$$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$

My progress: I have used substituon $u = \frac{dy}{dx}$ and arrived at $u^2 = \frac{(x+c)^2}{1-(x+c)^2}$ equation with $c$ - constant. After that I was unsure on whether it is allowed to consider separate cases for $u$ or not: maybe for some values of $x$, $u$ will have positive sign, on other occasions negative sign. In fact, any help related to such sign issues would be welcomed.

  • $\begingroup$ I suspect that $u^2 = \frac{(x+c)^2}{1-(x+c)^2}$ is false. Would you mind edit the steps of your calculus. The misunderstanding might be due to some ambiguities in the typing of your equation $\frac{d^2y}{dx^2}^2$ = $(1+\frac{dy^2}{dx^2})^3$. Is there $\left(\frac{d^2y}{dx^2}\right)^2$ or $\frac{d^2(y^2)}{dx^2}$ ? And what is the meaning of $\frac{dy^2}{dx^2}$ ? is it $\frac{d^2y}{dx^2}$ ? $\endgroup$
    – JJacquelin
    Commented May 2, 2020 at 8:33
  • $\begingroup$ I am sorry for $3/2$, I could not write the exponent in latex form. Now, DE is much clear. $\endgroup$
    – Snowball
    Commented May 2, 2020 at 8:40

4 Answers 4


Rearranging the terms, we get $$1 = \frac{(1+(y')^2)^{\frac 32}}{y''}$$ Which is the formula for radius of curvature. This DE satisfies an equation whose radius of curvature at every point is 1. This is a circle of radius 1. Hence, the solution would be $$(x-a)^2+(y-b)^2=1$$

  • $\begingroup$ the radius of curvature can also be negative $\endgroup$ Commented Nov 19, 2020 at 7:46
  • 1
    $\begingroup$ @AdityaDwivedi Please read the attached wiki page. Radius of Curvature cannot be negative. $\endgroup$ Commented Nov 19, 2020 at 11:15
  • 1
    $\begingroup$ in the wiki page modulus of the quantity is called radius of curvature $\endgroup$ Commented Nov 19, 2020 at 13:05
  • $\begingroup$ circles are not graphs of functions so they cannot be graphs of solutions. Sorry. $\endgroup$ Commented Nov 20, 2020 at 16:28

$$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$ I agree with you for $\left(\frac{dy}{dx}\right)^2 = \frac{(x+c_1)^2}{1-(x+c_1)^2}$ , then : $$\frac{dy}{dx}=\pm\sqrt{\frac{(x+c_1)^2}{1-(x+c_1)^2}}$$ $$y=\pm\int \sqrt{\frac{(x+c_1)^2}{1-(x+c_1)^2}}dx+\text{constant}$$ $$y+c_2=\pm\sqrt{1-(x+c_1)^2}$$ $$(x+c_1)^2+(y+c_2)^2=1$$


I think we need to be careful here, a maximal solution to this differential equation should be a function $\phi$ defined on some interval $I$, and circles are clearly not graphs of functions.

Now, consider a maximal solution $(\phi,I)$. The fact that, for $x\in I$, we have $$\phi’’(x)=(1+(\phi’(x))^2)^{3/2}>0$$ implies that $\phi$ is convex on $I$, or equivalently, that $\phi’$ is increasing on I. Further, $$\forall x\in I,\quad\left(\frac{\phi’(x)}{\sqrt{1+\phi’^2(x)}}\right)’=1$$ Thus, there exists some constant $a$ such that

$$\forall x\in I,\quad\frac{\phi’(x)}{\sqrt{1+\phi’^2(x)}}=x-a$$ In particular, for all $x\in I$ we have $x-a\in (-1,+1)$ that is $I\subset (a-1,a+1)$. Furthermore the sign of $\phi’(x)$ is the same as the sign of $x-a$. It follows that $$\forall x\in I,\quad\phi’(x)=\frac{x-a}{\sqrt{1-(x-a)^2}}$$ A further integration shows that there exists a real constant $b$ such that $$\forall x\in I,\quad\phi(x)=-\sqrt{1-(x-a)^2}+b.$$ Conversely, for any real numbers $a$ and $b$ the function $$\phi:(a-1,a+1)\to\mathbb{R}, \phi(x)=-\sqrt{1-(x-a)^2}+b$$ is a maximal solution to the proposed ode.$\qquad\square$

Remark. So, we note that the graphs of our solutions are in fact half unit circles.

  • $\begingroup$ a correct answer $\endgroup$ Commented Nov 21, 2020 at 14:38


$$ u = \frac{dy}{dx}$$

This is a first order homogeneous differential equation nonlinear. Because


alone and the square is risen to third order about a polynomial with a constant and the first order differential in seconder order.

Our new differential equation with $u$ is:


This can be solve by the very famous method of serepation of constants found everywhere in the literature for differential equation solution methodology. It is as simple as integrating:



Now we have x(u) but wanted x and u=y'. So we need to invert the equation for u(x). Squaring the equation and regrouping for u gives

$$u(x)=(\frac{(x+c)^2}{(1-(x+c)^2)})^\frac{1}{2}= \frac{dy}{dx}$$

This equation is first order and already integrable.

$$(\frac{(x+c)^2}{(1-(x+c)^2)})^\frac{1}{2}dx= dy$$

$$ y(x)= -(1-(x+c)^2))^\frac{1}{2} + C $$

Only this two integration step give us the allowance for two integration constants.

We can than rearrange the equation to the general representation of a unit circle:


that is shifted away from the origin in the plane. This is indeed a hard restriction we did had with the differential equation at the start. To cover all $ℝ^2$ we need to cover the space of ${c_1,c_2}$ in $ℝ^2$.

Only in the second step of this solution path there is a discussion necessary about complexes solution of the square root. For the second step to be real

$1-(x+c)^2)\geq0$ and $x+c\geq0$

are required. Can that be fullfilled for all $c\in ℝ$? That can be.

The question of interest, how can a nonlinear differential equation be so selective for a small part of values and can also be solvable for rather bigger number of solutions tuples $(x,y)$? Now the answer is the nonlinear differential equation is for the constants only. It does not matter on the tuples $(x,y)$ and in this case this is $ℝ^2$ each. In the case we leave $ℝ^2$ it gets not so restricted. The restriction arises from $ℝ^2$. We can define metrics on $ℝ^2$ and that is the point of investigation making this nonlinear differential equation of first order in the first derivative of $y$ so interesting. The metrics can be positive definite and therefore impose well defined orders.

It is necessary to formulate the equation with the square root for that result.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .