iteration in runge-kutta

hi,

how to calculate (w) for this matrix h=[0.5,0.1,0.7,0.01] in one script and keep data for each (h)

clc;clear all;close all;
h = 0.5;
w(1) = 0.5;
a = 0;
b = 2;
z = a:h:b;
t = zeros(1,length(z));
for i = 1:(length(z)-1)
    k1 = h*f(t(i),w(i));
    k2 = h*f(t(i) + h/2, w(i) + k1/2);
    k3 = h*f(t(i) + h/2, w(i) + k2/2);
    k4 = h*f(t(i) + h, w(i) + k3);
    w(i+1) = w(i) + ((k1 + 2*k2 + 2*k3 + k4)/6);
    t(i+1) = t(i) + h;
      
end

Answer 1

[KSSV]

clc; clear all ;
clc;clear all;close all;
H = [0.5,0.1,0.7,0.01]  ;

a = 0;
b = 2;

m = length(H) ;

T = cell(m,1);
W = cell(m,1);

for j = 1:m
    h = H(j) ;
    z = a:h:b;
    n = length(z) ;    
    w = zeros(1,n) ; 
    t = zeros(1,n) ; 
    w(1) = 0.5 ; 
    for i = 2:n
        k1 = h*f(t(i),w(i));
        k2 = h*f(t(i) + h/2, w(i) + k1/2);
        k3 = h*f(t(i) + h/2, w(i) + k2/2);
        k4 = h*f(t(i) + h, w(i) + k3);
        w(i+1) = w(i) + ((k1 + 2*k2 + 2*k3 + k4)/6);
        t(i+1) = t(i) + h;
    end
    T{j} = t ;
    W{j} = w ; 
end

[Torsten]

Note that n changes with h ...

[KSSV]

@Torsten Yes you are right....edited the answer.

[amir reza]

tnx for your answer

but if you run my code for h=0.5 and see (w) matrix, you realize that your answer is too wrong.

[Torsten]

Instead of

for i=2:n

use

for i=1:n-1