From Lagrangian to equations of motion I have a given Lagrangian: 
$$L= e^{st}\cdot\frac12\cdot(mv_y^2-ky^2)$$
And are asked to identify the equations of motions, the constants of motions and physical system. 
Without the exp-time-term, this wold be a harmonic oscillator with the solution
$$\frac{d^2y}{dt^2} = -\left(\frac km\right)\cdot y$$ and $$y(t)=\Re\left(Ae^{i\omega t}+Be^{-i\omega t}\right)$$
I'm wondering:


*

*How will the $\exp(st)$-term will effect my solution?

*What will happen to the Lagrangian if I do a transformation of variables $q=\exp(st)\cdot y$? 
 A: Why don't you just write the equations of motion? You will find that you get a second order lienar homogeneous DE with constant coefficients. Such an equation always has a closed form solution in terms of exponentials, so you can solve it; I guarantee the solution is not too bad.  You might even recognize the equation without having to solve it.

Edit: OK, more detail.
We have that $L = \frac12 e^{st} (m \dot{y}^2-ky^2)$. The Euler-Lagrange equation of motion is $\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} - \frac{\partial L}{\partial y} = 0$. When taking the partial derivatives, remember that $y$, $\dot{y}$ and $t$ are independent! When differentiating with respect to $y$, for example, you need to treat $\dot{y}$ and $t$ as constants. This is not the case when doing the total derivative $\frac{d}{dt}$. In that case, you treat everything, including $y$ and $\dot{y}$, as functions of $t$, and apply the chain rule as necessary.
Doing that, we get the following equation:
$$m\ddot{y} + sm\dot{y}+ky=0
$$
This looks similar to the harmonic oscillator but there is a term with $\dot{y}$. Rearranging, we can write this like so:
$$m\ddot{y}= -sm\dot{y}-ky$$
Now we can see that that term represents a force proportional to the velocity and opposite to it; in other words this is a damped oscillator. By doing similar calculations, we can see that the Lagrangian without the exponential term leads to the simple harmonic oscillator equation, as you noticed.
I'm not going to solve the damped oscillator equation in full since you can find the solution in lots of places (for example, Wikipedia). If you want to solve it yourself, use the usual method: "guess" a solution of the form $y = Ae^{\lambda t}$, plug that into your equation of motion and see what $\lambda$ should be. Depending on the value of $s$, you might need to consider solutions of the form $y = Ate^{\lambda t}$ also.
As for the substitution: if $q = e^{st}y$, then $y = e^{-st}q$ and then it's just a matter of doing the derivatives and sticking that into the Lagrangian. You will not get something very pretty. As has been suggested in the comments, the substitution $q = e^{-st/2}y$ may prove more fruitful.
