How to determine uniquely an extremal of a variational problem

Question

This is a question from a competitive examination:

Consider the functional $$ J(y) = \int_{a}^b (y+y^{'})dx $$ for admissible functions y, Then $J$ has:

1. No extremals.
2. Several extremals.
3. $y(x)=e^{-x}$ as an extremal.
4. $y(x)=$ constant as an extremal.

What I have tried:

Using the Euler's second equation: $$ \frac{d}{dx} \left(F - y^{'}\frac{\partial F}{ \partial y^{'} } \right) - \frac{\partial f}{\partial x}=0 $$

Since $F=y+y^{'}$ is independent of $x$, we get $$ F - y^{'}\frac{\partial F}{ \partial y^{'} } =c$$ $$ i.e. \hspace{2cm} y+y^{'}-y^{'}=c \implies \hspace{.5cm} y=c$$ for some constant c.

But if I use Euler's first equation namely $$ \frac{\partial F}{\partial y} -\frac{d}{dx}\left( \frac{\partial F}{\partial y^{'}} \right)=0 $$

I have $$ 1- \frac{d}{dx}(1)=0 \implies 1=0 $$ Thus no extremal exists.

The answer is no extremal exists.

So can someone please provide some insights into this nature of the extremal.