# 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.

• what is $(y y')' \; \; \; ? \; \;$ Feb 25, 2022 at 0:24
• Did you try solving the associated homogeneous equation? Feb 25, 2022 at 0:30
• @WillJagy oh- I’m sorry. I thought you were asking that question because you were confused. But now I realize the premise. Feb 25, 2022 at 0:37
• $y=x+k$ is not a solution. Feb 25, 2022 at 0:42
• @RadialArmSaw You're right. [I had put all but the +1 and it came out 1, but for a solution that should be $-1.$ Thanks. Feb 25, 2022 at 3:18

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 $$R=\frac{1}{2}\sqrt{b^2+4a}$$

$$x^2+y^2-bx-a=0$$

• Nice, +1. Formatting tip: use \pm to get $\pm$ (and, FYI, \mp for $\mp$)
– MPW
Feb 25, 2022 at 15:37

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.

• @Teepeemm Thanks for pointing it. You are totally right. Cheers :-( Feb 25, 2022 at 15:22

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.