Given a Newtonian system (with forces that may depend position, velocity, and time) with solutions $\vec{q}(t)$, can we always derive a Lagrangian system such that the following are satisfied:
- It yields the same trajectories $\vec{q}(t)$.
- It is possibly time-dependent.
- It has no generalized forces, so the solutions are generated by the action principle and they satisfy the original Euler-Lagrange equations $\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{i}}) - \frac{\partial L}{\partial q_{i}} = 0$.
If the answer is yes, is there a proof that demonstrates this to be the case? If the answer is no, is there an explicit counterexample that can be given?
For background info, I've looked over this post and this post, which explain that some Newtonian systems can be converted to Lagrangian systems with generalized forces, but this isn't what I'm looking for since it doesn't satisfy (3). I've also seen this interesting webpage, but the post operates under the restriction that the Lagrangian cannot be time-dependent, so the webpage doesn't satisfy (2).
Whether the notions of momenta, energy, and generalized forces in the Lagrangian system match the corresponding notions in Newtonian mechanics is not important here (they may fail to equal if the Lagrangian is time-dependent I believe). I am only interested in satisfying (1) (along with (2) and (3)), which is that the two systems generate the same trajectories $\vec{q}(t)$.
Edit: I thought maybe a damped oscillator would have fit as a counterexample, but apparently it does have a Lagrangian counterpart that produces the same trajectories. This is sort of why I want a concrete counterexample with a proof that it has no Lagrangian.