Using RK4 to solve an equation of one variable only

I need to solve the equation presented on the code with those initial condition, integration step and interval.

I am with this code at the moment and I need to create a loop.

On the commented section there's a code I found for two variables but I don't know to make it just for one. Can someone please explain?

I just started with MATLAB and Runge-Kutta so any help would be greatly appreciated.

clear; clc


h=0.1; 
tinitial=1; 
tfinal=5;


y(0)=1;

f=@(y) (1-2*y)*y^2

%rk4 loop - found this code for two variables and tested
%for i=1:ceil(tfinal/h)
    %update y
%    k1=f(y(i));
%    k2=f(t(i)+0.5*h,y(i)+0.5*k1*h);
%    k3=f(t(i)+0.5*h,y(i)+0.5*k2*h);
%    k4=f(t(i)+ h,y(i)+ k3*h);
%    y(i+1)=y(i)+h/6*(k1 + 2*k2 + 2*k3 + k4);
%end
    

Answer 1

[Alan Stevens]

The code you found is, in fact, for just one dependent variable (y). It is also updating the independent variable (t). Try the following:

h=0.1; 
tinitial=1; 
tfinal=5;
y(1)=1;
f=@(t,y) (1-2*y)*y^2;
t(1) = tinitial;
%rk4 loop 
for i=1:ceil(tfinal/h)
    %update y
   k1=f(t(i),y(i));
   k2=f(t(i)+0.5*h,y(i)+0.5*k1*h);
   k3=f(t(i)+0.5*h,y(i)+0.5*k2*h);
   k4=f(t(i)+ h,y(i)+ k3*h);
   y(i+1)=y(i)+h/6*(k1 + 2*k2 + 2*k3 + k4);
   t(i+1) = t(i) + h;
end
plot(t,y),grid
xlabel('time'), ylabel('y')

[Flint]

You were right about the loop but i don't understand what's t(1) doing there because you don't call it on the loop. Could you please explain? Also, the plot interval goes from [1,6] how do i change it do [1,5]?

[Walter Roberson]

y(1)=1 initializes y to exist and contain 1.

Inside the loop, on the first iteration, when i = 1, then y(i) is y(1) which will refer to the value you initialized to, y(1)=1

[Flint]

Thank you so much for the reply, I understand now. But still, how do I make the interval from [1,5] instead of [1,6]?

[Walter Roberson]

for i=1:ceil((tfinal-tinitial)/h)

[Flint]

Thank you so much sir.