Find Extremum of Functional with No Initial Conditions I have the following functional
$ J(x)=\int\limits_{a}^{b}[{y(x)^2 + \dot{y}(x)^2 + 2 y(x) e^x}] dx $
and I want to find the extremum of it. 
So I compute the Euler-Lagrange Equation and I get the result:
$ y^*(x) = c_1e^x+c_2e^{-x}-{e^x\over 4}+{xe^x\over 2} $
What can be done from here to find the $c_1$ and $c_2$ since I don't have any initial conditions ? Or at least the form that they should have ?
 A: Hints:


*

*Up to constant factors we may write OP's functional as
$$ S[y]~:=~\int_a^b \!dx ~L , \qquad L ~:=~ \frac{1}{2}\dot{y}^2 + \frac{1}{2}( y+e^x)^2 ~\geq~0 .\tag{1}$$

*EL equation is
$$ \ddot{y}-y~=~e^x.\tag{2}$$

*OP has already found the complete solution
$$ y^{\ast}(x)~=~\left(\frac{x}{2}-\frac{3}{4}+c_+\right)e^x + c_-e^{-x},\tag{3}$$
so that
$$\begin{align} \dot{y}^{\ast}(x)&~=~\left(\frac{x}{2}-\frac{1}{4}+c_+\right)e^x - c_-e^{-x}, \cr 
y^{\ast}(x)+e^x&~=~\left(\frac{x}{2}+\frac{1}{4}+c_+\right)e^x + c_-e^{-x}.\end{align}\tag{4}$$

*Therefore the Lagrangian becomes on-shell$^1$ equal to 
$$\begin{align}L^{\ast}(x)&~=~L^{\ast}_+(x)+L^{\ast}_-(x), \cr 
L^{\ast}_+(x)&~:=~ \left(\frac{x}{2}+c_+\right)^2e^{2x}~\geq~0, \cr
L^{\ast}_-(x)&~:=~\left(\frac{1}{4}e^x+c_-e^{-x}\right)^2~\geq~0 ,\end{align}\tag{5}$$
while the functional on-shell is
$$S[y^{\ast}]~=~S_+(c_+)+S_-(c_-), \qquad S_{\pm}(c_{\pm})~:=~\int_a^b \!dx ~L^{\ast}_{\pm}(x).\tag{6}$$

*Clearly it now just remains to minimize $S_{\pm}(c_{\pm})$ independently,
which is a straightforward exercise.
--
$^1$ The word on-shell means that the EL eq. is satisfied.
