Matlab code help on Euler's Method
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Sanjida Ahmed
am 11 Apr. 2016
Kommentiert: Ahmed J. Abougarair
am 20 Mär. 2024
I have to implement for academic purpose a Matlab code on Euler's method(y(i+1) = y(i) + h * f(x(i),y(i))) which has a condition for stopping iteration will be based on given number of x. I am new in Matlab but I have to submit the code so soon. I am facing lots of error in implementing that though I haven't so many knowledge on Matlab. If anyone provide me so easy and simple code on that then it'll be very helpful for me. Thank you.
1 Kommentar
Muhammad Tahir
am 24 Dez. 2023
Verschoben: Dyuman Joshi
am 26 Dez. 2023
y'=2x-3y+1, y(1)=5, y(1.2)=? MATLAB code using euler'method to obtain a four decimal and h= 0.1
Akzeptierte Antwort
James Tursa
am 11 Apr. 2016
Here is a general outline for Euler's Method:
% Euler's Method
% Initial conditions and setup
h = (enter your step size here); % step size
x = (enter the starting value of x here):h:(enter the ending value of x here); % the range of x
y = zeros(size(x)); % allocate the result y
y(1) = (enter the starting value of y here); % the initial y value
n = numel(y); % the number of y values
% The loop to solve the DE
for i=1:n-1
f = the expression for y' in your DE
y(i+1) = y(i) + h * f;
end
It is based on this link, which you have already read:
You need to fill in the values indicated, and also write the code for the f line. What is the DE you are trying to solve?
4 Kommentare
Ahmed J. Abougarair
am 20 Mär. 2024
% Euler's Method
% Initial conditions and setup
clc
clear
h = input('Enter your step size here :'); % step size
x = input('Enter the starting value of x :');
xend = input('Enter the ending value of xend :'); % the range of x
n = (xend-x)/h; % the number of y values
y = zeros(1,n); % allocate the result y
y(1) = input('Enter the starting value of y :'); % the initial y value
% The loop to solve the DE
for i=1:n
f(i) = 6- 2*(y(i)/x(i)); % dy/dx = 6-2y/x
y(i+1) = y(i) + h * f(i);
x(i+1)=x(i)+h;
end
[x' y']
Weitere Antworten (3)
mahmoud mohamed abd el kader
am 27 Okt. 2020
h=0.5;
x=0:h:4;
y=zeros(size(x));
y(1)=1;
n=numel(y);
for i = 1:n-1
dydx= -2*x(i).^3 +12*x(i).^2 -20*x(i)+8.5 ;
y(i+1) = y(i)+dydx*h ;
fprintf('="Y"\n\t %0.01f',y(i));
end
%%fprintf('="Y"\n\t %0.01f',y);
plot(x,y);
grid on;
4 Kommentare
James Tursa
am 3 Mär. 2021
Bearbeitet: James Tursa
am 3 Mär. 2021
@shireesha myadari Please delete this comment and open up a new question for this.
Ahmed J. Abougarair
am 20 Mär. 2024
% Euler's Method
% Initial conditions and setup
clc
clear
h = input('Enter your step size here :'); % step size
x = input('Enter the starting value of x :');
xend = input('Enter the ending value of xend :'); % the range of x
n = (xend-x)/h; % the number of y values
y = zeros(1,n); % allocate the result y
y(1) = input('Enter the starting value of y :'); % the initial y value
% The loop to solve the DE
for i=1:n
f(i) = 6- 2*(y(i)/x(i)); % dy/dx = 6-2y/x
y(i+1) = y(i) + h * f(i);
x(i+1)=x(i)+h;
end
[x' y']
Bakary Badjie
am 14 Jun. 2021
what is the Matlab function that implements Euler’s method
1 Kommentar
Chris Horne
am 31 Mär. 2022
Is the term 'forward Euler' the same as 'Euler' in terms of the algorithm? Here is my method for solving 3 equaitons as a vector:
% This code solves u'(t) = F(t,u(t)) where u(t)= t, cos(t), sin(t)
% using the FORWARD EULER METHOD
% Initial conditions and setup
neqn = 3; % set a number of equations variable
h=input('Enter the step size: ') % step size will effect solution size
t=(0:h:4).';%(starting time value 0):h step size
nt = size(t,1); % size of time array
%(the ending value of t ); % the range of t
% define the function vector, F
F = @(t,u)[t,cos(t),sin(t)]; % define the function 'handle', F
% with hard coded vector functions of time
u = zeros(nt,neqn); % initialize the u vector with zeros
v=input('Enter the intial vector values of 3 components using brackets [u1(0),u2(0),u3(0)]: ')
u(1,:)= v; % the initial u value and the first column
%n=numel(u); % the number of u values
% The loop to solve the ODE (Forward Euler Algorithm)
for i = 1:nt-1
u(i+1,:) = u(i,:) + h*F(t(i),u(i,:)); % Euler's formula for a vector function F
end
Rakshana
am 13 Nov. 2022
h=0.5; x=0:h:4; y=zeros(size(x)); y(1)=1; n=numel(y); for i = 1:n-1 dydx= -2*x(i).^3 +12*x(i).^2 -20*x(i)+8.5 ; y(i+1) = y(i)+dydx*h ; fprintf('="Y"\n\t %0.01f',y(i)); end %%fprintf('="Y"\n\t %0.01f',y); plot(x,y); grid on;
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