% Suppose a new problem of y'' = -ty^2, 
% with boundary conditions y(0) = -1 and y(1) = 2

% In the ODE from lecture 17 that would imply:
% p(t) = 0, q(t) = -t*y and r(t) = 0.

% Resulting in 
% A(t) = 1
% B(t) = -[2 + q(t)dt^2]
% C(t) = 1
% rhs = 0;

% Suppose we want this for N times tn, then we
% construct a matrix of size N-2:

N = 10;        % number of total points
n = N-2;       % number of interior points
dt = 1/(n+1);  % time step size
t = (0:dt:1)';

% Construct b including boundary conditions
b = zeros(n,1);
b(1) = 0 - 1*(-1);
b(end) = 0 - 1*2;

% Initialize x for the initial guess to the solution
xnp1 = ones(n,1);

At = zeros(n,n);

dx = ones(n,1);

while (norm(abs(dx)) > 10^(-3))
  xn = xnp1
  At = A(t,xn,n,dt); 
  f = F(xn,At,b);
  J = Jac(xn,At,b);
  dx = J\(-f);
  xnp1 = xn + dx
end


sol = zeros(N,1);
sol(1) = -1;
sol(end) = 2;
sol(2:N-1) = xn(1:n);
plot(t,sol,'g*-'); hold on;