Differential equation $y\cdot y'' - (y')^2 - 1 = 0$ I'm trying to solve this equation but at the end I'm stuck and can't reache the answer. I use the substitutions $y'=p$ and $y'' = p'\cdot p$:
$$y \cdot p'p - p^2 - 1 = 0 \implies y\cdot p \frac {dp}{dy} - (p^2 + 1) = 0 \implies \int \frac {p}{p^2 + 1}dp = \int \frac {dy}{y} \implies \frac{1}{2} \ln \left|C_1(p^2 + 1) \right | = \ln y \implies y = C_1 \sqrt{p^2+1} \implies p = \sqrt {C_1y^2 - 1} $$
Next step:
$$ y' = \sqrt {C_1 y^2 - 1} \implies \frac{dy}{dx} = \sqrt {C_1 y^2 - 1}  \implies \int \frac {dy}{C_1 \sqrt{ y^2 - \frac {1}{C_1^2} }} = \int dx \implies \\ C_1 \ln \left |y + \sqrt{y^2 - C^2_1} \right | = x + C_2 \implies y + \sqrt{y^2 - C_1^2} = \exp {\frac {x+C_2}{C_1}} $$ 
Here I don't know how to go on. The answer should be $ y = \frac {C_1}{2} \left ( \exp(\frac{x+C_2}{c_1}) + \exp(-\frac{x+C_2}{c_1}) \right )$
 A: Divide by $y^2$. Then you have $(y'/y)' = 1/y{^2}$. I'm not sure if this helps. The other way would be to take the original equation and differentiate. You get 
$y y'''  = y'y'' $. Rearranging, we have $y'/y = y'''/y''$. This gives $ln (y'') = ln (y) + c$, which gives $y'' = e^{c} y $. Now with $u^2 = e^{c}$, $y=a cosh (ut) + b sinh (ut) $ is a general solution. Also confirms @alex.jordan 's guess. Please excuse my fillers in between, I will learn laTex soon.
A: Maybe your $C_1$ and $C_2$ are different to the answer's. But let solve your equation:
$$
y+\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})\\
\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})-y\\
y^2-C_1^2=\exp(2\frac{x+C_2}{C_1})-2y\exp(\frac{x+C_2}{C_1})+y^2\\
y=\frac{1}{2}(\exp(\frac{x+C_2}{C_1})+C_1^2\exp(-\frac{x+C_2}{C_1}))
$$
A: You obtained :
$$y + \sqrt{y^2 - \frac{1}{C_1^2}} = \exp {\frac {x+C_2}{C_1}} $$
it remains to solve this equation for $y$
$$\sqrt{y^2 - \frac{1}{C_1^2}} = -y+\exp {\frac {x+C_2}{C_1}} $$
$$y^2 - \frac{1}{C_1^2} = \left( -y+\exp {\frac {x+C_2}{C_1}}\right)^2 $$
Then, simplify and express $y$.
