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Let us have the following equation of motion (it might not necessarily correspond to a physical system):

$$\dot{x} + a \cdot x + b \cdot x^2 + c=0.$$

I would like to deduce the corresponding Lagrangian. I have tried multiple combinations but I have not been able to obtain the $\dot{x}$ term.

All the systems I have worked with before had a $\dot{x}^2$ term instead of $\dot{x}$. I have unsuccessfully tried to compare the problem with a damped harmonic oscillator, but the $\dot{x}^2$ term is still missing.

I have solved the equation just in case:

$$x=\frac{A_1 \cdot e^{A_2 t} + A_3}{1 - A_4 \cdot e^{A_2 t}}.$$

Qmechanic
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TheAverageHijano
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    Problem is, this $\dot{x}$ is a friction term which does not conserve the total energy. You can take a look at the section "Extensions to include non-conservative forces" in https://en.wikipedia.org/wiki/Lagrangian_mechanics#Extensions_to_include_non-conservative_forces – Ewen Bellec Feb 12 '19 at 09:07
  • @E. Bellec thanks for your suggestion. Does that mean that the energy will not be conserved in this system? Is there any other magnitude that will remain constant over time? – TheAverageHijano Feb 12 '19 at 09:13
  • I'm not totally confident about this, but I stumble on the same question before. If you consider $\dot{x}$ as a force $F$. You can't write $F$ as the gradient of a potential energy $F\neq -\vec{\nabla}E_p$. Therefore, it's not a conservative force. It's similar to the damping term in the damped harmonic oscillator equation. – Ewen Bellec Feb 12 '19 at 09:39
  • This is your equations $\begin{aligned}\dfrac {d}{dt}\left( \dfrac {\partial L}{\partial \dot q}\right) +\dfrac {\partial L}{\partial q}=Q\ L=\dfrac {ax^{2}}{2}+\dfrac {bx^{3}}{3}+cx\ Q=-\dfrac {dx}{dt}\ q=x\end{aligned}$ – Eli Feb 12 '19 at 09:39
  • Possible duplicates: https://physics.stackexchange.com/q/340096/2451 and links therein. – Qmechanic Feb 12 '19 at 10:00

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