# how to developed numerical code for collocation method for linear differential equations

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Anup Negi on 6 Sep 2020
Commented: Anup Negi on 8 Sep 2020
our big problem to creat code in MATLAB
i can't undestand , how to write down the diifential equation and boundary condition in matlab
Anup Negi on 8 Sep 2020
Can you send your eamil I'd sr

Alan Stevens on 8 Sep 2020
Edited: Alan Stevens on 8 Sep 2020
I've just noticed you want to use a collocation method. So you proceed as follows
% For method of collocation choose a functional form for U
% Suppose, for simplicity you choose a simple parabola
% u(z) = a + b*z + c*z^2 (u is potential approximate version of U)
% dudz(z) = b + 2c*z
% d2ud2 = 2*c
% where a, b and c are constants to be determined.
%
% The actual ODE is
% d2Udz2 + A*dUdz + B = 0 where (presumably) you know A and B.
%
% so with your approximate solution this would become
% 2c + A*(b + 2c*z)+ B = 0
% choose a collocation point, z = zcol, say and this becomes
% 2c + A*(b + 2c*zcol)+ B = 0 ...(1)
%
% Assuming you know U on your boundaries, z = eta and z = 1 you also have
% U(eta) = a + b*eta + c*eta^2 ...(2)
% U(1) = a + b + c ...(3)
%
% You now have three equations with which to find a, b and c
% This can be done easily with
% M = [0 A 2+2*A*zcol; 1 eta eta^2; 1 1 1];
% V = [-B; U(eta); U(1)];
% [a; b; c] = M\V;
%
% so your approximate solution is: u(z) = a + b*z + c*z^2
%
% If you want to use more collocattion points you'll need to
% choose a higher order approximate equation.

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