Why does velocity function have velocity itself as a parameter?

kmcrane

You can think about it this way: the velocity function effectively says what the velocity of a solution should be. The velocity dq/dt is the velocity of any candidate function q whatsoever, whether or not it actually solves our ODE. We say that we've "solved" the ODE when we've found a function q such that dq/dt = f( q, dq/dt, t ). So there are no cyclical definitions here. :-)

kmcrane

@dsaksena: A proof of what? If you want to see that $be^{at}$ is the solution to that ODE, you need only differentiate:

Why does velocity function have velocity itself as a parameter?

You can think about it this way: the velocity function effectively says what the velocity of a solution

shouldbe. The velocity dq/dt is the velocity of any candidate function q whatsoever, whether or not it actually solves our ODE. We say that we've "solved" the ODE when we've found a function q such that dq/dt = f( q, dq/dt, t ). So there are no cyclical definitions here. :-)@dsaksena: A proof of what? If you want to see that $be^{at}$ is the solution to that ODE, you need only differentiate:

$$ \tfrac{d}{dt} u(t) = \tfrac{d}{dt} (be^{at}) = a(be^{at}) = au(t). $$

Sorry, I meant a proof for Euler-Lagrangian Equation it came up in review. This is the wrong slide for it.

where L = K - U and we have the equation to solve

Ah, ok. There's also a short proof on Wikipedia.