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function [ x ] = solve_tridiag(dU,odL,odU,b)
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
% dU: upper matrix diagonal
% odU: upper matrix diagonals, order 4, 8, 7
% odL: diagonal of lower matrix, order: 2, 6, 3
% b: 3x1 matrix
% for general case
upper_diag = diag(dU);
width = length(upper_diag);
if width == 3 % for function we made
L = eye(3) + diag(odL(1:2),-1) + diag(odL(3),-2); % creates L matrix
U = upper_diag + diag(odU(1:2),1) + diag(odU(3),2); % creates U matrix
% for forward substitution
for i = 1:width
x(i) = b(i);
for k = 1:i-1
x(i) = x(i) - L(i,k)*x(k);
end
x(i) = x(i)/L(i,i);
end
b = x';
% for backwards substitution
for i = width:-1:1
x(i) = b(i);
for k = i+1:width
x(i) = x(i) - U(i,k)*x(k);
end
x(i) = x(i)/U(i,i);
end
x = x';
else % for A matrices
U = diag(dU);
k = 1;
for i = 1:width-1
ofd = odU(k:k+width-i-1);
U = U + diag(ofd, i);
k = k + width - i;
end
% makign L matrix
L = eye(width);
k = 1;
for i = 1:width-1
ofdL = odL(k:k+width-i-1);
L = L + diag(ofdL, -i);
k = k + width - i;
end
% for forward substitution
for i = 1:width
x(i) = b(i);
for k = 1:i-1
x(i) = x(i) - L(i,k)*x(k);
end
x(i) = x(i)/L(i,i);
end
b = x';
% for backwards substitution
for i = width:-1:1
x(i) = b(i);
for k = i+1:width
x(i) = x(i) - U(i,k)*x(k);
end
x(i) = x(i)/U(i,i);
end
x = x';
end
end