Matlab code help on Euler's Method

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.

[Muhammad Tahir]

y'=2x-3y+1, y(1)=5, y(1.2)=? MATLAB code using euler'method to obtain a four decimal and h= 0.1

Answer 1

[James Tursa]

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:

http://www.mathworks.com/matlabcentral/answers/224319-euler-method-without-using-ode-solvers

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?

[Sanjida Ahmed]

Thank you Tursa.I don't know what will teacher give me to solve but I am now practicing to solve f=x+2y equation.I type exact same code you provide and my code is,

    % Euler's Method
% Initial conditions and setup
h = 0.01;  % step size
x = 0:h:1;  % the range of x
y = zeros(size(x));  % allocate the result y
y(1) = 1;  % the initial y value
n = numel(y);  % the number of y values
% The loop to solve the DE
for i=1:n-1
    f = (x(i)+2*(y(i)));
    y(i+1) = y(i) + h * f;
end

but it gives no answer in the editor.

[James Tursa]

After you enter this in the editor and save it, you need to run it either by typing the file name at the command prompt, or by pressing the green triangle Run button at the top of the editor. Since all of the lines end with a semi-colon ;, there will be no output to the screen when this runs. However, the variables are there. If you look in the Workspace list you will see them, or if you issue the whos command you also will see them. To see the result you could plot them. E.g.,

plot(x,y); grid on

[ATUL]

how many iterations, we will decide in this?

[Ahmed J. Abougarair]

% 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']

Answer 2

[mahmoud mohamed abd el kader]

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;

[Tendai Kufandirori]

Hi, I am trying to solve dy/dx = -2x^3 + 12x^2- 20x + 9 and am getting some errors when trying to use Euler's method. Do you know how to go about it please

[John D'Errico]

Was it necessary to post 3 identical answers, to an old question?

[James Tursa]

@shireesha myadari Please delete this comment and open up a new question for this.

[Ahmed J. Abougarair]

% 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']

Answer 3

[Bakary Badjie]

what is the Matlab function that implements Euler’s method

[Chris Horne]

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

Answer 4

[Rakshana]

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;