# ODE system with 2 degrees of freedom

7 views (last 30 days)
Nader Mohamed on 27 Oct 2021
Commented: Nader Mohamed on 28 Oct 2021
I'm trying to write this system of odes and then solve it with ode45, but I'm having trouble writing the function for the system:
k,J1,J2 and b are known. I tried to write this code but it doesnt work:
function dydt = ODEsystem(t,y)
k = 100;
b = 10;
J1 = 0.2;
J2 = 0.1;
dydt = zeros(6,1);
dydt(1) = y(2);
dydt(2) = y(3);
dydt(3) = k/J1 * dydt(1) - k/J1 * dydt(4);
dydt(4) = y(5);
dydt(5) = y(6);
dydt(6) = b/J2 * dydt(5) + k/J2 * dydt(4) - k/J2 * dydt(1);
end

James Tursa on 27 Oct 2021
Edited: James Tursa on 27 Oct 2021
I only see two variables, θ1 and θ2, and each is part of a 2nd order equation. So that would mean a 2 x 2 = 4 element state vector, not 6 element state vector as you are using. E.g., I would have the states defined as
y(1) =
y(2) =
y(3) =
y(4) =
which would mean the derivative function would be:
function dydt = ODEsystem(t,y)
k = 100;
b = 10;
J1 = 0.2;
J2 = 0.1;
theta1ddot = k/J1 * y(1) - k/J1 * y(2);
theta2ddot = b/J2 * y(4) + k/J2 * y(2) - k/J2 * y(1);
dydt = [y(3);y(4);theta1ddot;theta2ddot];
end
Nader Mohamed on 28 Oct 2021
Thank you so much!

David Goodmanson on 27 Oct 2021
Edited: David Goodmanson on 28 Oct 2021
you only need four variables, theta1, theta1dot, theta2, theta2dot (not six). Try
function dydt = ODEsystem(t,y)
k = 100;
b = 10;
J1 = 0.2;
J2 = 0.1;
% y = [theta1; theta1dot; theta2; theta2dot]
dydt = zeros(4,1);
dydt(1) = y(2);
dydt(3) = y(4);
dydt(2) = (k/J1)*(y(1)-y(3));
dydt(4) = (b/J2)*y(4) +(k/J2)*(y(3)-y(1));
Nader Mohamed on 28 Oct 2021
Thank you so much!

R2018b

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