Help with using the Runge-Kutta 4th order method on a system of 2 first order ODE's. The original ODE I had was $$ \frac{d^2y}{dx^2}+\frac{dy}{dx}-6y=0$$ with $y(0)=3$ and $y'(0)=1$. Now I can solve this by hand and obtain that $y(1) = 14.82789927$. However I wish to use the 4th order Runge-Kutta method, so I have the system:
$$
\left\{\begin{array}{l}
  \frac{dy}{dx} = z \\
  \frac{dz}{dx} = 6y - z
\end{array}\right.
$$
With $y(0)=3$ and $z(0)=1$.
Now I know that for two general 1st order ODE's $$ \frac{dy}{dx} = f(x,y,z) \\ \frac{dz}{dx}=g(x,y,z)$$ The 4th order Runge-Kutta formula's for a system of 2 ODE's are: $$ y_{i+1}=y_i + \frac{1}{6}(k_0+2k_1+2k_2+k_3) \\ z_{i+1}=z_i + \frac{1}{6}(l_0+2l_1+2l_2+l_3) $$ Where $$k_0 = hf(x_i,y_i,z_i) \\ k_1 = hf(x_i+\frac{1}{2}h,y_i+\frac{1}{2}k_0,z_i+\frac{1}{2}l_0) \\ k_2 = hf(x_i+\frac{1}{2}h,y_i+\frac{1}{2}k_1,z_i+\frac{1}{2}l_1) \\ k_3 = hf(x_i+h,y_i+k_2,z_i+l_2) $$ and $$l_0 = hg(x_i,y_i,z_i) \\ l_1 = hg(x_i+\frac{1}{2}h,y_i+\frac{1}{2}k_0,z_i+\frac{1}{2}l_0) \\ l_2 = hg(x_i+\frac{1}{2}h,y_i+\frac{1}{2}k_1,z_i+\frac{1}{2}l_1) \\ l_3 = hg(x_i+h,y_i+k_2,z_i+l_2)$$
My problem is I am struggling to apply this method to my system of ODE's so that I can program a method that can solve any system of 2 first order ODE's using the formulas above, I would like for someone to please run through one step of the method, so I can understand it better.
 A: A Matlab implementation is given below:
% It calculates ODE using Runge-Kutta 4th order method
% Author Ido Schwartz
% Originally available form: http://www.mathworks.com/matlabcentral/fileexchange/29851-runge-kutta-4th-order-ode/content/Runge_Kutta_4.m
% Edited by Amin A. Mohammed, for 2 ODEs(April 2016)

clc;                                               % Clears the screen
clear all;

h=0.1;                                             % step size
x = 0:h:1;                                         % Calculates upto y(1)
y = zeros(1,length(x)); 
z = zeros(1,length(x)); 
y(1) = 3;                                          % initial condition
z(1) = 1;                                          % initial condition
% F_xy = @(t,r) 3.*exp(-t)-0.4*r;                  % change the function as you desire
F_xyz = @(x,y,z) z;                                  % change the function as you desire
G_xyz = @(x,y,z) 6*y-z;

for i=1:(length(x)-1)                              % calculation loop
    k_1 = F_xyz(x(i),y(i),z(i));
    L_1 = G_xyz(x(i),y(i),z(i));
    k_2 = F_xyz(x(i)+0.5*h,y(i)+0.5*h*k_1,z(i)+0.5*h*L_1);
    L_2 = G_xyz(x(i)+0.5*h,y(i)+0.5*h*k_1,z(i)+0.5*h*L_1);
    k_3 = F_xyz((x(i)+0.5*h),(y(i)+0.5*h*k_2),(z(i)+0.5*h*L_2));
    L_3 = G_xyz((x(i)+0.5*h),(y(i)+0.5*h*k_2),(z(i)+0.5*h*L_2));
    k_4 = F_xyz((x(i)+h),(y(i)+k_3*h),(z(i)+L_3*h)); % Corrected        
    L_4 = G_xyz((x(i)+h),(y(i)+k_3*h),(z(i)+L_3*h));

    y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
    z(i+1) = z(i) + (1/6)*(L_1+2*L_2+2*L_3+L_4)*h;  % main equation

end

A: Although this answer contains the same content as Amzoti's answer, I think it's worthwhile to see it another way.
In general consider if you had $m$ first-order ODE's (after appropriate decomposition). The system looks like
\begin{align*}
\frac{d y_1}{d x} &= f_1(x, y_1, \ldots, y_m) \\
\frac{d y_2}{d x} &= f_2(x, y_1, \ldots, y_m) \\
&\,\,\,\vdots\\
\frac{d y_m}{d x} &= f_m(x, y_1, \ldots, y_m) \\
\end{align*}
Define the vectors $\vec{Y} = (y_1, \ldots, y_m)$ and $\vec{f} = (f_1, \ldots, f_m)$, then we can write the system as
$$\frac{d}{dx} \vec{Y} = \vec{f}(x,\vec{Y})$$
Now we can generalize the RK method by defining 
\begin{align*}
\vec{k}_1 &= h\vec{f}\left(x_n,\vec{Y}(x_n)\right)\\
\vec{k}_2 &= h\vec{f}\left(x_n + \tfrac{1}{2}h,\vec{Y}(x_n) + \tfrac{1}{2}\vec{k}_1\right)\\
\vec{k}_3 &= h\vec{f}\left(x_n + \tfrac{1}{2}h,\vec{Y}(x_n) + \tfrac{1}{2}\vec{k}_2\right)\\
\vec{k}_4 &= h\vec{f}\left(x_n + h, \vec{Y}(x_n) + \vec{k}_3\right)
\end{align*}
and the solutions are then given by
$$\vec{Y}(x_{n+1}) = \vec{Y}(x_n) + \tfrac{1}{6}\left(\vec{k}_1 + 2\vec{k}_2 + 2\vec{k}_3 + \vec{k}_4\right)$$
with $m$ initial conditions specified by $\vec{Y}(x_0)$. When writing a code to implement this one can simply use arrays, and write a function to compute $\vec{f}(x,\vec{Y})$
For the example provided, we have $\vec{Y} = (y,z)$ and $\vec{f} = (z, 6y-z)$. Here's an example in Fortran90:
program RK4
    implicit none   
    integer , parameter :: dp = kind(0.d0)
    integer , parameter :: m = 2 ! order of ODE
    real(dp) :: Y(m)
    real(dp) :: a, b, x, h
    integer :: N, i 

    ! Number of steps
    N = 10

    ! initial x
    a = 0
    x = a

    ! final x
    b = 1

    ! step size
    h = (b-a)/N

    ! initial conditions
    Y(1) = 3 ! y(0)
    Y(2) = 1 ! y'(0)

    ! iterate N times
    do i = 1,N
        Y = iterate(x, Y)
        x = x + h
    end do

    print*, Y


contains

    ! function f computes the vector f

    function f(x, Yvec) result (fvec)
        real(dp) :: x
        real(dp) :: Yvec(m), fvec(m)

        fvec(1) = Yvec(2) !z
        fvec(2) = 6*Yvec(1) - Yvec(2) !6y-z

    end function

    ! function iterate computes Y(t_n+1)

    function iterate(x, Y_n) result (Y_nplus1)
        real(dp) :: x
        real(dp) :: Y_n(m), Y_nplus1(m)
        real(dp) :: k1(m), k2(m), k3(m), k4(m)

        k1 = h*f(x, Y_n)
        k2 = h*f(x + h/2, Y_n + k1/2)
        k3 = h*f(x + h/2, Y_n + k2/2)
        k4 = h*f(x + h, Y_n + k3)

        Y_nplus1 = Y_n + (k1 + 2*k2 + 2*k3 + k4)/6

    end function

end program

This can be applied to any set of $m$ first order ODE's, just change m in the code and change the function f to whatever is appropriate for the system of interest. Running this code as-is yields
$   14.827578509968953    \qquad    29.406156886687729$
The first value is $y(1)$, the second $z(1)$, correct to the third decimal point with only ten steps.
A: A Fortran code shown below:

produces the following result

!Runge-Kutta Fourth Order Method

!For 2nd Order Differentiation Equation

!First you have to define the function

F(x,y,z) = z !dy/dx

G(x,y,z) = 6*y-z !dz/dx = d2y/dx2

INTEGER :: n,i

REAL :: k1,l1,k2,l2,k3,l3,k4,l4    !Most Important

Write (*,*) "Given Equation '(y2)-6(y1)+(y0)=0'"

Write (*,*) "Xo=0, Yo=3, Zo=Y'o=1, Xn=1, n=?"

Xo=0    !Given Condition

Yo=3    !Given Condition

Zo=1    !Given Condition

Xn=1    !Given Condition

read (*,*) n    !n=number of Intercept

h=(Xn-Xo)/n

do i=1,n    !you have to do the Calculation 'n' times

k1 = h*F(Xo,Yo,Zo)

l1 = h*G(Xo,Yo,Zo)

k2 = h*F(Xo+h/2,Yo+k1/2,Zo+l1/2)

l2 = h*G(Xo+h/2,Yo+k1/2,Zo+l1/2)

k3 = h*F(Xo+h/2,Yo+k2/2,Zo+l2/2)

l3 = h*G(Xo+h/2,Yo+k2/2,Zo+l2/2)

k4 = h*F(Xo+h,Yo+k3,Zo+l3)

l4 = h*G(Xo+h,Yo+k3,Zo+l3)

!Sum Up

Yn = Yo+(k1+2*k2+2*k3+k4)/6

Zn = Zo+(l1+2*l2+2*l3+l4)/6

!Operation for Next calculation

Xo=Xo+h     !(+h) than previous Term

Yo=Yn    !Now Yn becomes Yo

Zo=Zn    !Now Zn becomes Zo

End Do

Write (*,*) "Xn,Yn =",Xo,Yo

Stop

End

