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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 7

Lesson 5: Approximating solutions using Euler’s method

# Euler's method

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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• How X=1 be Y = 2? because y = e^x, if X = 1, then Y should be e^1 •   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`
• 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 ? • 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.
• Why does the y-values increment by half of the slope? •  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.
• For all Euler type problems, is the slope always equal to y? •  No, the slope is always equal to `dy/dx` (that is, after all, the definition of slope). In this example, `dy/dx = y`, but that is not general at all.
• At when x=1, why is that y is incremented by 'half' of dy/dx( 1.5 ) and not 1.5 itself? • Isn't this reminiscent of Riemann sums, but like for an arc length rather than an area? • Nice method! Is it based on the linear approximation principle y(x + Δx) ≈ y(x) + y'(x)*Δx ? • 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.   