MHB Converting ODE to a system of ODEs

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Ode Odes System
Dustinsfl
Messages
2,217
Reaction score
5
Given $x''-x+x^3+\gamma x' = 0$.

Is the below correct? Can I do this? The answer is yes.

Let $x_1 = x$ and $x_2 = x'$. Then $x_1' = x_2$.
\begin{alignat}{3}
x_1' & = & x_2\\
x_2' & = & x_1 - x_1^3 + \gamma x_2
\end{alignat}

Then I have the above linear system from the given ODE.
 
Last edited:
Physics news on Phys.org
dwsmith said:
Given $x''-x+x^3+\gamma x' = 0$.

Is the below correct? Can I do this? The answer is yes.

Let $x_1 = x$ and $x_2 = x'$. Then $x_1' = x_2$.
\begin{alignat}{3}
x_1' & = & x_2\\
x_2' & = & x_1 - x_1^3 + \gamma x_2
\end{alignat}

Then I have the above linear system from the given ODE.

Second equation should be
$$x_{2}'=x_{1}-x_{1}^{3}-\gamma x_{2}.$$
 
Ackbach said:
Second equation should be
$$x_{2}'=x_{1}-x_{1}^{3}-\gamma x_{2}.$$

Thanks typo. I trying to find the attraction basin for this system in another post. Are you familiar with that stuff?
 
dwsmith said:
Thanks typo. I trying to find the attraction basin for this system in another post. Are you familiar with that stuff?

I think your question is answered http://www.mathhelpboards.com/f17/damping-coefficient-2045/.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 0 ·
Replies
0
Views
664
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K