How I can check the system solution with a Matlab ODE function.

3 Ansichten (letzte 30 Tage)
dydt(1) = 1.3*(y(3) - y(1)) + 10400*exp(20.7 - 1500/y(1))*y(2);
dydt(2) = 1880 * (y(4) - y(2) * (1+exp(20.7 - 1500/y(1))));
dydt(3) = 1752 - 269*y(3) + 267*y(1);
dydt(4) = 0.1 + 320*y(2) - 321*y(4)
y(t0)= [50,0,600,1]

Akzeptierte Antwort

Jan
Jan am 11 Jun. 2018
function main
t0 = 0;
y0= [50,0,600,1]
[t,y] = ode45(@fcn, [t0, 7], y0);
plot(t, y);
end
function dydt = fcn(t, y)
dydt = zeros(4,1);
dydt(1) = 1.3*(y(3) - y(1)) + 10400*exp(20.7 - 1500/y(1))*y(2);
dydt(2) = 1880 * (y(4) - y(2) * (1+exp(20.7 - 1500/y(1))));
dydt(3) = 1752 - 269*y(3) + 267*y(1);
dydt(4) = 0.1 + 320*y(2) - 321*y(4);
end
  3 Kommentare
Jan
Jan am 11 Jun. 2018
I guessed the endpoint 7. This takes a long time, in fact. If you use 2, ODE45 can solve this in seconds. ODE45 is designed to integrate non-stiff ODEs. If your system is stiff, use e.g. ode23s.
tic
[t,y] = ode23s(@fcn, [t0, 7], y0);
toc
% Elapsed time is 0.045184 seconds.
Camilo Sánchez
Camilo Sánchez am 11 Jun. 2018
Thank for your answer. I already assumed it was something like that and I used ODE15S. Thanks for your support.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by