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}$?

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 $C_1 e^y=\sec(t+C_2)$. 
