How to solve and plot second order differential equation using ode45?

조회 수: 21 (최근 30일)
Studentguy
Studentguy 2023년 11월 23일
댓글: Fabio Freschi 2023년 11월 23일
Hey there
Im trying to solve and plot the following differential equation using ode45
x''=(-2k*x-2c*x'-r*omega*(cos(omega*t)+(r/L)*cos(2*omega*t))*m+x0*M)/M+m
where
M = 22
m = 0.9
k = 25000
c = 2
omega = 860
L = 0.2
r = 0.07
The starting conditions are :
x0=0
x'0=0
I've tried a bunch of different tutorials, but keep getting different error messages.
Any help is greatly appreciated, thank you!
  댓글 수: 4
Dyuman Joshi
Dyuman Joshi 2023년 11월 23일
Is the value of x0 that appears in the ODE the same as the value of x0 that is one of the initial conditions?
Also, you have defined the variable x_speed in the ODE function, but have not used it. Is there any particular use of that variable?
Studentguy
Studentguy 2023년 11월 23일
Ye, x0 is the same, its just part the equation i arrived at during calculations.
No x_speed isnt used anywhere.

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답변 (1개)

Fabio Freschi
Fabio Freschi 2023년 11월 23일
편집: Fabio Freschi 2023년 11월 23일
In your code there is a mistake in the definition of x(1) and x(2)
I have made a few stylistic changes (parameters outside the function, use of implicit function) and the correction of the equations. In addition, I suggest to let ode45 to choose the timestep and keep the t vector provided as output for the plot
% clear variables, close all
x0 = [0 0];
tspan = [0 10];
% params
M = 22;
m = 0.9;
k = 25000;
c = 2;
omega = 860;
l = 0.2;
r = 0.07;
x0_speed = 0;
x0_pos = 0;
% implicit function
% changes here
% | |
% V V
dxdt = @(t,x)[x(2); (-2*k*x(1)-2*c*x(2)-r*omega^2*(cos(omega*t)+(r/l)*cos(2*omega*t))*m+x0_pos*(M-m))/M];
[t,x] = ode45(dxdt,tspan,x0);
figure
plot(t,x)
figure
plot(t,x)
xlim([0 0.5])
  댓글 수: 4
Sam Chak
Sam Chak 2023년 11월 23일
I double-check. The original 2nd-order differential equation in your question
is slightly different from the state equation you defined in your code:
( - 2*k*x(1) - 2*c*x(2) - r*omega^2*( cos(omega*t) + (r/l)*cos(2*omega*t) )*m + x0_pos*(M - m) )/M;
Please clarify!
Fabio Freschi
Fabio Freschi 2023년 11월 23일
This is why I wrote my code according to the orignal equation (as stated in the comment)

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