Solving a first order ODE with Euler backwards method

I'm trying to solve the ODE below: F_ty = @(t,y) (2*t-4)*exp(-y); using Euler backwards method with the intial condition y(5) = 0. But I keep getting these two error messages, any help would greatly be appreciated, thanks in advance! The following two errors are:

Exiting fzero: aborting search for an interval containing a sign change

because NaN or Inf function value encountered during search.

(Function value at -799.643 is -Inf.)

Check function or try again with a different starting value.

Error using fzero (line 214)

Second argument must be finite.

Error in backwardeuler (line 22)

y(i+1) = fzero(@(Y) y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y, y(i), options);

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% y_true = log(t^2 -4*t-4); Initial condition y(5) = 0;
% F_ty = @(t,y) (2*t-4)*exp(-y);
dt = 0.01;
t0 = 0;
tf = 3;

t = t0:dt:tf;
y(5) = 0;

% using the formula for backward euler: y(i+1) = y(i) + dt*f(y(i+1),t(i+1))
% we get
%y(i+1) = y(i) + dt*((2*t(i+1)-4)*exp(-y(i+1)));
% setting the LHS equal to zero so we can use fsolve:
% 0 = y(i) + dt*((2*t(i+1)-4)*exp(-y(i+1))) - y(i+1);
% We define y(i+1) = Y, so that 
% 0 = y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y;

options = optimset('TolX',1e-06);

for i = 1:length(t)-1
    
    y(i+1) = fzero(@(Y) y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y, y(i), options);
    y_exact(i+1) = log((t(i+1))^2-4*t(i+1)-4);
end

Exiting fzero: aborting search for an interval containing a sign change
    because NaN or Inf function value encountered during search.
(Function value at -799.643 is -Inf.)
Check function or try again with a different starting value.

Error using fzero (line 214)
Second argument must be finite.

figure(1)
hold on 
plot(t,y,'bo')
plot(t,y_exact,'r-')
xlabel('time')
ylabel('y(t)')
title('Backward Euler method vs exact solution')
legend('Backward Euler', 'Exact')

Answer 1

[Alan Stevens]

Your y_true is only valid for t>= 5 (smaller values give imaginary results for y). So, try going from 5 to 8:

% y_true = log(t^2 -4*t-4); Initial condition y(5) = 0;
% F_ty = @(t,y) (2*t-4)*exp(-y);
dt = 0.01;
t0 = 5;
tf = 8;

t = t0:dt:tf;
y(5) = 0;

for i = 1:length(t)-1
    
    y(i+1) = fzero(@(Y) y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y, y(i));
    y_exact(i+1) = log((t(i+1))^2-4*t(i+1)-4);
end

figure(1)
hold on 
plot(t,y,'bo')
plot(t,y_exact,'r-')
xlabel('time')
ylabel('y(t)')
title('Backward Euler method vs exact solution')
legend('Backward Euler', 'Exact')