dTdt=−k(T−Ta) k=0.4,Ta=25∘, initial condition T(0)=T0=75∘C .how to solve using ode45 and report the temperature obtained at times t=1,2,3 and 4 . report T(1.0)

4 Ansichten (letzte 30 Tage)
ORIGINAL:
dTdt=k(TTa)
k=0.4,Ta=25C, k=0.4,Ta=25° and the initial condition T(0)=T0=75
The true solution of the ODE is given by:
T(t)=Ta+(T0Ta)ekt
Mathematically translated by @Sam Chak (in the original way) using LaTeX:
, °C, and the initial condition °C
The true solution of the ODE is given by:
  9 Kommentare
Chuguang Pan
Chuguang Pan am 13 Mär. 2024
I think the true solution of T should be , so the formula of T in your code should be:
T=@(t) Ta + (T0-Ta)*exp(-k*t);

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Antworten (1)

Udit06
Udit06 am 20 Mär. 2024
Hi Ramesh,
You can follow the following steps to solve a differential equation using "ode45" function
1) Define the ODE function:Create a function that represents the right-hand side of the ODE.
2) Solve the ODE using ode45: "ode45" function is used for solving initial value problems for ordinary differential equations. Give the function handle, time span and initial conditions as input to the function.
3) Extract the Temperatures at desired times: After solving the ODE, you can extract the temperature values at the desired times using interpolation
Here is the code implementation for the same:
% Define the parameters
k = 0.4;
Ta = 25;
T0 = 75;
% Define the ODE function
odefun = @(t, T) -k * (T - Ta);
% Time span (from t=0 to t=4)
tspan = [0 4];
% Initial condition
T_initial = T0;
% Solve the ODE
[t, T] = ode45(odefun, tspan, T_initial);
% Interpolate or find the closest values to t=1, 2, 3, and 4
T1 = interp1(t, T, 1.0); % Interpolating for T at t=1.0
% Display the result for T(1.0)
fprintf('The temperature at t=1.0 is approximately %.2f°C\n', T1);
The temperature at t=1.0 is approximately 58.52°C
You can refer to the following MathWorks documentation to understand more about "ode45" function.
I hope this helps.

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