TITLE: How do I solve $\frac{d^2y}{dx^2} = (1+(\frac{dy}{dx})^2)^{3/2}$?
QUESTION [9 upvotes]: $$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$
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.

REPLY [3 votes]: 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.