Runge - kutta 4th order method for two different steps

Why is h = 0.5 worst than h = 1 in my code ? I can't find where i am wrong.

clc, clear all, close all, format long

f = @(x,y) y;
h = [1 0.5];
y0 = 1;

syms Y(X)
Df = diff(Y) == Y;
Y = dsolve(Df, Y(0) == 1);
fplot(X,Y), hold on

for i = 1:length(h)
    
x = [0:h(i):4];
    
    for j = 1:length(x)-1

        y(1) = y0;
    
        K1 = f(x(j),y(j));
        K2 = f(x(j) + 0.5*h(i), y(j) + 0.5*h(i)*K1);
        K3 = f(x(j) + 0.5*h(i), y(j) + 0.5*h(i)*K2);
        K4 = f(x(j) + h(i), y(j) + h(i)*K3);
    
        y(j+1) = y(j) + (K1 + K4 + 2*(K2 + K3))/6;
    
    end

xlim([0 5]), ylim([0 55])
plot(x,y)

end
 

legend({'y(x) = e^x','Runge - Kutta 4th order with h = 1','Runge - Kutta 4th order with h = 0.5'},'Location','Best')

Answer 1

[Torsten]

        y(j+1) = y(j) + h(i)*(K1 + K4 + 2*(K2 + K3))/6;        

instead of

        y(j+1) = y(j) + (K1 + K4 + 2*(K2 + K3))/6;

[Left Terry]

Yes, you are right and I...need to visit the eye doctor. Thank you.

[Torsten]

And you should take the assignment

       y(1) = y0;       

out of the j-loop. Assigning once at the start is sufficient.