I want to find the extremum of
$$J(y)= \int_1^2 \frac{\sqrt{1+y'^2}}{x}dx, \ y(1)=0, \ \ y(2)=1$$
I thought to use the following theorem:
If $y$ is a local extremum for the functional $J(y)= \int_a^b L(x,y,y') dx$ with $y \in C^2([a,b]), \ y(a)=y_0, \ y(b)=y_1$ then the extremum $y$ satifies the ordinary differential equation of second order:
$$L_y(x,y,y')- \frac{d}{dx} L_{y'}(x,y,y')=0 (\text{ Euler's equation})$$
That's what I have tried so far:
$$L(x,y,y')= \frac{\sqrt{1+y'^2}}{x}$$
$$L_y(x,y,y')=0$$
$$L_{y'}(x,y,y')= \frac{2y'}{2x \sqrt{1+y'^2}}$$
So Euler's equation for this problem is:
$$- \frac{d}{dx} \left( \frac{2y'}{2x \sqrt{1+y'^2}} \right)=0 \Rightarrow \frac{y'}{x \sqrt{1+y'^2}}=c \Rightarrow y'=cx \sqrt{1+y'^2} \\ \Rightarrow y'^2=c^2 x^2 (1+y'^2) \Rightarrow \left( \frac{dy}{dx}\right)^2=c^2x^2+ c^2 x^2 \left( \frac{dy}{dx} \right)^2 \\ \Rightarrow \left( 1-c^2x^2 \right) \left( \frac{dy}{dx}\right)^2=c^2 x^2 \Rightarrow \left( \frac{dy}{dx} \right)^2=\frac{c^2 x^2}{1-c^2 x^2} \\ \Rightarrow \frac{dy}{dx}= \pm c \frac{x}{\sqrt{1-c^2 x^2}}$$
We set $\pm c= C$
Then: $\frac{dy}{dx}= \frac{Cx}{\sqrt{1-C^2 x^2}} \Rightarrow dy= \frac{Cx}{\sqrt{1-C^2 x^2}} dx \Rightarrow y(x)=- \frac{1}{c} \sqrt{1-c^2x^2}+k$
$$y(1)=0 \Rightarrow - \frac{1}{c} \sqrt{1-c^2}+k=0 \ (1)$$
$$y(2)=1 \Rightarrow -\frac{1}{c} \sqrt{1-4c^2}+k=1 \ (2)$$
Is it right so far? How could we continue?
EDIT:
$$(2)-(1) \Rightarrow \frac{1}{c} \sqrt{1-c^2}- \frac{1}{c} \sqrt{1-4c^2}=1 \Rightarrow \sqrt{1-c^2}- \sqrt{1-4c^2}=c \Rightarrow (\sqrt{1-c^2}- \sqrt{1-4c^2})^2= c^2 \Rightarrow \dots \Rightarrow 1-6c^2+9c^4=1-4c^2-c^2+4c^4 \Rightarrow 5c^4=5c^2 \Rightarrow c^2(c^2-1)=0 \Rightarrow c=0 \text{ or } c= \pm 1$$
EDIT 2: We reject $c=0$ since we dicide by $c$.
But from the relation $-\frac{1}{c} \sqrt{1-4c^2}+k=1$ we have also the restriction $1-4c^2 \geq 0 \Rightarrow 1 \geq 4c^2 \Rightarrow c^2 \leq \frac{1}{4} \Rightarrow - \frac{1}{2} \leq c \leq \frac{1}{2}$. So $c$ can't take the values $\pm 1$. So does this mean that there is no extremum? Or have I done something wrong?
EDIT 3: I must have done something wrong with the calculations. I retried them and I got $c= \frac{1}{\sqrt{5}}$.
So we get $k=\frac{1}{c} \sqrt{1-c^2}=\sqrt{5}\sqrt{1-\frac{1}{5}}=\sqrt{5} \sqrt{\frac{4}{5}}=\sqrt{4}=2$
So we deduce that $y(x)= - \sqrt{5} \sqrt{1-\frac{x^2}{5}}+2$ is a local minimum. Or am I wrong?