Hi Christella,
The primary issue in the code is the approach to finding (k) using the initial conditions and then using it to solve for (tfinal). Solving for (k) by equating two solutions of the differential equation at (t=0) does not mathematically make sense as it is trying to use the same differential equation with two different initial conditions separately, and then somehow equate these solutions to solve for (k). The correct approach is to use the information from both initial conditions together to solve for (k) in a single step, and then use this (k) to find when the temperature reaches (Ufinal).
Here's the modified code :
syms t k U(t);
% Initial Conditions
cond1 = U(0) == 165;
cond2 = U(10) == 150;
Ta = 35; % Ambient Temperature
Ufinal = 70; % Final Temperature
% Set the differential equation model
eqn = diff(U, t) == -k*(U - Ta);
% Solve the differential equation with the first initial condition to find a general solution
Usol = dsolve(eqn, cond1);
% Use the second condition to find k
kSol = solve(subs(Usol, t, 10) == 150, k);
% Substitute k back into the solution
UsolSub = subs(Usol, k, kSol);
% Solve for the time when the temperature is Ufinal
tfinal = solve(UsolSub == Ufinal, t);
% Prepare for plotting
fplot(subs(UsolSub, t), [0, double(tfinal) + 5], 'LineWidth', 2);
hold on;
title('Cake Temperature');
xlabel('Time (in Minutes)');
ylabel('Temperature (F)');
grid on;
plot(0, 165, 'ro'); % Initial condition point
plot(10, 150, 'ro'); % Second condition point
plot(double(tfinal), Ufinal, 'rx'); % Final condition point
legend('Temperature', 'Initial Condition', 'Condition at t=10', 'Ufinal', 'Location', 'Best');
hold off;
Hope this helps.
