solve differential equation $y'' = 1+ y'^2$ Solve $$y'' = 1+ y'^2$$
My attempt, let $p = y'$, $$ y''= \frac{dy'}{dx} = \frac{dp}{dy} \frac{dy}{dx} = p \frac{dp}{dy}$$
$$ p \frac{dp}{dy}=1+p^2$$
$$ \int \frac p{1+p^2} dp = \int dy $$
$$ \frac12 ln(1+p^2) = y + ln c_1,\ \sqrt{1+p^2} = c_1 e^y $$
$$ |p|=\sqrt{(c_1 e^y)^2 - 1}$$
I don't know how to continue to solve $ \frac{dy}{dx} = \pm \sqrt{(c_1 e^y)^2 - 1}$?
 A: I think you went in the wrong direction by trying to represent $y''$ in terms of $y$. An easier approach is as follows. You have
$$p'=1+p^2.$$
Separate variables:
$$\int \frac{dp}{1+p^2}=\int dt$$
Both integrals are easy; you get $F(p)=t+C$ for an explicit $F$. You can invert this $F$ to get $p=f(t+C)$. Then integrate this $f$ (which can be done explicitly) to get $y$.
Your approach works, though. You got to $y'=\sqrt{C^2 e^{2y}-1}$ (up to re-identifying constants). Now let $C e^y=\sec(\theta)$, then $y'=\tan(\theta)$ and $C e^y y'=\sec(\theta)\tan(\theta)=\sec(\theta)\tan(\theta)\theta'$. So $\theta'=1$, $\theta=t+C$, so you are left to solve $C_1 e^y=\sec(t+C_2)$. 
A: Integrate $$\frac{y''}{y'^2+1}=1$$
to get
$$\arctan(y')=x+C,$$ or
$$y'=\tan(x+C)$$ and
$$y=C'-\ln(\cos(x+C)).$$

Assuming the initial conditions $y(0)=y_0$ and $y'(0)=y'_0$, you have $y'_0=\tan(C)$ and $y_0=C'-\ln(\cos(C))$, then
$$y=y_0-\ln\left(\frac{\cos(x+\arctan(y'_0))}{\cos(\arctan(y'_0))}\right).$$
A: To balance physical dimensions let us  introduce a linear dimension $a$ instead of unity.
$$\int \frac{dp}{1+p^2}=\int dx /a $$
Integrate 
$$ \tan^{-1} p = x/a + c $$ 
General case is discussed by others already.An important particular case is discussed here. Boundary condition is chosen to have zero slope at $ x=0 , y'=0 $, the lowest point, so that  $ c=0 $. For a symmetrical situation 
$$ p = \tan x/a = dy/dx $$
Again integrate 
$$  y = a \log \sec \, (x/a)  $$
BTW, it is a catenary of increasing cable section dimension but uniform  strength.
EDIT 1:
To get back to the unsymmetrical reference axes we need to add the two constants of integration of second order differential equation as Cartesian shifts:
$$  \boxed {(y-k) = a \log \sec \, [(x-h)/a]}. $$
A: $$f''(x)=1+f'(x)^2$$
$$\frac{f''(x)}{1+f'(x)^2}=1$$
$$\int\frac{f''(x)}{1+f'(x)^2}dx=\int dx$$
Let $w=f'(x)$. Then $dw=f''(x)dx$. Therefore:
$$\int\frac{dw}{1+w^2}=x+c_1$$
$$\arctan w=x+c_1$$
$$f'(x)=\tan(x+c_1)$$
$$f(x)=-\ln|\cos(x+c_1)|+c_2$$
