Q:

Second order ODE with no $x$, problem with constants

I have an ODE
$1+(y')^2=2yy''$
for which I know the result must be 
$y(x)=\frac {(c_1^2+1)x^2}{4c_2}+c_1x+c_2$
Making substitution $v=f'(x)$ and solving the ODE
$ln|y|^2+c=ln|1+v^2|$, then
$\sqrt{c_1y-1}=v$
$v=\frac{dy}{dx}$
$dx=\frac {dy}{\sqrt{c_1y-1}}$
$x+c_2=\frac {-2}{c_1}\sqrt{c_1y-1}$
$\frac {c_1(x+c_2)}{-2}=\sqrt{c_1y-1}$
Ok, until now I think I didnt make any mistake, I think my mistake is in the next few steps with $c_1$ and $c_2$
$\frac {(c_1 x+c_1c_2)^2}{4}=c_1y-1$
$y=\frac {c_1(c_1 x+c_1c_2)^2 +4}{4c_1}$
Which is not the same answer i'm looking for (I expanded it and tried to play with it, but it can't be helped, it is not)

A:

This differential equation of x-axis co-directrixed parabolas  have been  recently derived. Instead of integration we follow the easier way it is generated.. as it indirectly includes all the needed integration including what you already started.
For a second order DE we should have only two arbitrary constants and we have to eliminate one of the three constants  in: 
$$ y=ax^2+bx +c \tag1 $$
Say we eliminate $c$.
Maximum/minimum (known) of point of parabola vertex is
$$ (x_m,y_m) = ( -b/2a , \, c- b^2/4a \,) \tag2$$
Definition of parabola locus with given initial radius of curvature using the above definition applicable for this differential equation we set distances to vertex and directrix equal.
$$ ( x+ (b/2a))^2  + (y- 1/(2a))^2 = y^2 \tag3$$
which simplifies to 
$$ y=ax^2+bx + \frac{1+b^2}{4a} \tag4 $$
If you want the parabola in a form 
$$ y= Ax^2 + c_1x + c_2  \tag5 $$
then compare coefficients in 4) and 5) to get:
$$ b= c_1,\quad A = \frac{1+c_1^2}{4c_2}. \tag6$$
EDIT2:
The quantity $ \dfrac{yy''}{ 1+(y')^2 }$ represents the ratio of principal curvatures if the curve is revolved about x-axis.
Let $ y' = \tan \phi $ be the substitution for integration then this ratio can be quite advantageously put into the form
$$ \dfrac{yy''}{ 1+(y')^2 }= \dfrac {d\,log( \cos \phi)}{d\,log( r)}=n, $$ 
then I had derived the following hopefully interesting cases.

$n =-2$  profile of a Parachute 
$n=-1$ Sphere
$ n=-\dfrac12 $  Cycloid
$n= 0 \,$  Cone/Cylinder
$n=+\dfrac12 $ Parabola  with directrix as x-axis
$n=+1$ Catenary
$n=+2 \,$ yet to find out; Are these Weingarten surfaces ?

