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)
조회 수: 4 (최근 30일)
이전 댓글 표시
ORIGINAL:
dTdt=−k(T−Ta)
k=0.4,Ta=25∘C, 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+(T0−Ta)e−kt
, °C, and the initial condition °C
The true solution of the ODE is given by:
댓글 수: 9
Chuguang Pan
2024년 3월 13일
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);
Sam Chak
2024년 3월 13일
답변 (1개)
Udit06
2024년 3월 20일
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);
You can refer to the following MathWorks documentation to understand more about "ode45" function.
I hope this helps.
댓글 수: 0
참고 항목
카테고리
Help Center 및 File Exchange에서 Ordinary Differential Equations에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!