reorganize plot for runge kutta

% It calculates ODE using Runge-Kutta 4th order method
% Author Ido Schwartz

clc;                                               % Clears the screen
clear all;
h=0.1;                                             % step size
x = 0:h:10;                                         % Calculates upto y(3)
y = zeros(1,length(x)); 
y(1) = 0.2;                                          % initial condition
F_xy = @(t) (0.32)+((0.24)*exp(-t))-2*(0.04*exp(-2*t));                 % change the function as you desire

for i=1:(size(x)-1)                              % calculation loop
    k_1 = F_xy(x(i),y(i));
    k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);
    k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
    k_4 = F_xy((x(i)+h),(y(i)+k_3*h));

    y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
end
% validate using a decent ODE integrator  
tspan = [0,1]; y0 = 0;
[tx, yx] = ode45(F_xy, tspan, 0.5);
plot(x,y,'o-', tx, 

How to make the equation start from zero then stop in the case steady state?

Answer 1

[Sulaymon Eshkabilov]

Hi,

In your code there are several flaws:

(1) In your RK code: y(0) = 0.2 and in validation ode45 (validation part) y0 = 0.5. This has to be fixed.

(2) In your RK code: you're simulating within [0, 10] and in validation ode45 (validation part) within [0, 1]. It is better to make them unique. Moreover, in validation part, ode45 takes a variable step, but for comparison puposes, it is better to have a fixed step of size 0.1 as you set h=0.1 in part - RK.

To start your simulation, already your starting at x =0 or you wish to set the initial condition set at 0, then y(0) has to be set at 0 not at 0.2 or 0.5. Moreover, to stop simulation at steady state, you need to intoruduce error tolerance calculation in your code and employ break operator.

Good luck.

[Mariam Gasra]

how can i correct it?

i am a new user in matlab