Extremize the functional $J[y]=\int\limits_1^2 (y'^{2}+2yy'+y^2)dx$ 
If $$J[y]=\int\limits_1^2 (y'^{2}+2yy'+y^2)dx, \ \ \ \ \ y(1)=1$$ and $y(2)$ is arbitrary. Find the extremal.

First I solve the Euler-Lagrange equation $F_y-\frac{d}{dx}F_{y'}=0$. Which gives $y=c_1e^x+c_2e^{-x}$ ($c_1,c_2$ are arbitrary constants). Next using the condition $y(1)=1$ and $y(2)=A$. I get the following system
$$c_1e^{}+c_2e^{-1}=1$$
$$c_1e^{2}+c_2e^{-2}=A$$
From this I get $c_2=\frac{e^2(A-e)}{(1-e^2)}$. I can't seem to solve for $c_1$ without getting a real messy equation. 
The answer provided is $y(x)=e^{1-x}$. My question is is there a different way to get this answer? How do I tackle this?
 A: The trick is to rewrite the functional as
$$ J[f]~=~\int_1^2 \! \mathrm{d}x~(y^{\prime}+y)^2~\geq~0. \tag{A}$$
Clearly a solution to $$y^{\prime}+y~=~0 \tag{B}$$ 
would minimize the functional (A). If we combine the first-order ODE (B) with the boundary condition
$$ y(1)~=~1\tag{C} $$
we get the sought-for solution
$$ y(x)~=~e^{1-x}. \tag{D} $$
A: Following the idea Rene gave us, we must add to the Euler-Lagrange equations some restriction. With
$$\int\limits_1^2 (y'^{2}+2yy'+y^2)dx=\int\limits_1^2L(x,y,y')dx$$
We must impose that $L_{y'}(2,y(2),y'(2))=0$, due to the variable value of $y(2)$
$$2y'(2)+2y(2)=0$$
The solution to the Euler-Lagrange equation is, as calculated by Miz,  $y=c_1e^x+c_2e^{-x}$ with $y'=c_1e^x-c_2e^{-x}$
Nevertheless, the system of equations is now:
$$\begin{cases}
   c_1e^{}+c_2e^{-1}=1\\
   c_1e^{2}+c_2e^{-2}=A\\
   c_1e^2+c_2e^{-2}+c_1e^{2}-c_2e^{-2}=0
\end{cases}$$
Leading to $c_1=0$ and $c_2e^{-1}=1$; $c_2=e$ and the bonus, $A=1/e$. Finally
$$y=ee^{-x}=e^{1-x}$$
