For $x=(x_1,x_2) \in \mathbb R^2$, we denote by $$L(x,\dot{x})= \dot{x_{1}}^{2}+x_{1}\, \dot{x_{1}}\, \dot{x_{2}}+\dot{x_{2}}$$ the Lagrangian associated to an Hamiltonian for an unbounded operator $L$.
I would like solve the Euler–Lagrange equation given by $$\frac{d}{ds}\frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x} .$$ Thank you in advance
Let us go step by step:
1) the term $\dot x_2$ is a total time-derivative, thus unimportant for the equations of motion. So I will consider the Lagrangian $\tilde L = \dot x_1^2 + x_1 \dot x_1 \dot x_2$
2) $x_2$ is cyclic and thus $$p_2 = \frac{\partial \tilde L}{\partial \dot x_2} = x_1 \dot x_1$$ is conserved
3) as the Lagrangian does not depend explicitly on time, we have that the "energy" $$ H = \sum_{j=1}^2\frac{\partial \tilde L}{\partial \dot x_j} \dot x_j - \tilde L = \dot x_1 (\dot x_1 + x_1 \dot x_2) = \dot x_1^2 +p_2 \dot x_2 $$ is conserved
Now we solve: 2) $$p_2 = x_1 \dot x_1$$ has the solution $$x_1 = \pm\sqrt{p_2 t + \alpha}. \tag{1}$$
Inserting into the expression 3) for $H$, we have the equation $$ H= \frac{p_2^2}{2 (p_2 t +\alpha)} + p_2 \dot x_2$$ with the solution $$ x_2 = \frac{H t} {p_2} + \beta -\frac12 \ln\left(p_2 t + \alpha \right) \tag{2} =\frac{H t} {p_2} + \beta - \ln x_1 .$$
The general solution is given by (1) and (2) where $\alpha,\beta,p_2,H$ are the 4 constants of integration.
Edit: in order that $x_2$ remains reals, we have to choose the `+' sign in (1). Here is the orbit for $H=p_2=1$ and $\alpha=\beta=0$.