Differential equation $y''+(y')^2+1=0$ I’m trying to solve this equation but at the end I’m stuck and can’t reach the answer. I use the substitutions $u=y'$ and $y''=du/dx$:
$$du/dx+u^2+1=0, \\ -du/(u^2+1)=dx, \\ -\arctan(u)=x+c$$
Here I don’t know how to go on. The answer should be $$y=\ln|\cos(c_1-x)|+c_2$$
 A: From $-\arctan(u) = x+c$ we find that $u = \tan(-x + c).$
Now replace $u$ with $y'$, i.e. $y' = \tan(-x+c)$.
We integrate to find $y$ (rewrite tan in terms of sin and cosine, make the substitution $v = \cos(-x+c)$).
This gives the desired result.
A: $$y''+(y')^2+1=0$$
$$u'+u^2+1=0$$
This is Riccati's differential equation substitute $y'=u=\dfrac {f'}{f}=(\ln f)'$:
$$f''+f=0$$
$$\implies f=c_1\cos x +c_2 \sin x$$
$$e^y=c_1\cos x +c_2 \sin x$$
As Lutz Lehmann pointed in the comment. You can  also multiply by $e^y$ the DE:
$$y''+(y')^2+1=0$$
$$e^yy''+e^y(y')^2+e^y=0$$
$$(e^y)''+e^y=0$$
This is a second order linear DE.
A: In order not to lose physical dimension of curvature we take $a=1$ in:
$$ \frac{y''}{1+y^{'2}}=\frac{-1}{a}$$
Let $ y'= \tan \phi, y^{''}= \sec^2 \phi\cdot \phi'~$ ; plugging in and integrating
$$ \phi^{'} = -\frac{1}{a},~\phi = c_1-\frac{x}{a},~ \tan\phi = \frac{dy}{dx}=\tan(c_1- \frac{x}{a});$$
$$ y= \log\sec(\frac{x}{a}-c_1) + c_2 $$
Setting arbitrary boundary condition  $~c_2= - \log ~y_1$ and for a symmetric boundary condition $c_1=0$ ,we have
$$ y=y_1 ~\log \sec \left( \frac{x}{a}\right)$$
that essentially agrees with the required result.
This is btw, the bounded shape taken by chainette cable of variable section and Uniform strength
