# 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)

• I can't open the picture, but your solution is true. Feb 15, 2017 at 4:40
• @MyGlasses edited, that's what wolfram throws Feb 15, 2017 at 4:50
• I could not match them. check the wolfram answer again. Feb 15, 2017 at 5:15

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 ?

• This was indeed a curvature radius problem wth distances to vertex and directrix equal, so although you didn't answer what I asked, this is the correct way of solving this kind of problem. Feb 15, 2017 at 17:17
• EDIT2 addresses your question. They could form a class of such questions on integration. Feb 17, 2017 at 0:41

struggling: More than one hour for one error!

The error occur in last line with a surplus $c_1$:

$\frac {(c_1 x+c_1c_2)^2}{4}=c_1y-1$

$y=\frac {\color{red}{c_1}(c_1 x+c_1c_2)^2 +4}{4c_1}$

From main answer we see $$y=\frac {(c_1 x+c_1c_2)^2 +4}{4c_1}$$ $$y=\frac {c_1}{4}x^2+\frac{c_1c_2}{2}x+\frac{c_1^2c_2^2+4}{4c_1}$$ $$y=\frac {c_1}{4}x^2+Dx+E$$ so \begin{cases} D=\dfrac{c_1c_2}{2}\\ E=\dfrac{c_1^2c_2^2+4}{4c_1}=\dfrac{4D^2+4}{4c_1}=\dfrac{D^2+1}{c_1} \end{cases} then $$c_1=\dfrac{D^2+1}{E}$$ and answer is $$y=\dfrac{D^2+1}{4E}x^2+Dx+E$$

• Ok im dumb, thanks for that. Feb 15, 2017 at 17:16