i.e. Euler-Lagrange equations. The Legendre transformation to the Hamiltonian formulation is traditionally only defined for a variational formulationof eq. (42). So there is no conventional to the Hamiltonian formulation is of the EOMform
$$\begin{align} \frac{dq^j}{dt}~=~&\frac{\partial H}{\partial p_j},\cr \frac{dp_j}{dt} +\frac{\partial H}{\partial q^j}~=~&Q_j, \cr j~\in~&\{1,\ldots, n\},\end{align}\tag{5}$$
see (1)this Phys.SE post for details.
Trick with exponential factor$^1$: Define for later convenience the function $$ e(t)~:=~\exp(\frac{\beta t}{m}). \tag{5}$$$$ e(t)~:=~\exp(\frac{\beta t}{m}). \tag{6}$$ A possible variational formulation (4) of Lagrange equations is then given by the Lagrangian $$\begin{align} L(q,\dot{q},t)~:=~&e(t)L_0(q,\dot{q},t), \cr L_0(q,\dot{q},t)~:=~&\frac{m}{2}\dot{q}^2-V(q,t).\end{align}\tag{6}$$$$\begin{align} L(q,\dot{q},t)~:=~&e(t)L_0(q,\dot{q},t), \cr L_0(q,\dot{q},t)~:=~&\frac{m}{2}\dot{q}^2-V(q,t).\end{align}\tag{7}$$ The corresponding Hamiltonian is $$ H(q,p,t)~:=~\frac{p^2}{2me(t)}+e(t)V(q,t).\tag{7}$$$$ H(q,p,t)~:=~\frac{p^2}{2me(t)}+e(t)V(q,t).\tag{8}$$ One caveat is that the Hamiltonian (78) does not represent the traditional notion of total energy. Another caveat is that this unconventional approach cannot be generalized to the case where two coupled sectors of the theory require different factors (56), e.g. where each coordinate $q^j$ has individual friction-over-mass-ratios $\frac{\beta_j}{m_j}$, $j\in\{1, \ldots, n\}$. For this unconventional approach to work, it is crucial that the factor (56) is an overall common multiplicative factor for the Lagrangian (67). This is an unnatural requirement from a physics perspective.
Imposing EOMs via Lagrange multipliers $\lambda^j$: A variational principle for the EOMs (1) is $$\begin{align}L ~=~& m\sum_{j=1}^n\dot{q}^j\dot{\lambda}^j\cr &-\sum_{j=1}^n\left(\beta\dot{q}^j+\frac{\partial V(q,t)}{\partial q^j}\right)\lambda^j.\end{align}\tag{8}$$$$\begin{align}L ~=~& m\sum_{j=1}^n\dot{q}^j\dot{\lambda}^j\cr &-\sum_{j=1}^n\left(\beta\dot{q}^j+\frac{\partial V(q,t)}{\partial q^j}\right)\lambda^j.\end{align}\tag{9}$$ (Here we have for convenience "integrated the kinetic term by parts" to avoid higher time derivatives.)
Classical Schwinger/Keldysh "in-in" formalism: The variables are doubled up. See e.g. eq. (20) in C.R. Galley, arXiv:1210.2745. Ignoring boundary terms$^2$ the Lagrangian reads $$\begin{align} \widetilde{L}(q_{\pm},\dot{q}_{\pm},t) ~=~&\left. L(q_1,\dot{q}_1,t)\right|_{q_1=q_+ + q_-/2}\cr ~-~&\left. L(q_2,\dot{q}_2,t)\right|_{q_2=q_+ - q_-/2}\cr ~+~&Q_j(q_+,\dot{q}_+,t)q^j_-\end{align}\tag{9}. $$$$\begin{align} \widetilde{L}(q_{\pm},\dot{q}_{\pm},t) ~=~&\left. L(q_1,\dot{q}_1,t)\right|_{q_1=q_+ + q_-/2}\cr ~-~&\left. L(q_2,\dot{q}_2,t)\right|_{q_2=q_+ - q_-/2}\cr ~+~&Q_j(q_+,\dot{q}_+,t)q^j_-\end{align}\tag{10}. $$ The initial conditions $$\left\{\begin{array}{rcl} q^j_+(t_i)&=&q^j_i,\cr\dot{q}^j_+(t_i)&=&\dot{q}^j_i\end{array}\right.\tag{10} $$$$\left\{\begin{array}{rcl} q^j_+(t_i)&=&q^j_i,\cr\dot{q}^j_+(t_i)&=&\dot{q}^j_i\end{array}\right.\tag{11} $$ implement the system's underlying initial values. The final conditions $$\begin{align}\left\{\begin{array}{rcl} q^j_-(t_f)&=&0\cr \dot{q}^j_-(t_f)&=&0 \end{array}\right. & \cr\cr\qquad\Downarrow&\qquad\cr\cr \left.\frac{\partial \widetilde{L}}{\partial \dot{q}^j_+}\right|_{t=t_f}~=~&0 \end{align}\tag{11} $$$$\begin{align}\left\{\begin{array}{rcl} q^j_-(t_f)&=&0\cr \dot{q}^j_-(t_f)&=&0 \end{array}\right. & \cr\cr\qquad\Downarrow&\qquad\cr\cr \left.\frac{\partial \widetilde{L}}{\partial \dot{q}^j_+}\right|_{t=t_f}~=~&0 \end{align}\tag{12} $$ implement the physical limit solution $q_-^j= 0$. The doubling trick (910) is often effectively the same as introducing Lagrange multipliers (89).
Gurtin-Tonti bi-local method: See e.g. this Phys.SE post.
$^2$ The variational problem (910)+(1011)+(1112) needs an appropriate initial term, which might not always exist! In particular, since we already imposed $4n$ boundary conditions (1011)+(1112), it would be too much to also impose the initial condition
$$ q^j_-(t_i)~=~0. \qquad (\leftarrow\text{Wrong!})\tag{12}$$$$ q^j_-(t_i)~=~0. \qquad (\leftarrow\text{Wrong!})\tag{13}$$
Example: If
$$L~=~\frac{1}{2}m\dot{q}^2,\qquad Q~=~0,\tag{13}$$$$L~=~\frac{1}{2}m\dot{q}^2,\qquad Q~=~0,\tag{14}$$
then
$$\widetilde{L}~\stackrel{(13)+(9)}{=}~m\dot{q}_+\dot{q}_-,$$$$\widetilde{L}~\stackrel{(14)+(10)}{=}~m\dot{q}_+\dot{q}_-,\tag{15}$$
and one should add an initial term $\left. m\dot{q}_+q_-\right|_{t=t_i}$ to the action
$$\widetilde{S}~=~\left. m\dot{q}_+q_-\right|_{t=t_i}+
\int_{t_i}^{t_f}\!dt~\widetilde{L}
,\tag{14}$$$$\widetilde{S}~=~\left. m\dot{q}_+q_-\right|_{t=t_i}+
\int_{t_i}^{t_f}\!dt~\widetilde{L}
,\tag{16}$$
so that an infinitesimal variation becomes
$$ \delta \widetilde{S}~\stackrel{\rm IBP}{=}~\text{bulk terms} ~+~ \text{boundary terms},\tag{15}$$$$ \delta \widetilde{S}~\stackrel{\rm IBP}{=}~\text{bulk terms} ~+~ \text{boundary terms},\tag{17}$$
where the boundary terms
$$\begin{align}\text{boundary terms}
~=~&\left. mq_-\delta\dot{q}_+ \right|_{t=t_i}
+\left. m\dot{q}_+\delta q_-\right|_{t=t_i}
+\left[m\dot{q}_{\pm}\delta q_{\mp}\right]_{t=t_i}^{t=t_f}\cr
~=~&\left. mq_-\delta\dot{q}_+ \right|_{t=t_i}
+\left. m\dot{q}_+\delta q_-\right|_{t=t_f}
+\left[m\dot{q}_-\delta q_+\right]_{t=t_i}^{t=t_f}\cr
~\stackrel{(10)+(11)}{=}&~0.\end{align}\tag{16}$$$$\begin{align}\text{boundary terms}
~=~&\left. mq_-\delta\dot{q}_+ \right|_{t=t_i}
+\left. m\dot{q}_+\delta q_-\right|_{t=t_i}
+\left[m\dot{q}_{\pm}\delta q_{\mp}\right]_{t=t_i}^{t=t_f}\cr
~=~&\left. mq_-\delta\dot{q}_+ \right|_{t=t_i}
+\left. m\dot{q}_+\delta q_-\right|_{t=t_f}
+\left[m\dot{q}_-\delta q_+\right]_{t=t_i}^{t=t_f}\cr
~\stackrel{(11)+(12)}{=}&~0.\end{align}\tag{18}$$
vanish, as they should.
