How to make a function that uses Runge-Kutta Method

Hello, I am trying to create a function that can take in a function and solve it using Runge-Kutta's method. For example, I should be able to input dy/dx = x+y , y(0) = 1 and get an answer from the funtion. I've been working with this equation for a while, I just cannnot figure out how to format this into a function. Here is what I have.

function [OutputX,OutputY] = FunctionBeta_Executor(h,x,y,FunctionA)
    
    h=1.5;                                             % step size
    x = 0:h:3;                                         % Calculates upto y(3)
    y = zeros(1,length(x)); 
    y(1) = 5;                                          % initial condition
    FunctionA = F_xy;               % change the function as you desire
    for i=1:(length(x)-1)                             
        i=1:(length(x)-1)                              % calculation loop
        k1 = F_xy(x(i),y(i));
        k2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k1);
        k3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k2));
        k4 = F_xy((x(i)+h),(y(i)+k_3*h));
        y(i+1) = y(i) + (1/6)*(k1+2*k2+2*k3+k4)*h;  % main equation
        
    end
end
% For example, FunctionA might be dy/dx = x+y , with inital conditions y(0)=1. 

Answer 1

[Sulaymon Eshkabilov]

Hi,

Here is the corrected code:

function [x, y] = FunctionBeta_Executor(F)
% Note that F function expression is defined via Function Handle: F = @(x, y)(x+y)               
% change the function as you desire
h=0.15;                                             % step size (smaller step size gives more accurate solutions)
x = 0:h:3;                                          % x space
y = zeros(1,length(x));                             % Memory allocation
y(1) = 5;                                           % initial condition

for i=1:(length(x)-1)                             
% i=1:(length(x)-1)                              % calculation loop
    k1 = F(x(i),y(i));
    k2 = F(x(i)+0.5*h,y(i)+0.5*h*k1);
    k3 = F((x(i)+0.5*h),(y(i)+0.5*h*k2));
    k4 = F((x(i)+h),(y(i)+k3*h));
    y(i+1) = y(i) + (1/6)*(k1+2*k2+2*k3+k4)*h;  % main equation
end
figure, plot(x, y)  % To see the solution results
end

Now you can test the above function with the followings, e.g. in the command window:

>> F = @(x, y)(x+y);
>> [OutputX,OutputY] = FunctionBeta_Executor(F);

Good luck.

[OvercookedRamen]

Thank you so much! You're a life saver.

[Sulaymon Eshkabilov]

It is a pleasure that I could be of some help.