ode45 for Higher Order Differential Equations
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I'm learning Matlab and as an exercise I have following:
+ 6 = 1 -
on the time interval [0 20]and with integration step < 0.01. Initial conditions are x(0) = 0.44; = 0.13; = 0.42; = -1.29
To solve it I wrote the code:
f = @(t, y) [y(4); y(3); y(2); 1 - y(1)^2 - 6*y(3)];
t = [0 100];
f0 = [-0.44; 0.13; 0.42; -1.29];
[x, y] = ode45(f, t, f0);
plot(x,y, '-');
grid on
However I'm not really sure that it's correct even it launches and gives some output. Could someone please have a look and correct it if it's wrong. Also where is integration step here?
1 Kommentar
Star Strider
am 18 Jun. 2022
Bearbeitet: Star Strider
am 18 Jun. 2022
The integration is performed within the ode45 function. To understand how it works to do the integration, see Algorithms and the Wikipedia article on Runge-Kutta methods.
Akzeptierte Antwort
Sam Chak
am 18 Jun. 2022
Bearbeitet: Sam Chak
am 18 Jun. 2022
Hi @d
A little fix on the system of 1st-order ODEs.
f = @(t, x) [x(2); x(3); x(4); 1 - x(1)^2 - 6*x(3)];
tspan = 0:0.01:20;
x0 = [0.44; 0.13; 0.42; -1.29];
[t, x] = ode45(f, tspan, x0);
plot(t, x, 'linewidth', 1.5)
grid on, xlabel('t'), ylabel('\bf{x}'), legend('x_{1}', 'x_{2}', 'x_{3}', 'x_{4}', 'location', 'best')
2 Kommentare
Sam Chak
am 18 Jun. 2022
You are welcome, @d. The link suggested by @Star Strider has many good examples and 'tricks' to learn.
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