I think for people who are having trouble understanding this (me included) intuitively. Think about this, in the formula for eulers method y_old + Δx (dy/dx). This part of Δx (dy/dx) is just giving us the value of Δy which is then added to the y_old to give us our new y value. It can be derived from Δy/Δx ≈ dy/dx, multiplying by Δx to both sides. But the reason we multiply by Δx is because think about rise over run. For every change in Δx, Δy changes by a specific amount relative to the slope of the line. In the senario when Δx = 0.5 and dy/dx = 1, we are think about how much does "y" change when x changes by 0.5 when the slope is one, giving us Δy = 0.5, this can be applied when dy/dx is equal to different values.
To people who got confused by the colouring: To match the table, the line segment from x = 0 to x = 1 should be purple (since it has a slope of 1), the next one green (slope 2), and the third one pink (slope 4). He hasn't really got to the orange line segment yet, which goes from x = 3 to x = 4.
If for some real number x and this ODE for y=e^x, we apply the method n times with the increment x/n for any natural number n, starting at zero, we will get (1+x/n)^n, which converges to e^x as n goes to infinity.
I used the four x,y points that you found in the ∆x=1 .. and put them in a table and found ∆y, ∆^2y, ∆^3y and used the taylor expansion and got (1+x+x^2/2!+x^3/3!) which is exactly e^x .. so for that big ∆x how did i reach that accuracy?
it's because the slope at x=0.5 is 1.5 (dx/dy), so remember that he is stepping by 0.5 and not by 1, then "y" evaluated should be equal to "y" before plus 0.5 * "slope", which is 1.5, so... y = 1.5 + 0.5 * "1.5" = 2.25.
Delta X is 0.5, so after 0.5 "steps" in x axis he goes up in y axis half of the previous slope and then the new slope is equal to that y value. y values were 1, 1.5 and 2.25 because the slopes were 1 and 1.5 .
LizzieAthey from the slope formula slope=y2-y1/(x2-x1). you slope is equal 1.5 and (x2-x1)=0.5 and y1=1.5 so that will be equal: 0.5*1.5=y2-1.5.......and y2=2.25.
The only problem with this is the exercises that come after. On the quiz for this topic, it solely relies on the formula for Euler’s method, which is not mentioned a single time in the video, so having the quiz be about the formula makes absolutely no sense and should be changed instantly
Gleb Khachatryan look at his initial conditions for x and y. for every time x increases by a particular factor, i.e delta x, y would increase by that factor plus 1
That's only the case when the line goes through the origin. In our case, we're currently at (2, 4), and since the slope is now 4, y increases by 4 when x increases from 2 to 3, making the new y value 8.
I am sort of new at this stuff. It would have been nice if he used the formula for Euler's method to do the calculations along with the graph. Using the formula I got the following results: Can anybody tell me where I went wrong?Xo=0 Yo=1X1=1, Y1= 1+1(0+1)=2X2=2, Y2= 2+1(1+2)=5X3=3, Y3= 5+1(2+5)=12X4=4, Y5= 12+1(3+12)=27 and so on.
what we more or less get from this is, a ) a good local approximation and b) a terrible global approximation. global one is bad because 1.) often times, solutions to differential equations might not be unique and if we consider lim to infty or smth they often diverge differently (hence things like weather forevast etc are terrible for more than like a week) 2.) the error term if were lipschitz and the solution is thus unique contains an exp(x-x_0) times a constant depending on our problem functions max curvature and the delta_x we chose. so even if we dont curve a lot, we get a lot if error if x is far away from x_0... this is also the reason the local approximation is pretty good, because we get an exp term of pasically one and the other term is pretty small if were feedinf the problem to a computer which calculates it which a small delta_x, keeping the error minimal. tl:dr for unique solitions local approximation is good using small delta_x but the error grows exponentially so after some time the quality drops massively
dy/dx is the change in the y direction when x changes. so if we increase x by 1, "dy/dx = 2" dictates that we should increase y by 2. the statement dy/dx=2 reads as "per change in x, there is a change of 2 in y".
Lmao at people disliking this video because he doesn’t plug-and-chug using a preexisting formula. The derivation for that formula is incredibly simple, especially with this video as a reference. Think harder!
+Swati Chow dy/dx denotes the slope. Given that y=dy/dx, and y precedes n increases in x as well as dy/dx. We are not always increasing by the slope (dy/dx).
Analytic is basically when you can use formulas to get a generalized answer. From there you can input any value and get an accurate answer. However, since that is not always possible you can use may numbers and find out the answer (in this case the shape the graph) numerically.
Simo2009BORO simple enough, y(x) = constant That's the solution, dy/dx is the rate of change of y with respect to x, if the rate of change is zero, the function must be a constant..
CHOOSE THE CORRECT OPTION: The first approximate value of dy/dx =1-y ,y(0)=0 find value of y at x=0.1 using Euler's method is A)0.9 B)1.2 C)1.0 D)0.1 Please reply quickly because tomorrow is my exam
The formula to get the next y value is this: y + (dy/dx * DX) y is the previous y value dy/dx is the previous dy/dx value DX is how much you’re changing by x
Good question. I think he's doing it completely wrong. If you plug 1 into x, y is e, and since dy/dx = y, dy/dx is e as well. He isn't even approximating the solution correctly.
No he isn't. He just picked big values of Δx so that the example is comprehensive, avoiding unnecessary calculations. If you pick Δx=1 you can see even from the first step that it is a bad approx since it gives e≈2. If you want better approximations you pick smaller Δx as he mentions in the video. His purpose is to explain, not to approximate.
Euler's method can only solve first order equations.. Actually I have derived a technique that can solve any differential equation numerically, no matter what order it is.. Is there any known method that does so, or is it something new I have invented ???
The only way to know is to try publishing it in a journal and have it peer reviewed. Also, if you can find a college professor, try asking them to check your mathematical proof.
