Converting ODE to a system of ODEs

[Dustinsfl]

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.

 

[Ackbach]

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
$$x_{2}'=x_{1}-x_{1}^{3}-\gamma x_{2}.$$

 

[Dustinsfl]

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?

 

[Sudharaka]

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/.

 

[blue_raver22]

Yes, your approach is correct. Converting a single second-order ODE into a system of first-order ODEs is a common technique in solving differential equations. By defining $x_1 = x$ and $x_2 = x'$, you have effectively created a system of two first-order ODEs, which can be solved using standard methods. Your resulting linear system is also correct.