Solve 2nd order ODE using Euler Method

VERY new to Matlab...
Trying to implement code to use Euler method for solving second order ODE.
Equation:
x'' + 2*z*w*x' + w*x = 2*sin(2*pi*2*t)
z and w are constants. "t" is time.
Any help would be great.
Thanks!

5 Kommentare

John D'Errico
John D'Errico am 27 Sep. 2022
Bearbeitet: John D'Errico am 27 Sep. 2022
If you need to solve that ODE, then why in the name of god are you writing an Euler's method to solve the ODE. Use ODE45. Do not write your own code. Since the only reason you need to use Euler's method is to do this as a homework assignment, then you need to write your own code. But Answers is not a service where we do your homework with no effort shown by you.
Matt
Matt am 27 Sep. 2022
This is a HW assignment. No need to be a jerk about it. If you don't want to help then don't post.
Matt
Matt am 4 Okt. 2022
Verschoben: James Tursa am 4 Okt. 2022
So, I've been going at it a different way but still getting errors...
%Euler solution for ODE function
HW2_Parameters % Load initial conditions into the workspace
t0 = 0 ;
n1 = (tn-t0)/h ;
for i=1:n1
t(i+1) = t0 + i*h ;
x_dd(i) = (2*sin(2*pi*2*t(i))) - (2*zeta*omega_n*x_d(i-1)) - (omega_n*x(i-1));
x_d(i) = x_d(i-1) + (x_dd(i)*h);
x(i) = x(i-1) + (x_d(i)*h);
end
Any thoughts?
James Tursa
James Tursa am 4 Okt. 2022
@Matt - FYI, when you get errors, it is best to post the entire error message along with your code. Regardless, see my answer below ...
Matt
Matt am 4 Okt. 2022
Will do. Really new here so just learning the ways about this. I'll be back! :-)

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 Akzeptierte Antwort

James Tursa
James Tursa am 4 Okt. 2022
Bearbeitet: James Tursa am 4 Okt. 2022

0 Stimmen

You start your loop with i=1, but that means your x_d(i-1) will be x_d(0), an invalid index, hence the error. You need to set initial values for x_d(1) and x(1), and then have your starting loop index be 2. E.g.,
x(1) = initial x value
x_d(1) = initial xdot value
for i=2:n1 % start loop index at 2
x_dd(i-1) = use (i-1) indexes on everything on rhs
x_d(i) = use (i-1) indexes on everything on rhs
x(i) = use (i-1) indexes on everything on rhs

Weitere Antworten (1)

Davide Masiello
Davide Masiello am 27 Sep. 2022
Bearbeitet: Davide Masiello am 27 Sep. 2022

0 Stimmen

Hi Matt - a second order ODE can be decomposed into two first order ODEs.
The secret is to set 2 variables y as
The you have
An example code is
clear,clc
tspan = [0,1]; % integrates between times 0 and 1
x0 = [1 0]; % initial conditions for x and dx/dt
[t,X] = ode15s(@odeFun,tspan,x0); % passes functions to ODE solver
x = X(:,1);
dxdt = X(:,2);
plot(t,x)
function dydt = odeFun(t,y)
z = 1;
w = 1;
dydt(1,1) = y(2);
dydt(2,1) = 2*z*w*y(2)-w*y(1)+2*sin(2*pi*2*t);
end

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am 27 Sep. 2022

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am 4 Okt. 2022

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