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Euler's method

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.C (LO)
,
FUN‑7.C.4 (EK)
Euler's method is a numerical tool for approximating values for solutions of differential equations. See how (and why) it works.

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  • leaf blue style avatar for user Scott Jang
    How X=1 be Y = 2? because y = e^x, if X = 1, then Y should be e^1
    (40 votes)
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    • male robot hal style avatar for user Yamanqui García Rosales
      You are right, the correct point is y(1) = e ≅ 2.72; Euler's method is used when you cannot get an exact algebraic result, and thus it only gives you an approximation of the correct values. In this case Sal used a Δx = 1, which is very, very big, and so the approximation is way off, if we had used a smaller Δx then Euler's method would have given us a closer approximation.

      With Δx = 0.5 we get that y(1) = 2.25
      With Δx = 0.25 we get that y(1) ≅ 2.44
      With Δx = 0.125 we get that y(1) ≅ 2.57
      With Δx = 0.01 we get that y(1) ≅ 2.7
      With Δx = 0.001 we get that y(1) ≅ 2.72
      (146 votes)
  • blobby green style avatar for user lpkool
    I understand the concept behind Euler's method and it is quite interesting , but I don't get how sal got his values for y ?
    (53 votes)
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    • aqualine ultimate style avatar for user Dema Saad Aldeen
      for the first table:
      Δx=1
      Δy/Δx=y
      Δy=y (how much we add to get a new y)
      we can write it like: Δy=y(old)
      y(new)=y(old)+Δy=2y(old)
      -if y=1 then y(new)=2y(old)=2(1)=2
      -then y=2 then y(new)=2y(old)=2(2)=4
      -then y=4 then y(new)=2(4)=8
      for the second table:
      Δx=0.5
      Δy/Δx=y
      Δy/0.5=y
      Δy=y/2
      Δy=y(old)/2
      y(new)=y(old)+Δy=3/2 y(old)
      -if y=1 then y(new)=3/2 y(old)=3/2(1)=1.5
      -then y=1.5 then y(new)=3/2 y(old)=3/2(1.5)=2.25
      -then y=2.25 then y(new)=3/2 y(old)=3/2(2.25)=3.375
      ...and so on
      hope this helps.
      (17 votes)
  • leaf green style avatar for user Maryam
    Why does the y-values increment by half of the slope?
    (22 votes)
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    • blobby green style avatar for user Funky Bibimbap
      I assume you are talking about the second case. The slope dy/dx tells us that for a given number of steps on the x axis, we must take a certain number of steps on the y axis. So you should read dy/dx = 1.5 as dy/dx = 1.5/1, which means that for one step on the x axis, we go one step and a half on the y axis. We can also say dy/dx = 1.5/1 = 3/2, for every two steps on the x axis, we take three steps on the y axis, this is equivalent.
      Lastly we also have dy/dx = 1.5/1 = 0.75/0.5. So when we take half a step on the x axis, we must take 0.75 (three quarters) steps on the y axis.
      (22 votes)
  • leaf grey style avatar for user cstievater
    For all Euler type problems, is the slope always equal to y?
    (14 votes)
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  • male robot hal style avatar for user Pranjal  Bhumij
    At when x=1, why is that y is incremented by 'half' of dy/dx( 1.5 ) and not 1.5 itself?
    (8 votes)
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  • male robot hal style avatar for user Sean
    Isn't this reminiscent of Riemann sums, but like for an arc length rather than an area?
    (10 votes)
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    • male robot donald style avatar for user Aaron Hargrove
      I was thinking the same thing. As the x intervals get smaller and smaller they approach what is essentially dx, then evaluate it infinitely many times just as you do in an integral. That could be demonstrated using this example, but it may not work with more complicated differential equations that can't be integrated as easily.
      (7 votes)
  • aqualine tree style avatar for user CodeLoader
    Nice method! Is it based on the linear approximation principle y(x + Δx) ≈ y(x) + y'(x)*Δx ?
    (10 votes)
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  • primosaur ultimate style avatar for user Sam Mastnardo
    Can some please help me with this method? I'm stuck on this I need some help or explaining to. I'm only in 6th grade but I've watched this video 6 times pausing to take notes and I keep getting stuck on the skill.
    (5 votes)
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    • spunky sam blue style avatar for user Samuel Jett
      You should probably get strong foundations of high school math (algebra- precalc) and basic calculus before doing this. Calc BC is a more advanced calculus. If you already know high school math, do calc AB, then this will all make sense. I’m glad you’re interested in the subject, though! The world needs more people like that.
      (1 vote)
  • purple pi pink style avatar for user Carolyn Dewey
    At , Sal says "If we stepped by 0.0001 we would get even closer and closer and closer." What would happen if we took the limit as Δx approaches 0? (A reasonable assumption is that we would get the exact solution if we took the limit, but how would we take that limit?)
    (4 votes)
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  • leafers seed style avatar for user Shraddha
    Can Euler's method be used for higher order differential equations ?
    (2 votes)
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Video transcript

- [Voiceover] We've already seen that if we were to start with the differential equation, the derivative of Y with respect to X is equal to Y and we have the initial condition that Y of zero is equal to one, but the particular solution to this given these initial conditions, is Y of X is equal to E to the X. Or I guess we can just Y is equal to E of the X if we didn't want to write it with the function notation. And that's all fair and well and this works out well. This is a separable differential equation and we can integrate things quite easily. But as you will see as you go further in the world of differential equations, most differential equations are not so easy to solve. In fact, many of them are impossible to solve using analytic methods. And so given that, what do you do? We've nicely described some phenomena, modeled some phenomena using the differential equations but if you can't solve it analytically do you just give up? And the answer to that question is no. You do not just give up because we now have computers, and computers are really good at numerical methods. Numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that? Well, in this video we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so I'm gonna draw a little table here. So, a little table here. Actually let give myself. I'm gonna do it over here on the left hand side. A little table. So, X and then Y. XY. And then DY, DX. And you could set up a table like this to create a slope field. You could just pick all the ... You could sample X as in Y in the XY plane, and then figure out for our first order differential equation like this, what is the slope going to be at that point and you could construct a slope field. And we're gonna do something kind of related but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that Y of zero is equal to one. We know that the particular solution of this differential equation contains this point. So, we're gonna start with that point. So we're gonna start with X is equal to zero and let me do this in a different color. We're gonna start with X is equal to zero, Y is equal to one. Which is that point right over there. And we're gonna say, well, okay what is the derivative at that point? Well, we know the derivative at any point that's for any solution to this differential equation the derivative is going to be equal to the Y value. So in this case, the derivative is going to be equal to Y. It's going to be equal to one. And in general, if the derivative just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of Xs and Y of Xs, then you can figure out what the slope of the tangent line will be at that point. And so, you say okay, there's a slope of one at that point so I can depict it like that. And instead of just keep doing that with a bunch of points we'll say okay, well let's just ... We know that the slope is changing or it's probably changing for most cases. But let's just assume it's fixed until our next X and then use that assumption to estimate what the next Y would be. So, what am I talking about here? So, when I talk about the next X we're talking about well, let's just step. Let's just say for the sake of simplicity, we're gonna have a delta X of one. A change in X of one. So we're gonna step from X equals zero now. We're gonna now step from that to X is equal to one. So we're now gonna go to ... Actually I may not use that. I used that yellow color already for the actual graph or for the actual E to the X. So now let's say X is equal to one. Our delta X is one. So we've just added one here. And what we can do in our little approximation scheme here is well, let's just assume that that slope was constant over that interval. So where does that get us to? Well, if Y was at one and if I have a slope of one for one more, for one increase in X, I'm gonna increase my Y by one. So then Y is going to increase by one and is going to get to two. And we see that point right over there and you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to Y. The slope of the tangent line is going to be equal to Y. So, in this case, the slope of the tangent line is now going to be equal to two. And we could depict that. Let me depict that in magenta here. So, it is going to be two. It's gonna look ... So the slope of the tangent line there is going to be two. And so, what does that tell us? Well if we step by our delta X one more. So now our X is equal to two. What should the corresponding Y be? Well, let's see. Now for every one that we increase in the X direction we should increase two in the Y direction because the slope is two. So, the very next one should be four. Y is equal to four. So, we could imagine we have now kind of had a constant slope when we get to that point right over there. And now we can do the same thing. Well if we assume DY, DX based on the differential equation it has to be equal Y, okay, the slope of the tangent line there is going to be the same thing as Y. It's going to be four. And so, if we step our X up by one, if we increment our X by one again, once again, we just decided to increment by one. We could have incremented by 10, we could have incremented by .01. And you could guess which one's going to give you a more accurate result. But if we step up by one now and our slope is four, well, we're gonna increase by ... If we increase X by one we're gonna increase Y by four. So we are going to get to eight. And so, we are at the point three comma eight which is right over here. And so, for this next stretch, the next stretch is going to look like that. And as you can see just by doing this, we have been able to approximate what the particular solution looks like and you might say, "Hey, so how do we know "that's not so good of an approximation?" And my reply to you is well, yeah I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand I took fairly large delta X steps. If I wanted a better approximation I could have lowered the delta X and let's do that. So let's take another scenario. So let's do another scenario where instead of delta X equal one, let's say delta X equals 1/2. So once again, X, Y and the derivative of Y with respect to X. So now let's say I want to take ... So we know this first point. We're given this initial condition. When the X is zero, Y is one and so the slope of the tangent line is going to be one. But then if we're incrementing by 1/2 so then when X is, I'll just write it as 0.5. 0.5. What is our new Y going to be? Well we're gonna assume that our slope from this to this is this slope right over here. So our slope is one, so if we increase X by 0.5 we're gonna increase Y by 0.5 and we're going to get to 1.5. So, we can get 0.5, 1.5. We get to that point right over there. Actually you're having trouble seeing that. This stuff right over here is this point right over here and now our new slope is going to be 1.5. Which is going to look like. Which is going to look like actually not quite that steep. I don't want to overstate how good an approximation is and it's starting to get a little bit messy but it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment X by another 0.5 where you get to one. So now if you increment by 0.5 and your slope is 1.5, your Y is going to increment by half of that by 0.75 and so, you're gonna get to 2.25. So now you get to one, 2.25 which is this point right over here. Once again, this is a better approximation. Remember, in the original one Y of one you know should be equal to E. Y of one in the actual solution should be equal to E. 2.7 on and on and on and on and on. Now in this one, Y of one got us to two. In this one Y of one got us to 2.25. Once again, closer to the actual reality, closer to E. Instead of stepping by 0.5, if we stepped by 0.1 we would get even closer. If we stepped by 0.0001 we would get even closer and closer and closer. So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations gets solved. You know it's not the exact same solution or the same method that the idea that most differential equations are actually solved or I guess you can say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, "Hey, well what method is "this one right over here called?" Well, this right over here is called Euler's. Euler's Method after the famous Leonhard Euler. Euler's Method. And not only actually is this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular you can actually even use this to find E with more and more and more precision. Anyway, hopefully you found that exciting.