Symmetry for a Lagrangian system Given the Lagrangian
$$L(x,y,\dot{x},\dot{y},t)=y^2\dot{x}^2+x^2\dot{y}^2+x+y+\dot{x}t+\dot{y}t
$$
in the $x,y$ coordinates, I want to find some conserved quantities. Since there are no cyclic coordinates, I am searching for a symmetry for the system but I'm not sure how to do that. Is there any method I could use?
 A: To find conserved quantities there is nothing better then derive the canonical equations of Hamilton. So I have the hint
Hint: Find the Hamiltonian and the canonical equations of motion for the hamiltonian system

$$\frac{\partial \mathcal{H}}{\partial q} = -\dot{p} \hspace{2cm} \frac{\partial \mathcal{H}}{\partial p} =  \dot{q}$$

You should of course find the equations of motion for the Lagrangian, i.e., the Euler-Lagrange equations. 
If you are searching specifically for a symmetry there is the most beatifull theorem of physics (my personal opinion), so the other hint should be
Hint: Use Noether's Theorem to find a symmetry. If you read the link you should see that, in the notation used,

$$\sum_\alpha \frac{\partial \mathcal{L}}{\partial \dot{q_\alpha}}\gamma_\alpha = \text{constant}$$

The theorem says that for a specific symmetry you always can find  conserved quantities. See that if you change $x \rightarrow y$ you get the same Lagrangian.

A: First of all, we can write the lagrangian in this form:
$$L(x,y,\dot{x},\dot{y},t)=y^2\dot{x}^2+x^2\dot{y}^2+x+y+\dot{x}t+\dot{y}t=$$
$$=y^2\dot{x}^2+x^2\dot{y}^2+\dfrac{\mathrm d}{\mathrm dt}(xt+yt)$$
We can drop the total derivative with respect to time because it doesn't affect the equations of motion:
$$\tilde L(x,y,\dot{x},\dot{y},t)=y^2\dot{x}^2+x^2\dot{y}^2$$
Now, as the lagrangian doesn't depend explicitly of time, the energy is conserved. So is,
$\dot x\dfrac{\partial\tilde L}{\partial\dot x}+\dot y\dfrac{\partial\tilde L}{\partial\dot y}-\tilde L=K$
$\dot x\dfrac{\partial\tilde L}{\partial\dot x}+\dot y\dfrac{\partial\tilde L}{\partial\dot y}-\tilde L=\dot xy^22\dot x+\dot yx^22\dot y-y^2\dot x^2-x^2\dot y^2=y^2\dot x^2+x^2\dot y^2$
$y^2\dot x^2+x^2\dot y^2=K$
I tink this lagrangian is simply the one for the free particle in dimension two, but in a weird coordinate system.
