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Inhomogeneous means that that right hand side is not zero, we are going to have to take the Laplace transform of that as well because it is not just coming out of the zero.
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But still the arithmetic kind of work out quite similar to the previous equations, let us see how that goes, we will need L(Y′), Y″ and we had an identity for that at the beginning of this lecture which was s² L(y) - sy(0)- Y′(0).
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We have numbers for y(0) and Y′(0), I will go ahead and fill those in, s² L(y) - sy(0), that is 4s - Y′(0) is -1 and then we also have an identity for L (Y′).
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Again from the beginning lecture these the two identities that never change s x L (y) - y(0) that is the identity from beginning of lecture that never changes is equal to sL(y) and my y(0) we are given that as our initial condition that was 4.
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I'm going to plug those back into the differential equation taking the Laplace transform of both sides, I see L(Y″), I'm going to g s² - L(y) - 4s - 1.
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Now -2(Y′) that is minus 2L(Y′) is SL(y) - 4 + y, plus L(y) is equal to 4e⁺t , this is the Laplace transform of 4e⁺t.
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If you go back and look at the first lecture on Laplace transforms, we figured out what the Laplace transform of e⁺t is and it is 1/s-1.
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Basically the Laplace transform of e⁺at, go back and look at the first lecture on Laplace transform, I think it was two lectures ago here in the differential equations lectures series on www.educator.com.
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It is 1/s - a, person, in this case we have a 4 as a coefficient, I'm going to write that as 4/s -1 and now just like the other problems, we are going to think of L(y) as our variable, we are going to try to solve this equation for L(y).
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It is going to involve collecting some algebra, I have some terms for L(y) and I also have some constant terms, let me see what I have for L(y), I see I have an s² here, I see I have a -2s here, and here is an L(y), that is +1.
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Let me collect the other constant terms, I see -4s, that is the only term of s I see, -4s, -1 and this -2 x -4 would give me a +8, 8-1 is 7.
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That is combining a couple constant terms there and this is still equal to 4/S -1 and let me move these terms over the other side, the 4s + 7, if I move those over to the other side, on the left I will get s² -2s + 1 x L(y)=4/s-1 + 4s - 7.
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I like to combine all if that over a common denominator, I'm going to take that 4s -7 and multiply by S -1 / S -1 and if I expand out that multiplication the 4s -7 x S -1, that is 4s² -7s - 4s, that is -11s.
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Now I have -7 x -1 is + 7, I combine these terms together, I saw I have an S -7 in the denominator that was a whole point of combining them and I get 4 + 4s² -11s + 7.
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On the left hand side, I got an L(y) and I think I'm going to factor that s² - 2s + 1, I know that factors as S -1² and if I keep working on the right I can combine this 4 and this 7 into 11.
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I got 4s² -11s + 11 and on the bottom I have S -1 but I could also divide by S -1² combine those two together and I get S -1³ for my L(y) that is just a pile of algebra, all with the goal of solving for L(y) into a function of s.
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The point of this is that I can now take the inverse Laplace transform, take the inverse Laplace transform and remember that all the examples in this lectures are synced up to the examples from the previous lecture.
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I worked out all the inverse Laplace transforms in the previous lecture, you can go back and check those out if you have not seen it recently but this was example 4 in the previous lecture where we worked out the inverse Laplace transform of this function.
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It is 4s² -11s +11/ s -1³, it was messy that is why I do not want to do the work again but we figured out that the original y as a function of t is 4e⁺t -3et⁺t + 2t² e⁺t.
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That was some work in the previous lecture to calculate that out but it worked and if it does not make sense to anymore, maybe go back and look the previous lecture and work through example 4 and you will see how we got from this step to this step.
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Let me remind you of each of the steps there, in case anything is still a little fuzzy, we wanted to take the Laplace transform of both sides, we need the Laplace transform of Y″ and we have an identity for that.
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That is a universal identity that we learned that at the beginning of lecture and it is the same in every problem, s² L(y) - sy(0) - Y′(0) and then what changes in each problem is the values of y(0) and Y′(0).
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We took the 4 for y(0), plug that in and we took the 1 for Y′(0) and plug that in, and then we have an L(Y′), that by our identity from the beginning of lecture is SL(y) - y(0).
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We plug in y(0)=4 and then we plug those back into the original differential equation, there is my L(Y″) and there is my L(Y′) just copying that from above but of course with a coefficient of -2.
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Now from the differential equation there is my L(y) but since the right hand side is not zero, we have to take the Laplace transform of the right hand side as well, that is L(4e⁺t).
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And something we learned in the first lecture on Laplace transforms is the L(e⁺at) is 1/s -a, if you are little rusty on where that comes from, check out the first lecture on Laplace transforms it is still in this differential equations lecture series.
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It is a couple of lectures ago and you will see where that, I forget my s here in transforms.
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You will see where that L(4e⁺t) turns out to be 4/ S -1, now we have a slightly nasty algebra problem, what we want do is to solve for L(y), you think of L(y) as being the variable.
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That is why I collected all my terms multiplied by L(y), collected all the extra terms, that 7 by the way came from this -1 and this -2 times this -4, -1+ 8.
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That is equal to 4/s - 1 and I move this 4s -7 over to the other side here, put it over a common denominator which meant I had to multiply it by S -1, that is me expanding out that multiplication right there.
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When you combine it, the 4 and the 7, that 4 and that 7 combined and gave you that 11, meantime on the left hand side I have factored S² - 2s + 1and s-1² and then divide both sides by S -1².
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That is why you end up getting S -1³ on the denominator on the right and now we have to take the inverse Laplace transform which is another batch of algebra which we worked out in the previous lecture.
00:35:42.000 --> 00:36:00.000
The previous lecture example 4, we start with exactly this function and then we did partial fractions on this function, expanded out into the inverse Laplace transform and we converted it into 4e⁺t - 3te⁺t + 2t² e⁺t.
00:36:00.000 --> 00:36:23.000
That is our answer to the initial value problem, on our last example here, we got the following initial value problem Y″ + 4y=ae⁻2t and then we have a couple initial values y(0)=4, Y′(0)=-4.
00:36:23.000 --> 00:36:37.000
We are going to take the Laplace transform of both sides there, on the left we are going to need to know L(Y″), we have an identity for that from the beginning of this particular lecture.
00:36:37.000 --> 00:37:03.000
That was s² L(y) - s x y(0) -Y′(0) and we can fill in the numbers that we have here, this is s² L(y) and we cannot fill in anything for that yet, but s x y(0), y(0) is 4t, that is -4s and -Y′(0).
00:37:03.000 --> 00:37:14.000
Y′(0) is -4, that is + 4, it looks like we are not going to need the Laplace transform of Y′, I'm going to go ahead and plug this right into the differential equation.
00:37:14.000 --> 00:37:39.000
We get s² L(y) -4s + 4 + 4L(y), that term right there is coming from here because if we took Laplace transform both sides and then on the right, we have to take the Laplace transform of 8e⁻2t.
00:37:39.000 --> 00:37:56.000
Let me remind you of the Laplace transforms we did back 2 lectures ago here in the differential equation series, in the Laplace transform lecture you will see that in the list of lectures over there.
00:37:56.000 --> 00:38:17.000
In the Laplace transform lecture we figured out that the Laplace transform of e⁺at is 1/s - a, check that out in the original Laplace transform lecture if you do not remember where that comes from.
00:38:17.000 --> 00:38:39.000
That is just a couple lectures ago here on the www.educator.com and e⁻2t our a is -3 a is -2, that is 1/s + 2 and we still have an 8 here coming along as the coefficient.
00:38:39.000 --> 00:39:05.000
We really got an algebraic problem we are trying to solve for L(y), think of that as being the variable and solve for that, I'm going to collect my terms there multiplied by L(y), I see we have an w² + 4 x L(y) - 4s + 4=8/s + 2.
00:39:05.000 --> 00:39:26.000
I'm trying to solve for L(y), I'm going to take the -4s + 4 and move it over to the other side, that will turn into + 4s - 4, in order to put that over a common denominator I have to multiply that by S + 2 / S + 2.
00:39:26.000 --> 00:39:53.000
I need to expand that out, this 4s -4 and the s + 2, if I expand that out and I will get 4s² - 4s + 8s, that is a + 4s and then -4 times +2 is -8, if I combine that all together, I have a -8 and a +8 here, those will cancel each other out.
00:39:53.000 --> 00:40:13.000
I will just get for 4s² + 4s/s + 2 and that is still equal to s² + 4 x L(y) and remember we are solving for L(y), I'm going to divide by the coefficient of L(y), s² + 4.
00:40:13.000 --> 00:40:31.000
L(y)=4s² + 4s/ s² + 4 x s + 2.
00:40:31.000 --> 00:40:52.000
That is the end of that step of algebra, the next step is now that I know what the Laplace transform of y is in terms of s, I need to take the inverse Laplace transform.
00:40:52.000 --> 00:41:15.000
I synced these examples up to the examples in the previous lecture, this was example 5 in the previous lecture, see example 5 in the previous lecture which was called inverse Laplace transforms.
00:41:15.000 --> 00:41:37.000
What we did was we started with this function of s and we did partial fractions on it, we worked backwards and we finally found out a function for y in terms of t,e⁻2t + 3 cos(2t) - sin(2t).
00:41:37.000 --> 00:41:47.000
That is really exploiting a result from the previous lecture, example 5 from the previous lecture if you want to look it up and see where that came from, see how we did that.
00:41:47.000 --> 00:41:56.000
We worked out all the arithmetic back there, it all worked out do not worry that is really the end of that problem, let me show you the steps we went through to get there.
00:41:56.000 --> 00:42:11.000
We took the Laplace transform of both sides which means we had figure out L(Y″), we had this identity from the beginning of lecture which always works s² L(y) - sy(0) - Y′(0).
00:42:11.000 --> 00:42:29.000
We fill in our values for y(0) and Y′(0) which is 4 for y(0) and -4 for Y′(0) of course the -4 turns into a +4 because that negative sign cancel that negative sign and gave me a positive there.
00:42:29.000 --> 00:42:40.000
We plugged that back into the differential equation s² L(y) -4s + 4 and then that 4y gave me that term right there.
00:42:40.000 --> 00:42:52.000
Then we also have this right hand side, we had to take the Laplace transform of 8e⁻2t and we were called from the original Laplace transform lecture, not the previous one but the one before that.
00:42:52.000 --> 00:43:07.000
We first started talking about Laplace transforms, we figured out the Laplace transform of e⁺at is 1/s -a, in this case our a is -2 and we get 1/s +2 and then there is also an 8 there.
00:43:07.000 --> 00:43:23.000
We just bring that along too and what we find here is kind of a big algebraic expression, you want to think L(y) as the variable and solve for that, just kind of sort out all the s's and just get a nice expression for L(y) in terms of s.
00:43:23.000 --> 00:43:41.000
That is what we are doing here, we moved the -4s + 4 over to the other side, that is where that came from, I want to put it over a common denominator, I saw and s + 2 here, I multiply top and bottom here by s + 2.
00:43:41.000 --> 00:43:59.000
That is a little messy I had to expand 4s - 4 x s + 2, that expanded out into 4s² + 4s -8 and when I combined that with this 8, the 2 8 cancel each other out which is why there is no constant over here.
00:43:59.000 --> 00:44:20.000
We just get the 4s² + 4s still over S + 2, in the meantime I still have s² + 4 x L(y) on the left and when I divide that across that joins the denominator here and we get L(y) is 4s² + 4s/s + 2 x s² + 4.
00:44:20.000 --> 00:44:34.000
We had to take the inverse Laplace transform which will be another really healthy dose of algebra here but it is exactly the example 5 in the previous lectures, go back and check that out, work that through.
00:44:34.000 --> 00:44:48.000
What we figured out is the inverse Laplace transform of that is exactly this e⁻2t +3 cos 2t - sin (2t), I see that I wrote that is if it were not exponents.
00:44:48.000 --> 00:44:58.000
Let me write that a little nicer down here, 3cos(2t) - sin(2t) is our final answer there.
00:44:58.000 --> 00:45:09.000
That is the end of our lecture on using Laplace transforms to solve initial value problems and that is actually the end of this chapter on Laplace transforms, I really appreciate you watching.
00:45:09.000 --> 00:45:15.000
My name is Will Murray. You are watching the differential equations lectures series here on www.educator.com, thanks for joining us.