Asked by Arlene
Consider the action S = ∫L(x, y(x), y'(x))dx. In class, we showed that the action is stationary δS = 0 for y(x) satisfying the Euler-Lagrange equations
∂L/∂y = d/dx ∂L/∂y' (∂L/∂y - d/dx ∂L/∂y' = 0)
Show that if the Lagrangian L doesn’t depend explicitly on x (that is, if ∂L/∂x = 0), then d/dx(L - ∂L/∂y'*y') = 0.
∂L/∂y = d/dx ∂L/∂y' (∂L/∂y - d/dx ∂L/∂y' = 0)
Show that if the Lagrangian L doesn’t depend explicitly on x (that is, if ∂L/∂x = 0), then d/dx(L - ∂L/∂y'*y') = 0.