solving 2nd order nonlinear ode Numeric solution by using ode45

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.

Answer 1

[Alan Stevens]

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).

[Aseel Alamri]

so this part a

how or what is the formela that I can use to calculate the the time between peaks?

and calculate the frequency ?

[Alan Stevens]

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]

great , What about part (b)

[Alan Stevens]

Try playing with k.