finite difference method code

% set domains limits and boundary conditions
xo = pi/2; xf = pi; yxo = 1; yxf = 1; N = 10;

% compute interval size and discrete x vector 
  dx = (xf-xo)/N;  dx2 = dx*dx; x = (xo+dx):dx:xf;
  
  % analytica solution (exact)
    xe = linspace(xo,xf,N);
    ye = (pi./(2*xe)).*(sin(xe) - 2*cos(xe));
    
    % arranging the matrix a
    %node 1
      a(1,1)=dx2-2; a(1,2)=1+(dx/(xo+dx));  b(1)= ((yxo*dx) /(xo*dx))-yxo;  
    
    
    for i = 2:N-1
        a(i,i-1) = (1-(dx/x(i)));
        a(i,i) = dx2-2;
        a(i,i+1) = (1+(dx/x(i)));
        b(i)=0;
    end
        
    a(N,N-1)=(2*xf+2*dx)/xf;  a(N,N-2)=-1;  b(N)=yxf*dx2+yxf+((2*yxf*dx)/xf);
    
    yi=a\b;

i keep getting the following error code

finite_1

Error using \

Matrix dimensions must agree.

Error in finite_1 (line 26)

yi=a\b

Answer 1

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/mldivide.html

• If A is a rectangular m-by-n matrix with m ~= n, and B is a matrix with m rows, then A\B returns a least-squares solution to the system of equations A*x= B.

Your A is 10 x 10, and your b is 1 x 10, which has 1 row, rather than the 10 rows needed to match the 10 rows of a

It would be legal to use a\b' but you will need to decide whether that is meaningful for your situation.