Need help with matlab code

Need to numerically solve the differential equation below and plot T vs t which goes like radiative cooling curve.

dT/dt= -Q(T^4-T0^4)/3*L*T where T =1000K, T0=300K, Q=1.38e-23 and L=1e-4 m.

Thank you in advance!

Answer 1

[Star Strider]

Tv=1000; % K
T0=300; % K 
Q=1.38e-23; 
L=1e-4; % m.
syms T(t)
DE = diff(T) == -Q*(Tv^4-T0^4)/3*L*T;
DEs = dsolve(DE, T(0)==T0)

figure
fplot(DEs, [0 1E+16])
grid
xlabel('$t$',  'Interpreter','latex')
ylabel('$T(t)$',  'Interpreter','latex')
title(['$T(t) = ' latex(DEs) '$'], 'Interpreter','latex')

Or, use ode45.

[Emma K.]

Hi, T and Tv are the same but when I do that, the graph turns to be kind of linear instead of a curve. Again why do you have 1E+16 in the fplot command line? Thank you

[Walter Roberson]

If T0 and Tv are the same then Tv^4 - T0^4 would be 0, and the right hand side of the differential equation would be 0, which implies a linear equation.

[Emma K.]

No,I wrote T and Tv are the same not T and T0.

[Star Strider]

‘... T and Tv are the same ...’

Not really. Here, ‘T’ is the function and ‘Tv’ is one value (as is ‘T0’).

When plotted, it appears linear over relatively short intervals for ‘t’, while over longer intervals (such as I use here, and the answer to your other question about it), it has the expected exponential decay.

To do a true numerical integration, you will likely want to use ode45. This code demonstrates a symbolic integration, and what the correct plot looks like.

[Emma K.]

Thank you Strider, a little tweak helped resolved the problem.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

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