How to solve this functional problem? I am new to calculus of variations, till now I know how to get the extremal functions for a given functional using Euler-Lagrange equation.
What if I have a functional but I am not looking for minimizing/maximizing it, but instead solving equations involving functionals, say:
$$I = \int_{x_{1}}^{x_2}{F(x,y,y') \,\mathrm{d}x}=\alpha\quad \,,\text{for }\alpha\in \mathbb{R}$$ 
How to solve for $y(x)$ that satisfy this equation? can I transform it to a classical problem then solve it using Euler-Lagrange equation?
EDIT:
for example say, we have the following problem :
$$I = \int_{0}^{1}{\left(f(x)+2f'(x)\right) \, \mathrm{d}x}= 1/2$$
I appreciate any ideas,
Thank you
 A: You can rig this to have unique solutions for very particular $\alpha$ (e.g. $F = (y-f)^2$, $\alpha = 0$ has unique solution $y=f$) but in general you should expect a large family of solutions. 
Intuitively this is because the space of functions is much bigger than the space of possible values of the functional $J(y)=\int F[y]$, so $J$ can't be anything close to injective. Making this rigorous isn't quite as simple as a cardinality argument (since e.g. $C^1([0,1])$ has the same cardinality as $\mathbb R$), but reasonable requirements on $F$ will make $J$ a differentiable map when restricted to a finite-dimensional space of functions. Sard's theorem then tells you that $J(y)=\alpha$ has multiple solutions (or no solution) for almost every $\alpha$, even amongst a 2-parameter family of functions $y$.
This shouldn't be too surprising - prescribing the value of the functional (a single real number) is much less information than prescribing the derivative of the functional (an element of some infinite-dimensional function space).
Another way of putting it: You're prescribing some kind of average of $y$, but not the local behaviour.
A: Suppose we have a point particle of mass $m$ moving along the $y$ axis. The Lagrangian is the difference between the kinetic ($K$) and potential ($U$) energies
$$\mathcal{L} (y, \dot y) := K - U = \frac{1}{2} m \dot y^2 - m g y$$
and the action is the functional
$$S (y) := \int_0^T \mathcal{L} (y, \dot y) \, \mathrm{d}t$$
The Euler-Lagrange equation gives us the 2nd order differential equation $\ddot y = -g$. However, suppose we have the equality constraint
$$\int_0^T \mathcal{L} (y, \dot y) \, \mathrm{d}t = S_0$$
which can be rewritten in the form
$$\frac{S_0}{T} = \left(\frac{1}{T} \int_0^T \frac{1}{2} m \dot y^2 \, \mathrm{d}t \right)- \left(\frac{1}{T} \int_0^T m g y \, \mathrm{d}t \right) = \langle K \rangle - \langle U \rangle$$
where $\langle K \rangle$ and $\langle U \rangle$ are the average kinetic and potential energies, respectively. Thus, the equality constraint merely imposes a constraint on the average kinetic and potential energies.
