Finite difference method 1st degree

I have to solve this problem using a first degree finite difference method. I am really struggling to understand what is not working with my code. If someone could take a look at ir and explain to me what<s wrong I would really appreciate it. Here is the assignement and the code below

function[y]=problimite(N,P,Q,R, a, b, alpha, beta)
h=(b-a)/(N+1);
S=-1+P(1:end-1)*h/2;
D=2+Q(1:end)*h^2;
I=-1-P(2:end)*h/2
B=[(-R(1)*(h^2)+(1+P(1)*h/2))*alpha,-R(2:end-1).*h^2,-R(end)*(h^2)+(1+P(end)*h/2)*beta];
disp(S)
y = tridiagonal(N, S, D, I, B);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
function y = tridiagonal(N, S, D, I, B)

y(1)=B(1)/D(1);

for i=2:N
    S(i-1)=(S(i-1))/D(i-1);
    D(i)=D(i)-I(i-1)*S(i-1);
    y(i)=(B(i)-I(i-1)*y(i-1))/D(i);
end
x(N)=y(N);

for i=N-1:1
    x(i)=y(i)-S(i)*x(i+1);
end
%%script%%%%
b=1;
alpha=0;
beta=1;
X=[0.9:0.01:1];
p=@(x) -1/x;
q=@(x) 0;
r=@(x) (1.6)/(x^4);
N=9;
P=arrayfun(p,X);
R=arrayfun(r,X);
Q=arrayfun(q,X);
%h=.01;
%I=[-1-P(2:end)*h/2]
%D=[2+Q(1:end)*h^2]
%S=[-1+P(1:end-1)*h/2]
%display(P);
%display(R);
%display(Q);

plot(problimite(N,P,Q,R, a, b, alpha, beta))
hold on
%% partie 2 %%
X=[0.9:0.005:1];
p=@(x) -1/x;
q=@(x) 0;
r=@(x) (1.6)/(x^4);
N=19;
P=arrayfun(p,X);
R=arrayfun(r,X);
Q=arrayfun(q,X);
%display(P);
%display(R);
%display(Q);
plot(problimite(N,P,Q,R, a, b, alpha, beta))

[darova]

Can't you use bvp4c? It can solve this task

[J. Alex Lee]

@darova, this looks like a homework assignment requiring hand-coding; it even appears to have an analytical solution for verification.

@Patirick, what do you mean by "first degree" finite difference method? Do you mean first order accuracy? If so I'm not sure there is a first order accurate difference formula for second order derivatives...You can break up your problem into a pair of coupled first order ODEs and use a pair of first order accurate finite differencing schemes. Perhaps you can explain in words the strategy that your code is implementing before jumping into the coding.