functional analysis - Minimum curve for the distance between two points at the plane

The problem is to determine the curve y=y(x) in the plane, the lenght of which is given by the functional:

$$\begin{equation} I(y)=\int_{x_1}^{x_2}\sqrt{1+(y')^2}dx=\int_{x_1}^{x_2}F(x,y,y')dx \end{equation}$$
which will make the distance between two points of the plane minimum.
In other words, we need to determine the curve for which $I(y)$ is minimum.

Now I just want to ask the following perhaps naive question, but I am not sure about it:

The Euler-Lagrange equation becomes:

$$\begin{equation} \frac{d}{dx}\left(\frac{\partial{F}}{\partial(y')} \right)=0 \end{equation}$$
since the functional $F(x,y,y')=\sqrt{1+(y')^2}$ actually depends only on $(x,y')$, which means that $\partial{F}/{\partial{y}}=0$

Also I can see that (if I am right):
$$\begin{equation} \frac{\partial{F}}{\partial(y')}=F_{y'}=\frac{y'}{\sqrt{1+(y')^2}} \end{equation}$$

but the book ends up with the DE: $y''=0$ and finally the curve extremizing $I(y)$ is:

$$\begin{equation} y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1 \end{equation}$$

I cannot see why $y''=0$ from Euler Lagrange and how I end up with that solution?

Thank you!

Answer

Well, I solved it. Here it goes:

I do not end up with $y''=0$ as the author of this example does but the Euler Lagrange will give:

$$\begin{equation} \frac{d}{dx}\left(\frac{\partial{F}}{\partial(y')} \right)=0 \Leftrightarrow \frac{d}{dx}\left( \frac{(y')^2}{1+(y')^2} \right)=0 \Leftrightarrow \frac{(y')^2}{1+(y')^2}=c \\ (y')^2=\frac{c^2}{1-c^2} \end{equation}$$
for some $c \in \mathbb{R}$ such that $0

$$\begin{equation} y(x)=\lambda x+\mu \end{equation}$$
for $\lambda=\pm \sqrt{c/(1-c^2)}$ and $\mu \in \mathbb{R}$.

Now if we apply the initial conditions as $y(x_1)=y_1$ and $y(x_2)=y_2$ we get the final result:

$$\begin{equation} y(x)= \left( \frac{y_2-y_1}{x_2-x_1} \right) (x-x_1)+y_1 \end{equation}$$

a straight line as promised.