Can we find a Lagrangian for the the following first order equation?
\begin{equation} \dot x=x^2,\tag{1} \end{equation} where the above equation is the Euler-Lagrange equation of that Lagrangian?.
For second order equations, it can be found easily for some class of equations.
On
(Too long for a comment)
In the same spirit as Jorge's great answer, I wanted to complete the discussion by making the following remark : the Lagrangian formalism has not been really designed for tackling second-order equation of motion, whence the present difficulty to find a suitable Lagrangian. On the contrary, the Hamiltonian mechanics does it pretty well.
Indeed, recalling that the equations of motion generated by a Hamiltonian $H(q,p)$ are given by $$ \begin{cases} \displaystyle \dot{q} = +\frac{\partial H}{\partial p} \\ \displaystyle \dot{p} = -\frac{\partial H}{\partial q} \end{cases} $$ and setting $\dot{p} = p^2$, we thus have $H(q,p) = -p^2q + f(q)$, where $f(q)$ is an arbitrary function, and the work is done. Note that the conjugate variable obeys consequently the following equation of motion : $\dot{q} = -2pq$. Of course, we could have inverted the roles of the two conjugate variables.
In addition, since $p = -\frac{\dot{q}}{2q}$, the associated Lagrangian would be given by $$ L(q,\dot{q}) = p\dot{q} - H = -\frac{\dot{q}^2}{2q} - \left(-\frac{\dot{q}^2}{4q}+f(q)\right) = -\frac{\dot{q}^2}{4q}-f(q). $$ Let's check its functional derivative : $$ \frac{\delta L}{\delta q} = \left(\frac{\partial}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial}{\partial\dot{q}}\right)\left(-\frac{\dot{q}^2}{4q}-f(q)\right) = -f'(q), $$ which fails to produce a differential equation. One could argue that we should have impose the equation of motion for the other conjugate variable at the very beginning, i.e. $\dot{q} = q^2$, but then the situation would get even worse, because the corresponding Lagrangian would be of the form $L = -\frac{1}{2}q^3 + g(q)$, with $g$ another arbitrary function.
In conclusion, the Lagrangian formalism fails to handle first-order equations of motion.
On
Yes, if $x\in\mathbb{C}\backslash\{0\}$ is a non-zero complex variable$^1$. Define $$z~\equiv~q+ip~\equiv~-1/x.\tag{A}$$ Then OP's first order equation (1) becomes $$\dot{z}~=~1.\tag{B}$$ Eq. (B) has a Lagrangian $$\begin{align}L~\equiv~&\frac{i}{2}(\bar{z}\dot{z}-z\dot{\bar{z}})-\frac{z-\bar{z}}{2i}\cr ~\equiv~&p (\dot{q}-1)\cr ~\equiv~&{\rm Im}(1/x) \left({\rm Re}\frac{d(1/x)}{dt}+1\right). \end{align}\tag{C}$$
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$^1$ There is always the trivial possibility to implement OP's equation (1) via a Lagrange multiplier. Here we assume that we are not allowed to introduce new variables.
I think that the answer is negative, at least for Lagrangians of the form $\mathcal{L}(\theta,\dot \theta)$. Suppose such a mythical Lagrangian exists. Thus, Euler Lagrange equations is of the form. $$ \ddot \theta \mathcal{L}_{\dot \theta \dot\theta} + \dot \theta \mathcal{L}_{\dot \theta \theta} = \mathcal{L}_{\theta} $$ Thus, $\mathcal{L}_{\dot \theta \dot\theta} =0$. So $\mathcal{L} = \dot \theta f(\theta) + g(\theta)$. But then, Euler Lagrange equation is $$ \dot \theta f'(\theta) = \dot \theta f'(\theta) + g'(\theta) \Rightarrow g=C $$ Then, up to a constant $\mathcal{L} = \dot\theta f(\theta)$, and Euler Lagrange equations are trivial.
If $\mathcal{L}(\theta,\dot \theta, \mathcal t)$, then $\mathcal{L}=\dot \theta f(\theta, t) + g(\theta,t)$. But then Euler-Lagrange reduce to $$ \partial_t f(\theta,t) = \partial_\theta g(\theta,t)$$
If $f$ and $g$ are smooth enough, $f=\partial_\theta F$ and $g=\partial_tG$. But then $F(\theta,t) = G(\theta,t) + i(\theta) - j (t)$, Let $H(\theta,t) = F(\theta,t) + j(t) = G(\theta,t) + i(\theta)$. Then $\mathcal{L}=\dot \theta \partial_\theta H+\partial_t H$.