Hi community, i request assistance in getting the code for this particular question.
I tried watching yotube and looking around matlab answer but i still don't understand the approach in solving it.
Please advice, thanks.

 Réponse acceptée

Per Hyldahl
Per Hyldahl le 19 Fév 2020

0 votes

Hi,
You need to treat your 2nd order differential equation as a system of two 1st order equations and arrange them in a vector, like:
f = [y; y']
such that
f' = [y'; y'']
Then you can obtain the solution using the following code
clc; clear; close all
[t,y] = ode45(@deriv, [0, 25], [0, 0]);
plot(t, y(:,1), t, y(:,2))
legend('y(t)', 'y''(t)')
function f_prime = deriv(t,f)
f_prime = zeros(2,1);
f_prime(1) = f(2);
f_prime(2) = 3*cos(t) -1.5*sin(t) - 3*f(2) - 3.25*f(1);
end
I myself also had problems to wrap my head around this approach when i learned it :)
// Per

5 commentaires

David Lee
David Lee le 19 Fév 2020
I know [0,0] is from the initial condition
But may I know how you derived the number [0, 25]
Per Hyldahl
Per Hyldahl le 19 Fév 2020
The array [0, 25] is the time vector and I just chose som arbitrary number; hence starting at t= 0s and endning at t=25s
David Lee
David Lee le 19 Fév 2020
Then how does the function at the 2nd part of the script contribute to the plot when its made after the plot of the graph?
Per Hyldahl
Per Hyldahl le 20 Fév 2020
Hi,
The sub-routine 'deriv' is a function which evaluates the differential equation, and is given to the ode45 as an input argument.
As Hiro-San suggested in a different answer, you should read the documentation of the ODE-solver suite; e.g. for ODE45: https://www.mathworks.com/help/matlab/ref/ode45.html.
Especially, the section regarding input arguments.
/ Per
David Lee
David Lee le 22 Fév 2020
Thanks!

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Hiro Yoshino
Hiro Yoshino le 19 Fév 2020

0 votes

You can walk through this:
I guess this is what you work on - read, understand and apply to your homework.

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