Solve 2nd order ODE using Euler Method

2022 9 27
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更新时间：2022 10 4
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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!

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John D'Errico
John D'Errico 2022-9-27
编辑：John D'Errico 2022-9-27
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 2022-9-27
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 2022-10-4
移动：James Tursa 2022-10-4
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 2022-10-4
@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 2022-10-4
Will do. Really new here so just learning the ways about this. I'll be back! :-)

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 采纳的回答

James Tursa
James Tursa 2022-10-4
编辑：James Tursa 2022-10-4

0 个投票

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

更多回答（1 个）

Davide Masiello
Davide Masiello 2022-9-27
编辑：Davide Masiello 2022-9-27
Ran in:

0 个投票

Ran in:
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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