1. ## Euler-Lagrange Equation

Let $\displaystyle f = \frac{\dot{x}^2}{t^2}$

The Euler-Lagrange Equation is

$\displaystyle \frac{\partial f}{\partial x} - \frac{d}{dt} \Big(\frac{\partial f}{\partial \dot{x}} \Big) = 0$

I get

$\displaystyle 0 - \frac{d}{dt} \Big(\frac{2 \dot{x}}{t^2} \Big) = 0$....(Eqn A)

$\displaystyle \frac{2 \ddot{x}}{t^2} = c$

$\displaystyle \ddot{x} = \frac{1}{2}ct^2$, $\displaystyle \dot{x} = \frac{1}{6} ct^3$, $\displaystyle x = \frac{1}{24}ct^4 + k$

but this is wrong. According to solutions

$\displaystyle x = \frac{1}{3}ct^3 + k$ where k is a constant.

I can't see where I have gone wrong from eqn A.

2. Originally Posted by MrJack1990
Let $\displaystyle f = \frac{\dot{x}^2}{t^2}$

The Euler-Lagrange Equation is

$\displaystyle \frac{\partial f}{\partial x} - \frac{d}{dt} \Big(\frac{\partial f}{\partial \dot{x}} \Big) = 0$

I get

$\displaystyle 0 - \frac{d}{dt} \Big(\frac{2 \dot{x}}{t^2} \Big) = 0$....(Eqn A)

$\displaystyle \frac{2 \ddot{x}}{t^2} = c$

$\displaystyle \ddot{x} = \frac{1}{2}ct^2$, $\displaystyle \dot{x} = \frac{1}{6} ct^3$, $\displaystyle x = \frac{1}{24}ct^4 + k$

but this is wrong. According to solutions

$\displaystyle x = \frac{1}{3}ct^3 + k$ where k is a constant.

I can't see where I have gone wrong from eqn A.
Integrate both sides wrt t: $\displaystyle \displaystyle \frac{2 \dot{x}}{t^2} = A \Rightarrow \dot{x} = Bt^2$ etc.

3. $\displaystyle \frac{2 \ddot{x}}{t^2} = c$
Should this be

$\displaystyle \frac{2 \dot{x}}{t^2} = c$

4. Originally Posted by MrJack1990
Should this be

$\displaystyle \frac{2 \dot{x}}{t^2} = c$
Well yes (why would it not?).