Let $q(t)$ be a function $\reals \to \reals$ $\dot{q} = \mderiv{t}{q}$ and $L = L(q, \dot{q}, t)$ Begin with the time-relative Euler-Lagrange condition

\begin{equation}\label{el} \mpderiv{q} L - \mderiv{t}\left(\mpderiv{\dot{q}} L\right) = 0. \end{equation} If $\mpderiv{t}L = 0$ then the Euler-Lagrange condition reduces to

\begin{equation} L - \dot{q}{\mpderiv{\dot{q}} L} = C, \end{equation}which is the Beltrami identity. In the calculus of variations, the ability to use the Beltrami identity can vastly simplify problems, and as it happens, many physical problems have $\mpderiv{t}L = 0$

In space-relative terms, with $q' \defined \mderiv{x}q$ we have \begin{equation} \mpderiv{q} L - \mderiv{x}{\mpderiv{q'} L} = 0. \end{equation} If $\mpderiv{x}L = 0$ then the Euler-Lagrange condition reduces to

\begin{equation} L - q' {\mpderiv{q'} L} = C. \end{equation} To derive the Beltrami identity, note that \begin{equation}\label{step2} \mderiv{t}\left(\dot{q}\mpderiv{\dot{q}}L\right) = \ddot{q}\mpderiv{\dot{q}}L + \dot{q}\mderiv{t}\left(\mpderiv{\dot{q}}L\right) \end{equation}Multiplying (1) by $\dot{q}$ we have \begin{equation}\label{step3} \dot{q}\mpderiv{q} L - \dot{q}\mderiv{t}\left(\mpderiv{\dot{q}} L\right) = 0. \end{equation}Now, rearranging (5) and substituting in for the rightmost term of (6), we obtain \begin{equation}\label{step4} \dot{q}\mpderiv{q} L + \ddot{q}\mpderiv{\dot{q}}L - \mderiv{t}\left(\dot{q}\mpderiv{\dot{q}}L\right) = 0. \end{equation}Now consider the total derivative \begin{equation}\label{step1} \mderiv{t}L(q, \dot{q}, t) = \dot{q}\mpderiv{q}L + \ddot{q}\mpderiv{\dot{q}}L + \mpderiv{t}L. \end{equation}If $\mpderiv{t}L = 0$ then we can substitute in the left-hand side of (8) for the leading portion of (7) to get \begin{equation} \mderiv{t}L - \mderiv{t}\left(\dot{q}\mpderiv{\dot{q}}L\right) = 0. \end{equation}Integrating with respect to $t$ we arrive at \begin{equation} L - \dot{q}{\mpderiv{\dot{q}} L} = C, \end{equation}which is the Beltrami identity.