Euler's method code
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hello , how can I code this : y'(t)=cos(t + y) y(0)=0 t[0,3] (exact solution y(t)=-t + 2arctan(t)) using Euler's method like the code below?
clc
clear all
close all
h=0.01;
N=100; % number of steps
y(1)=1;
for n=1:N
y(n+1)=y(n)+h*(-6*y(n));
x(n+1)=n*h;
end
plot(x,y)
hold on
x=0:0.01:1;
y=exp(-6*x); % exact solution
plot(x,y)
0 commentaires
Réponses (1)
Paul
le 18 Juin 2021
The code that is posted solves a differential equation of the form
y'(x) = f(x,y), y(0) = y0
and compares it to the exact solution
y(x) = some function of x
over the interval
x [0 1]
As it happens, your problem has the exact same form, with the independent variable being t instead of x and the endpoint of the interval is different.
2 commentaires
Voir également
Catégories
En savoir plus sur Logical dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!