Error in Finite Difference Method for ODE Code

I'm working on an ODE Finite Difference Method. I think I have it down (so I deleted my previous question), but I am getting this error:

Array indices must be positive integers or logical values.
Error in FiniteDifference (line 18)
p(X) = (1 + 0*X); q(X) = (2 + 0*X); r(X) = cos(X);

Here is my code:

 function [A,B,W] = FiniteDifference(N)

 %equation
 %y'' = y' + 2y + cosx,   0<x<pi/2
 %y(0) = -0.3, y(pi/2) = -0.1

 %breaking it down:
 % a = 0; alpha = -0.3; b = pi/2; beta = -0.1; 
 % p(x) = 1; q(x) = 2; r(x) = cosx;

 %to check: plot and residue
 A = zeros(1,N+1);  B = zeros(1,N+1); C = zeros(1,N+1);  D = zeros(1,N+1);  
 L = zeros(1,N+1);  U = zeros(1,N+1); Z = zeros(1,N+1);  W = zeros(1,N+1);
 aa = 0;    alpha = -0.3;
 bb = pi/2; beta = -0.1;
 h = (bb-aa)/(N+1); %Step Size
 X = aa + h;
 p(X) = (1 + 0*X); q(X) = (2 + 0*X); r(X) = cos(X);
 A(1) = 2+h^2*q(X);
 B(1) = -1+0.5*h*p(X);
 D(1) = -h^2*r(X)+(1+0.5*h*p(X))*alpha;

 for i = 2:N-1
    X = aa + i*h;
    A(i) = 2+h^2*q(X);
    B(i) = -1+0.5*h*p(X);
    C(i) = -1-0.5*h*p(X);
    D(i) = -h^2*r(X);
 end

 X = bb-h;
 A(N) = 2 + h^2*q(X);
 C(N) = -1 - 0.5*h*p(X);
 D(N) = -h^2*r(X) + (1 - 0.5*h*p(X))*beta;

 L(1) = A(1);
 U(1) = B(1)/A(1);
 Z(1) = D(1)/L(1);

 for i = 2:N-1
    L(i) = A(i) - C(i)*U(i-1);
    U(i) = B(i)/L(i);
    Z(i) = (D(i) - C(i)*Z(i-1))/L(i);
 end

 L(N) = A(N) - C(N)*U(N-1);
 Z(N) = (D(N) - C(N)*Z(N-1))/L(N);

 W(N) = Z(N);

 for j = 1:N-1
    i = N-j;
    W(i) = Z(i) - U(i)*W(i+1);
 end
 i = 0;

 for i = 1:N
    X = aa + i*h;
 end
 plot(X,W);
 end

Answer 1

[Torsten]

Maybe you mean

p(X) = @(X)(1 + 0*X); q(X) = @(X)(2 + 0*X); r(X) = @(X)cos(X);

?

Best wishes

Torsten.

[Mel Hernandez]

I completely forgot to add that. Thank you! However, it's still giving me the same error. I'm not sure what else I missed.

[Torsten]

Sorry, should have been

p = @(X)(1 + 0*X); q = @(X)(2 + 0*X); r = @(X)cos(X);

Best wishes

Torsten.

[Jan]

r = @(X)cos(X)? What about r=@cos? Anonymous functions are much slower than function handles. q = @(X)(2 + 0*X)? Looks like: q = 2. Do I oversee something?

[Mel Hernandez]

Tortsen, thank you so much!!!