What is the Lagrangian and how does it relate to classical mechanics?

[Char. Limit]

So I type a differential equation into Wolfram Alpha, like so:

And one of the things that W-A outputs is the "Lagrangian" of that equation, which is so:

My question is, what is this Lagrangian, what does it describe, and how do I find it?

 

[Lavabug]

Yeah I've noticed that too. Look closely, if you apply Euler-Lagrange's equation to that Lagrangian, it'll spit out the original differential equation you put into it.

Tried putting in the differential equation for a pendulum y''(x) + (g/l)siny = 0 (y and x are theta and t respectively) but it didn't give me a Lagrangian. It would be cool if it did.

Edit: And what it describes, well in classical mechanics the definition is the kinetic minus potential energy in terms of the generalized coordinates.

(huge arm-waving here) Hamilton's principle states that if you integrate the lagrangian over time is always such that the resulting integrand is minimum (or maximum). It just so happens that the Euler-Lagrange equation is how you get the integrand (the lagrangian) to be stationary (be it a maximum or a minimum). I know it translates directly into the principle of least time in classical optics and it has a more general definition, but I don't know enough about it. I saw it early on in my mechanics course and is immensely important, but its still quite a mystery to me.

I hope someone reads this and corrects any (likely) mistakes I made in that last statement, but I think that's more or less the general idea.

 

[I like Serena]

My question is, what is this Lagrangian, what does it describe, and how do I find it?

The Lagriangian is from classical mechanics and is defined as

L = T - V

where T is the kinetic energy of the system and V the potential energy.

If for instance a point mass has a vertical coordinate y at time t, the following will hold:

$$L(y', y, t) = T - V = \frac 1 2 m y'^2 - m g y$$

According to Lagrangian mechanics we have:

$$\frac d {dt} \frac {\partial L} {\partial y'} - \frac {\partial L} {\partial y} = 0$$

which would work out as:

$$m y'' + m g = 0$$

Note that this looks a lot like your formula, if only for a different system.Note also that in your case the applicability of the Lagrangian is zero.