How do I solve $\frac{d^2y}{dx^2} = (1+(\frac{dy}{dx})^2)^{3/2}$?

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 $$(x-a)^2+(y-b)^2=1$$

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