Solve the differential equation with adams

My assingment is this: The following second-order differential equation y′′(x) = (1 − y(x)^2) y(x)′ - y(x), y (0) = 1, y ′ (0) = 1 convert it to a system of first-order differential equations and solve it with the given method. Solve the differential equation with adams for h = 1/3 and h = 1/6. How many times does the approximation error decrease (calculate the error coefficient err (h = 1/3) / err (h = 1/6)), for y (10). For the exact value, take e.g. the one obtained with ode45 for a small enough step. This is how I solve it but I do not get the correct answer, especially I got the error1 = NaN

function [x,y] = RungeKutta(f, a, b, y0, h)
x = a:h:b; 
y(:,1) = y0;
A = [0 0 0 0 0 ; 1/3 1/3 0 0 0 ; 2/3 -1/3 1 0 0 ; 1 1 -1 1 0 ;0 1/8 3/8 3/8 1/8];
d = length(y0); 
k = zeros(d,length(A)-1); 
for z = 1:length(x)-1 
    for i=1:length(A)-1 
        sum2 = 0; 
        for j=2:length(A) 
            sum2 = sum2 + A(i,j)*k(:,j-1);
        end
        k(:,i) = h*f(x(z)+A(i,1)*h,y(:,z)+sum2);
    end
    sum3 = 0; 
    for i = 2:length(A)
        sum3 = sum3 + A(length(A),i)*k(:,i-1); 
    end
    y(:,z+1) = y(:,z) + sum3; 
end
end

function [x,y] = adams(f,a,b,y0,h)
    x = a:h:b; 
    y = zeros(length(y0),length(x)); 
    y(:,1) = y0;
    [~,yr] = RungeKutta(f,a,b,y0,h);
    y(:,2) = yr(:,2);
    for i=1:length(x)-2
        y(:,i+2) = y(:,i+1)+h*(3/2*f(x(i+1),y(:,i+1))-1/2*f(x(i),y(:,i)));
    end
end

F = @(x,y) ([y(2);(1-y(1).^2).*y(2)-y(1)]);
y0 = [1;1];

[xa1, ya1] = adams(F,0,10,y0,1/3);
[xo1,yo1] = ode45(F,0:1/3:10,y0, odeset('RelTol', 10^(-11), 'AbsTol', 10^(-11)));
[xa2, ya2] = adams(F,0,10,y0,1/6);
[xo2,yo2] = ode45(F,0:1/6:10,y0, odeset('RelTol', 10^(-11), 'AbsTol', 10^(-11)));
error1 = abs(ya1(1,end)-yo1(end,1));
error2 = abs(ya2(1,end)-yo2(end,1));
result2 = error1/error2;

[Torsten]

You didn't define "F" anywhere.

You didn't define "y0" anywhere.

You didn't define "vsota2" anywhere.

[Torsten]

You "unformatted" your code. But the settings for the above parameters are still missing.

[Nikodin Sedlarevic]

Now I defined it. I hope it works now

[Torsten]

Except for the line

[~,yr] = RungeKutta(f,a,b,y0,h);

which should read

[~,yr] = RungeKutta(f,a,a+h,y0,h);

your Adams method is coded correctly. But the step size h=1/3 is too large for Adams: it diverges.

Thus the error coefficient cannot be evaluated.

[Nikodin Sedlarevic]

So how can I correct then this code, other then RungeKutta?

[Torsten]

The code is correct - the assignment is not well-thought-out.

You can't change anything - only your teacher can.

You don't need to call ode45 twice - it has an adaptive step size control, and the tspan interval you supply is for output purposes only. It does not influence the error of integration. The tolerated error can only be influenced by the prarameters AbsTol and RelTol which you chose equal in both cases.