Confusion about what the Euler-Lagrange equation says I roughly understand the concept of the Lagrangian $L = T - V$ as well as the idea of stationary action $\delta \mathcal{S} =0$. However, I am confused what the Euler-Lagrange equation actually says. 
Consider the Euler-Lagrange equation: 
\begin{equation}
\frac{\partial L}{\partial q} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\right) = 0 
\end{equation} 
Here's my confusion:
To me, this looks like an empty mathematical exercise. If I know the partial derivatives $\frac{\partial L}{\partial q}$ and $\frac{\partial L}{\partial \dot{q}}$ and can take the derivative with respect to $t$ of the latter, what is the use of plugging all that into this setup? That's like saying after I show $2 + 3 = 5$, then show $x + y = z$, where $x=2,y=3,z=5$. In short, an empty exercise since it would be the same proof showing $x + y = z$ as showing $2 +3 = 5$ in this case. 
Can someone explain what I am misunderstanding?
 A: The reason algebra ($x + y = z$) is more useful than arithmetic ($2 + 3 = 5$) is that it allows you to treat large classes of problems in the same way, and to obtain very general results about them without having to do a separate calculation for every individual case. Same goes for the stationary action principle (a.k.a. action minimization). It allows you to treat a large class of problems - in this case, all of physics - in the same way, by solving $\delta S = 0$. Different choices of $S$ correspond to different areas of physics, but once you have $S$, the procedure for solving is always the same, namely the Euler-Lagrange equations.
Perhaps you've only been exposed to one formula for the action, $S = \int (T - V)\,\mathrm{d}t$ (or equivalently $L = T - V$), which is the action for a Newtonian system, and in that case it's understandable that the stationary action principle seems kind of useless. Similarly, if you had only been exposed to the one example of $x + y = z$ where $x = 2$, $y = 3$, and $z = 5$, it would seem pretty pointless to use letters. But it makes more sense once you see how general the formalism is.
A: Euler-Lagrange equations are a necessary condition if the action integral reaches an extremal through the trajectory described by $q(t)$ and you should obtain an expression for $q(t)$ for a given Lagrangian $L$ by solving Euler-Lagrange equations. Take for example the simple Lagrangian for a free particle with constant mass: $L=m\dot{q}^2/2$ then E-L equations are
\begin{equation}
 \frac{d}{dt}\left(m\dot{q}\right)=m\ddot{q}=0
\end{equation}
which has for solution
\begin{equation}
 q(t)=at+b
\end{equation}
a straight line, something you already knew from Newton first law of motion.
Try solving E-L equations with the Lagrangian $L=m\dot{z}^2/2-mgz$ for the particle in free fall (taking partial derivatives with respect to $z$ and $\dot{z}$ instead of $q$ and $\dot{q}$) to obtain the well known trajectory $z(t)=gt^2/2+\dot{z}_0t+z_0$, where $z_0$ and $\dot{z}_0$ are the initial position and velocity.
