What Values of \( n \) Make the Given Lagrangian a Total Derivative?

[Ziggy12]

Homework Statement

We have the Lagrangian $$L=\frac{1}{2}\dot q^2-\lambda q^n$$
Determine the values for n so that the Lagrangian transform into a total derivative
$$\delta q = \epsilon (t\dot q - \frac{q}{2})$$

Homework Equations

The theorem says that if the variation of action
$$\delta S = \int_{t_1}^{t_2} dt\hspace{0.1cm} \delta q \left(\frac{\partial L}{\partial q}- \frac{d}{dt}\frac{\partial L}{\partial \dot q}\right) + \left[\delta q \frac{\partial L}{\partial \dot q}\right ]_{t_1}^{t_2}$$
there is a delta q, and a function lambda such as
$$\delta S = \int_{t_1}^{t_2} dt \frac{d}{dt}\Lambda(q,\dot q)$$
Then $$\epsilon Q(q,\dot q) = \delta q \frac{\partial L}{\partial \dot q} - \Lambda(q,\dot q)$$
is a constant of motion
(At least from what I understood)

The Attempt at a Solution

So I derived the needed derivatives:
$$\frac{\partial L}{\partial q} = -\lambda n q^{n-1}$$
$$\frac{\partial L}{\partial \dot q} = \dot q$$
$$\frac{d}{dt}\frac{\partial L}{\partial \dot q} = \ddot q$$

After some calculations this just gives me formula:
$$\delta S = \epsilon\int_{t_1}^{t_2} \lambda n (\frac{q^n}{2}-t\dot q q^{n-1})+\frac{\dot q^2}{2}+t\ddot q \dot q$$

Getting a function lambda out of this seems impossible, am I doing something wrong here?
Thanks for the help

 

[BigJDubs]

.A:The method of Lagrange multipliers works perfectly fine. The equation of motion is$$ \ddot q +\lambda n q^{n-1}=0. $$Let$$ \delta q = \epsilon (t\dot q - \frac{q}{2}) $$then$$ \delta \dot q = \epsilon (\ddot q +t \dot q). $$Plug into the Euler-Lagrange equation and solve for $\lambda$$$ \lambda = -\frac{2 \epsilon}{n q^{n-1}}. $$Note that when $\epsilon=0$ we recover the original equation of motion.