How is this differential equation solved? I have a differential equation as follows:
$$ y \cdot y'' +(y')^2 +1=0.$$
I'm interested in how to solve it. So far I have found a few solutions like $y=\sqrt{r^2-(x-k)^2}$ and $y=x+k.$ [In these, $r$ and $k$ are real constants.]
It came up while looking at a property of surface area of revolutions which held for spheres and certain cones. I am wondering if a general soution can be found, and in particular whether only circles and certain lines are solutions.
Note: As user RadialArmSaw noted, $y=x+k$ is not a solution.
 A: This equation is simple to solve if you start switching variables
$$y \, y'' +(y')^2 +1=0 \implies -y\frac{x''}{[x']^3}+\frac{1}{[x']^2}+1=0$$ that is to say
$$-y \,x''+(x')^3+x'=0$$ Redution of order $p=x'$ gives
$$-y\,p'+p^3+p=0$$ which is separable
$$\log \left(\frac{p}{\sqrt{p^2+1}}\right)=\log(y)+c_1\implies p=x'=\pm\frac{c_1\, y}{\sqrt{1-c_1^2\, y^2}}$$
$$x+c_2=\pm\frac{\sqrt{1-c_1^2 \,y^2}}{c_1}$$ Squaring leads to
$$\big[c_1(x+c_2)\big]^2+c_1^2\,y^2=1$$ which is the equation of a circle.
A: Noting that $y y'' + (y')^2 + 1 = (y y' + x)'$, the original equation reduces to $y y' + x = c_1$, which is an equation with separable variables whose solution is implicitly defined by the relation
$$
\frac{y^2}{2} = c_1 x-\frac{x^2}{2}+c_2
$$
that can be rewritten as
$$
y^2=k_1 x-x^2+k_2 \Leftrightarrow (x-\alpha_1)^2+y^2 = \alpha_2.
$$
Given initial or boundary conditions, we can compute (when possible) the constants $\alpha_1, \alpha_2$. When the initial/boundary data is compatible with the differential equation the solution is a circumference, as noted by others.
A: You can observe that
$(y^2)’’=(2yy’)’=2(yy’’+(y’)^2)$
Thus you get
$(y^2)’’=-2 $
$(y^2)’-(y^2)’(0)=-2x$
$y^2=y^2(0)+ (y^2)’(0)x-x^2$
Set $a= y^2(0)$, $b= (y^2)’(0)$, then
$y^2=a+bx-x^2$ so that all the possible solutions are of kind
$y=+/-\sqrt{a+bx-x^2}$ for every $a,b\neq 0$
The function it makes sense only for $ \frac{b-\sqrt{b^2+4a}}{2}\leq x\leq \frac{b+\sqrt{b^2+4a}}{2}$
and it’s the solution of your problem (I.e. the maximal definition domain of your solution is) for
$\frac{b-\sqrt{b^2+4a}}{2}< x< \frac{b+\sqrt{b^2+4a}}{2}$
Another way to see the solution is the following one:
The graph of $y$ is the circle (minus the two points $(\frac{b+/-\sqrt{b^2+4a}}{2} ,0)$ ) of center $(\frac{b}{2}, 0)$ and radius $R=\frac{1}{2}\sqrt{b^2+4a}$
$x^2+y^2-bx-a=0$
