Equivalence of two lagrangians While preparing to the exam of Analytical Mechanics I found an exercise I wasn't able to do on my own:
Prove that following Lagrangian functions
$$L( \dot{x} ,\dot{y}, x, y  )= \dot{x}x^2 + \dot{y}^2 + x^2 - y^3$$ ,
and
$$\tilde{L}(\dot{x} ,\dot{y}, x, y  )= x^2 + \dot{y}^2-y^3(1-\dot{y})$$ ,
are equivalent.
The main question here is if exists any technique how to check if lagrangians are equivalent as I couldn't find some clear explanation anywhere. The only usefull thing i found here : https://www.emis.de/journals/HOA/JAM/Volume2012/860482.pdf which says:
theorem 1
I guess the problem here is also that we have two variables which complicate the problem and make me totally confused.
I can see that L and $\tilde{L}$ are very similar and $L+\dot{x}x^2=\tilde{L}-\dot{y}y^3$.
 So, As my suggestion I would say that we have to consider the variables separately and maybe use another theorem from the page I provided which states theorem 2.
If anyone has some experience with this topic or some ideas how to solve the problem I would be really happy. Or maybe someone knows at least some technique to check if two Lagrangians are equivalent, that wold be really helpful as well
UPD: The problem is solved with two different and as I can judge correct approaches and now I now feel much better about the upcoming exam, thank you everyone for your contribution.
 A: Two Lagrangian are said to be equivalent if they differ by a total derivative of time or
$$
\tilde{L}=L+\frac{dF}{dt}.
$$
In this case $\tilde{L}$ and $L$ give rise to the same set of Euler Lagrange equations. 
In your case it is easy to see that such a function $F$ can be one of the infinite family of functions
$$
F(x,y)=\frac{1}{4}y^4-\frac{1}{3}x^3+k,\quad k\in\mathbb{R}.
$$
A: With regards to physics, when they say the Lagrangians are equivalent, it is meant that they produce the same physics, i.e. the two Lagrangians have the same equations of motion. This can be shown by applying the Euler-Lagrange equation for both of the Lagrangians
$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x_i}\right)-\frac{\partial L}{\partial x_i}=0$$
Applying this to both Lagrangians yield the same equations of motion being
$$x=0\qquad \&\qquad 2\ddot y+3y^2=0$$
Turns out the coordinate for $x$ is in a steady state for these systems.

More rigorously:
$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\right)-\frac{\partial L}{\partial x}=2x\dot x-2x\dot x-2x = 0 \implies x=0$$
$$\frac{d}{dt}\left(\frac{\partial \bar L}{\partial \dot x}\right)-\frac{\partial \bar L}{\partial x} = -2x=0 \implies x=0$$
$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot y}\right)-\frac{\partial L}{\partial y} = 2\ddot y+3y^2=0$$
$$\frac{d}{dt}\left(\frac{\partial \bar L}{\partial \dot y}\right)-\frac{\partial \bar L}{\partial y} = 2\ddot y+3y^2\dot y+3y^2-3y^2\dot y=0 \implies 2\ddot y+3y^2=0$$
