% Here we are going to use the euler method to find an approximation of y(1.5)
    % given that the function y(x) satisfies the differential equation:
    %                    dy/dx = 2xy, y(1)=1

    function [y_n] = euler1(h)
    format long 
    % x is an array of n elements where n = length (x) . Initial value is 1
    % and final value is 1.5
    
     x =[1:h:1.5];
     
    % The loop will iterate n times where n is the number of elements in
    % x-array.
    
    n = length(x);

    % The y_n array will store the values of y_n(1), y_n(2), y_n(3) etc. The first value
    % y_n(1) equals to 1.
 
    y_n(1) = 1;  
      
    % The first element of the y_n array is set to 1 (y_n = 1).
    
    for i = 1:(n-1)
        y_n(i+1) = y_n(i)+h*2*x(i)*y_n(i);
    end
   
    % In figure 1 we plot the approximate solution
    
    %figure(1);
    %plot(x,y_n,'b')
    %hold on
    %a=y_n(i+1);