Asked by Jane
Let dy/dx=xy . Use Euler's method with step size of 0.5 and the initial condition of y(0) = 3 to estimate y(2).
I really am not sure if I did this right, but I got 3.4568 for y(2). I actually feel like it is wrong.
I really am not sure if I did this right, but I got 3.4568 for y(2). I actually feel like it is wrong.
Answers
Answered by
Anonymous
initial x = 0
Y(0) = 3
delta y = x y delta x
= (0)(3)(.5) = 0
so y(.5) = 3 + 0 = 3
Y(.5) = 3
delta y = (.5)(3)(.5) = .75
Y(1) = 3 + .75 = 3.75
Y(1) = 3.75
delta y = 1 (3.75)(.5) = 1.875
Y(1.5) = 3.75+1.875 = 5.625
Y(1.5) = 5.625
delta y = 1.5(5.625).5 = 4.22
Y(2) = 5.625 + 4.22 = 9.84
Y(0) = 3
delta y = x y delta x
= (0)(3)(.5) = 0
so y(.5) = 3 + 0 = 3
Y(.5) = 3
delta y = (.5)(3)(.5) = .75
Y(1) = 3 + .75 = 3.75
Y(1) = 3.75
delta y = 1 (3.75)(.5) = 1.875
Y(1.5) = 3.75+1.875 = 5.625
Y(1.5) = 5.625
delta y = 1.5(5.625).5 = 4.22
Y(2) = 5.625 + 4.22 = 9.84
Answered by
Reiny
This is tedious work
I found an excellent youtube for you where they solve a problem very similar to yours.
I assume your text or your notes contain the basic original formula for Euler's Method
I suggest you pause the video at critical points and make the necessary changes for your problem
(There are other youtubes on the right column for similar questions)
https://www.youtube.com/watch?v=nYXclUjpFaw
I found an excellent youtube for you where they solve a problem very similar to yours.
I assume your text or your notes contain the basic original formula for Euler's Method
I suggest you pause the video at critical points and make the necessary changes for your problem
(There are other youtubes on the right column for similar questions)
https://www.youtube.com/watch?v=nYXclUjpFaw
Answered by
Damon
That Anonymous is me, Damon.
By the way there are lots of Euler methods. I took the first order one by default because you did not specify.
By the way there are lots of Euler methods. I took the first order one by default because you did not specify.
Answered by
Damon
now let's see how we did
dy/dx = x y
dy/y = x dx
ln y = x^2/2
y = C e^(x^2/2)
when x = 0, y = 3
so
y = 3 e^(x^2/2)
when x = 2
y = 3 e^2 = 22.1
dy/dx = x y
dy/y = x dx
ln y = x^2/2
y = C e^(x^2/2)
when x = 0, y = 3
so
y = 3 e^(x^2/2)
when x = 2
y = 3 e^2 = 22.1
Answered by
Damon
graph all that and you will see that Euler slopes slant way below the real exponential curve. You need a higher order method to do this.
Answered by
Writeacher
There are many of these videos at http://www.khanacademy.org, too.
http://www.google.com/search?q=euler+methods+khanacadem.org&oq=euler+methods+khanacadem.org&aqs=chrome..69i57.6983j0j7&sourceid=chrome&es_sm=119&ie=UTF-8
http://www.google.com/search?q=euler+methods+khanacadem.org&oq=euler+methods+khanacadem.org&aqs=chrome..69i57.6983j0j7&sourceid=chrome&es_sm=119&ie=UTF-8
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