solving 2nd order nonlinear ode Numeric solution by using ode45
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the eqution
d^2u/dt -k(1-u^2)du/dt+au = 0
initial condition
u(0)=2 (dimensionless); du/dt (0)=0
question
(a) With 𝑘𝑘 = 1.0 s-1, determine the value of 𝑎𝑎 that would give a heart rate of 1.25 beats/second and Graphically display 𝑢𝑢(t) for this value of 𝑎𝑎 and 0 ≤ t≤ 5 𝑠𝑠 . (25 points).
(b) Graphically display 𝑢𝑢(t) for your chosen values 𝑘𝑘 and 𝑎𝑎 and 0 ≤ t ≤ 5 𝑠𝑠 . Interpret the results.
采纳的回答
Alan Stevens
2020-10-4
This is the basic structure for solving the ode.
u0 = 2;
v0 = 0;
tspan = [0 5];
k = 1;
a = 25;
[t,U] = ode45(@odefn, tspan, [u0 v0],[],k,a);
u = U(:,1);
v = U(:,2);
plot(t,u),grid
xlabel('t'),ylabel('u')
function dUdt = odefn(~,U,k,a)
u = U(1);
v = U(2); % v = du/dt
dvdt = k*(1-u^2)*v - a*u;
dUdt = [v;
dvdt];
end
You could investigate fzero to get the value of k that gives 1.25 beats/sec, or adjust it manually (as I did here to get an approximate value).
4 个评论
Aseel Alamri
2020-10-4
Alan Stevens
2020-10-4
1.25 beats per second means there will be four beats in five seconds. Since you are asked to plot for 5 seconds, adjust a until you get exactly four peaks on the graph in that time. Frequency is just the inverse of the period.
Aseel Alamri
2020-10-4
great , What about part (b)
Alan Stevens
2020-10-4
Try playing with k.
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