How to define k in bcfun?

function dYdy = odefun( x, Y, s, d )
T = Y(1);
dTdy = Y(2);
k = 0.001 * T + 1.2;      % function of x, T, dTdx, etc.
f = -s/d;                         % function of x, T, dTdx, etc.
dYdy = [ dTdy / k; f ];

end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                                                        %
% Function evaluating the boundary conditions: res = F( ya, yb, ... )    %
%                                                                        %
% For this problem: y = [ y1 y2 ], y1 = T, y2 = k*dT/dy                    %
%                                                                        %
%                                                                        %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function  res = bcfun( ya, yb, h, q0, Ta,)

res = [ ya(2) + q0/k ;  

          yb(2) + h*(yb(1) - Ta)/k ];
    
end

Answer 1

[Walter Roberson]

In bcfun, for ya, you sould calculate k based on the x that would be used for the first boundary condition, and for yb you should calculate calculate k based on the x that would be used for the second boundary condition.