Let $P_1=(a,y_a),P_2=(b,y_b), y\in C^1 (a,b), y_a>0,y_b>0$
And the area integral: $\int^b_a y(x) \sqrt{1+y'(x)}dx$
From the Euler differential-equation we obtain:
$y'=1/\alpha \sqrt{y^2-\alpha^2}, \quad \alpha\in \mathbb R_0$
Now the Author Book (Hans Sagan,Introduction to the calculus of Variations) concludes "Separation of Variables and substitution of $y=\alpha \cosh(t)$ yield
$\alpha t+\beta =x$ and hence,
$y=\alpha \left(\cosh\left(\frac{(x-\beta)}{\alpha}\right)\right)$
I can't reproduce this results, maybe you can see where i do go wrong:
$y'=1/\alpha \sqrt{y^2-\alpha^2}, \quad \alpha\in \mathbb R_0$
I do separate the variables:
$\int \frac{dy}{\sqrt{y^2-1}}=\int dx$
$\Longleftrightarrow \operatorname{arccosh}(y/\alpha)+\beta=x$, $\beta\in \mathbb R$ now theres obviously an $\alpha$ missing
but lets just go on and substitute like he does:" $y=\alpha \cosh(t)$"
$\Longleftrightarrow \mbox{arccosh}(\alpha \cosh(t/\alpha )\alpha)+\beta=\mbox{arccosh}(t)+\beta=x$
$\Longleftrightarrow t=\cosh(x-\beta)$ Which is wrong.
Also shouldn't he calculate the solution with the help of the limits a,b of the integral? I guess he just uses $\alpha$ and $\beta$ , because thats more convient for him in this case.