Solve IVP with Taylor method of order
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Minjae Cho
am 23 Jun. 2022
Bearbeitet: Minjae Cho
am 24 Jun. 2022
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.
1 Kommentar
Torsten
am 24 Jun. 2022
So what's your numerical method to solve the IVP ?
y_(n+1) = y_n + dx*y_n' + dx^2/2 * y_n''
?
Akzeptierte Antwort
Walter Roberson
am 23 Jun. 2022
https://www.mathworks.com/matlabcentral/answers/1746100-whole-derivation-of-two-variable-differential-function#comment_2229660 already shows you how to change to a different variable yx
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.
1 Kommentar
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!