Does symmetric Lagrangian and boundary condition imply symmetric solution? Assume I have a Lagrangian $L(x,y,y')$ such that $L(x,y,y')=L(-x,y,y')$. Moreover, assume that I am interested in the minimization of the action corresponding to $L$ under the boundary condition $y(a)=y(-a)=c$ for some $a,c$. Must a solution to the corresponding Euler Lagrange equation (under the given boundary conditions) satisfy $y(x)=y(-x)?$
If not, are there any additional requirements that can guarantee this?
I know that there are analogous theorems for initial value problems, but since we do not have existence and uniqueness here, I am not sure about the answer in this case.
Thanks in advance.
Edit:
If it helps, my particular Lagrangian is of the form:
$$L=y+a\cdot\delta(x)y$$
And I am trying to minimize:
$$\int_{-b}^{b}L(x,y,y')dx$$
Under the condition $y(-b)=y(b)=c$
Can anything be said about this particular case?
 A: Notation. I will use $t$ as the independent variable, whereas you used $x$.

The answer is negative; there may be solutions $y=y(t)$ to the Euler-Lagrange equation that do not satisfy $y(-t)=y(t)$.
This happens even in the simplest example of the harmonic oscillator $\ddot{y}+y=0$, which is the Euler-Lagrange equation for
$$
L(y, \dot y)=\frac{(\dot y)^2}{2}-\frac{y^2}{2}.$$
On the interval $[-\pi, \pi]$, the most general solution to
$$
\begin{cases}
\ddot{y}+y=0, \\
y(-\pi)=y(\pi),
\end{cases}$$
is $y(t)=A\cos(t)+B\sin(t), $ which needs not satisfy $y(-t)=y(t)$.

However, this example is specific of the interval $[-\pi, \pi]$. It is not generic. If $a$ is such that $\sin(a)\ne 0$, then the most general solution to
$$
\begin{cases} 
\ddot{y}+y=0, \\ 
y(-a)=y(a), 
\end{cases}
$$
is $y(t)=A\cos (t)$, which does satisfy $y(-t)=y(t)$.

I would be tempted to say that this is a general phenomenon; that "most of the times" the solutions to the Euler-Lagrange equations will have the same symmetries as the Lagrangian. But this is just informal chatting. I have no real facts to support this vague conjecture.

The specific example
The OP asks about the Lagrangian
$$L(y, t)=y +a\delta(t), $$
which I find weird, because the Euler-Lagrange equation in this case reduces to $0=0$ (see the other answer). The action reduces to
$$
S(y)=\int_{-b}^b L(y(t), t)\, dt= \int_{-b}^b y(t)\, dt +a, $$
which is not bounded below, as we can take, say,
$$
y_n(t)=-n+\frac{c+n}{b^2}t^2,\qquad n=1, 2, 3, \ldots$$
which is a sequence of functions that satisfy the boundary condition $y_n(b)=y_n(-b)=c$ and is such that $S(y_n)\to -\infty$.
A: More generally, if an action functional is invariant under a group $G$, then the corresponding Euler-Lagrange (EL) equations are also invariant, but their solutions may not be, cf. e.g. this related Phys.SE post. The latter phenomenon is known as spontaneous symmetry breaking (SSB), which happens in many physical systems.
A: If Lagrange is T-symmetric (invariant under $x\to -x$), the Euler-Lagrange equation is of course T-symmetric, i.e.
\begin{equation}
0=\frac{d}{dx}\frac{\partial L(x)}{\partial(dy/dx)}-\frac{\partial L(x)}{\partial y}=\frac{d}{d(-x)}\frac{\partial L(-x)}{\partial(dy/d(-x))}-\frac{\partial L(-x)}{\partial y}.
\end{equation}
Thus with the symmetric boundary condition, the solution must satisfies $y(-x)=y(x)$.
The same conclusion can be extended to T-odd $L$, i.e. $L(-x)=-L(x)$. Generally, if you have $L(-x)=(-1)^s L(x)$ with a symmetric boundary, you should have $y(-x)=y(x)$.
