Surfaces of Revolution with Constant Gaussian Curvature

Surface $S$ is parametrized by $$X(u,v) = (\varphi(v) \cos{(u)}, \varphi(v) \sin{(u)}, \psi{(v}))$$ with everywhere-constant Gaussian curvature $K$. Let $v$ be the arc length of the generating curve $(\varphi(v), \psi(v))$ of $S$; thus, $(\varphi')^{2} + (\psi')^{2} = 1$; fix $a \in \mathbb{R}$.

Prove that $\varphi$ satisfies $$K \varphi + \varphi'' = 0$$

Prove that $\psi = \int (\sqrt{a-(\varphi')^{2}}) dv$, where the domain of $v$ is such that this integral makes sense. $u \in (0, 2 \pi)$; why?

Let us fix a plane with origin $O$ and directions $x$, $y$; denote this plane $\langle x,y \rangle$ by name $xOY$. Fix $K=1$ and let $S$ intersect $xOy$ perpendicularly. Prove that, for some $C = \varphi(0) \in \mathbb{R}$, $\varphi(v) = C cos{(v)}, \psi(v) = \int_{0}^{v} (\sqrt{1-C^{2} sin{(v)}^{2}}) dv$.

Show that all surfaces $S$ of revolution of fixed everywhere-constant Gaussian curvature $K=-1$ and parametrized by a function $X$ of $u$ and $v$ (in that order), wherein $v$ is the arc length of the generating curve of $S$ from a fixed point, are described by the following (up to Euclidean motion):

(a:) $$\varphi(v) = C \cosh{(v)}, \psi(v) = \int_{0}^{v} (\sqrt{1-(C \sinh{(v)})^{2}}) dv$$

(b:) $$\varphi(v) = C \sinh{(v)}, \psi(v) = \int_{0}^{v} (\sqrt{1-(C \cosh{(v)})^{2}}) dv$$

(c:) $$\varphi(v) = \mathbb{e}^{v}, \psi(v) = \int_{0}^{v} (\sqrt{1-\mathbb{e}^{2v}}) dv$$

I am thinking differential equations, but I am not sure how to set them up.

Calculating the coefficients of the first and second fundamental forms we have:

$$E = \varphi^2$$ $$F = 0$$ $$G = (\varphi')^2 + (\psi')^2 = 1$$

and the second form:

$$L = -\varphi\psi'$$ $$M = 0$$ $$N = \varphi' \psi'' - \psi'\varphi''$$

Derivating $(\varphi′)^2 + (\psi′)^2 = 1$. We have $\varphi'\varphi'' + \psi'\psi'' = 0$

Thus, $N$ can be written as $N = \varphi''/\psi'$.

Finally, being $K = \frac{LN - M²}{EG - F^2}$

$$K = \frac{LN}{EG} = \frac{-\varphi\varphi''}{\varphi^2} = - \frac{\varphi''}{\varphi}$$

Then we get, $$K\varphi + \varphi'' = 0$$

$$Q.E.D.$$

do we get K φ - φ′′= 0 ??

The sign is important to choose between elliptic/hyperbolic cases using above standard Gauss definition for K.