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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.

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    $\begingroup$ what is $(y y')' \; \; \; ? \; \; $ $\endgroup$
    – Will Jagy
    Feb 25, 2022 at 0:24
  • $\begingroup$ Did you try solving the associated homogeneous equation? $\endgroup$ Feb 25, 2022 at 0:30
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    $\begingroup$ @WillJagy oh- I’m sorry. I thought you were asking that question because you were confused. But now I realize the premise. $\endgroup$ Feb 25, 2022 at 0:37
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    $\begingroup$ $y=x+k$ is not a solution. $\endgroup$ Feb 25, 2022 at 0:42
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    $\begingroup$ @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. $\endgroup$
    – coffeemath
    Feb 25, 2022 at 3:18

3 Answers 3

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

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

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

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

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