Solve $(y')^2=(y/c)^2-1$ Can someone help me solve $(y')^2=(y/c)^2-1$?  WolframAlpha is giving me $\frac 12(c^2 e^{(x/c)-k}+e^{k-(x/c)})$.  One book I have is giving me $y=c\cdot \cosh(\frac {x+b}c)$ -- but that one won't work for the HW problem I'm solving because with the conditions $y(0)=y(D)=0$ it gives $y$ as identically $0$ -- so it must not be the most general form (though it does clearly solve this).  I'm also seeing elsewhere that the answer should be $y_0 + A\cosh(k(x-x_0))$, which would work for these boundary conditions, but I'm having trouble verifying that it actually solves this ODE.
I'm just not at all good at solving nonlinear ODEs.
 A: $$\dfrac{dy}{dx} = \sqrt{\left(\frac{y}{c}\right)^2-1} \implies \int{\dfrac{dy}{\sqrt{\left(\dfrac{y}{c}\right)^2-1}}} =\int1dx$$
we let $u=\dfrac{y}{c}$, so $du = \dfrac{dy}{c}$
$$c\int{\dfrac{du}{\sqrt{u^2-1}}} =x+k$$
With $k$ a constant. Now, we know this is a "classic" integral: arccosh
$$c\int{\dfrac{du}{\sqrt{u^2-1}}} =x+k \implies \cosh^{-1}u=\dfrac{x+k}{c}\implies u = \cosh\dfrac{x+k}{c} \implies$$
$$\dfrac{y}{c} = \cosh\dfrac{x+k}{c} \implies y = c\cosh\dfrac{x+k}{c} $$
A: With $ y'(x) = \tan\phi $ substitution the equation reduces to $  y = c \sec(\phi) $ which is the standard  ODE of Catenary:
$$ y = c \cosh (x/c) \tag {1*} $$
$$ c \dfrac{dy}{dx} = \sqrt{y^2-c^2} $$ transpose, integrate
$$ x/c = \cosh^{-1}\frac{y}{c} + C_1 $$ Choosing symmetric BC, $ C_1$ can be made to vanish getting (1*).
Else, choosing general  BC: 
$$ y = c \cosh ((x- C_2)/c) \tag {1*} $$
btw, it is sketched often with its involute, the Tractrix. 
A: "Elementary" solution (The Sh. Holmes type "elementary")
From the form of the equation we see that it is a hyperbolic relation 
$$
1=(y/c)^2-(y')^2
$$
that can be parametrized as $y(x)/c=\cosh(u(x))$ and $y'(x)=\sinh(u(x))$. Consistency is reached when the derivative of the first coincides with the second, 
$$
\sinh(u(x))=y'(x)=c·\sinh(u(x))·u'(x)
$$
which gives $u'=1/c$ or $u(x)=x/c+k$. Thus
$$
y(x)=c·\cosh(x/c+k)
$$
for some $k\in\Bbb R$.

$y(0)=0$ is not possible in the real numbers since $(y')^2=-1$ is not solvable. However, $\cosh(\pm i\frac\pi2)=\cos(\pm \frac\pi2)=0$
