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)

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Ramesh 2024-3-13

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回答： Udit06 2024-3-20

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

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:

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Chuguang Pan 2024-3-13

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Hi @Sam Chak,

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 2024-3-13

Thanks @Chuguang Pan. @Ramesh could have checked all equations before posting the problem.

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Udit06 2024-3-20

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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