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
= 0.13; = 0.42; = -1.29To 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 个评论
Star Strider
2022-6-18
编辑：Star Strider
2022-6-18
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.
采纳的回答
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 个评论
d
2022-6-18
Sam Chak
2022-6-18
You are welcome, @d. The link suggested by @Star Strider has many good examples and 'tricks' to learn.
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