Solve IVP with Taylor method of order

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Minjae Cho
Minjae Cho 2022 年 6 月 23 日
編集済み: Minjae Cho 2022 年 6 月 24 日
I wanna implement this into a code.
My code is followed by :
  • syms x y(x)
  • f = y(x) - x^3 + x + 1
  • df = diff(f, x)
  • f = subs(df, diff(y(x), x), f)
and it gives OUTPUT
  • f = x + y(x) - 3*x^2 - x^3 + 2
What I am trying to do is change y(x) (symfun) to new y variable
so that I can use the function of f(x,y) = x + y - 3*x^2 - x^3 + 2; to plug f(a,b) into x and y variable.
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Torsten
Torsten 2022 年 6 月 24 日
So what's your numerical method to solve the IVP ?
y_(n+1) = y_n + dx*y_n' + dx^2/2 * y_n''
?

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Walter Roberson
Walter Roberson 2022 年 6 月 23 日
If you really really need it to be in terms of y and no other name will do then you can follow with
syms y
subs(sol, yx, y)
The "syms y" will destroy the association between the name y and the symbolic function y(x) allowing a substitution as a name instead of a function.
There are ways to do this without using a temporary variable name such as the "yx" that I showed.
But I already showed you exactly how to substitute in numeric values.
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Minjae Cho
Minjae Cho 2022 年 6 月 24 日
Thanks a lot!
I was gonna do it recursively for Taylor method of order.
It was a lot of help thanks!

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