MHB Converting ODE to a system of ODEs

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The discussion focuses on converting a second-order ordinary differential equation (ODE) into a system of first-order ODEs. The original equation is $x'' - x + x^3 + \gamma x' = 0$, which is correctly transformed by defining $x_1 = x$ and $x_2 = x'$. The resulting system is $x_1' = x_2$ and $x_2' = x_1 - x_1^3 - \gamma x_2$. A typo in the second equation is acknowledged, and the user expresses interest in exploring the attraction basin for this system in a separate post. The conversation also references a related question about damping coefficients.
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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:
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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/.
 

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