Getting back the original curve having only the curvature More specifically, I'm trying to prove that all plane curves with curvature $k(s) = \cfrac{1}{as + b}$ are logarithmic spirals and trying to describe all the ones such that $k(s) = \cfrac{1}{\cosh(s)}$. I'm aware that this might be done with integrals (there is a constructive proof that all plane curves can be reparameterized by arclength that I thought could help me, but it turns out to be far too complicated). I thought about solving the following system of differential equations, but it didn't get me anywhere: 
$x'' = -ky'$
$y'' = kx'$
Any ideas on how to do this an easier way (or even how I mentioned above, if there is something I'm missing and it's easier than I'm making it out to be)? I'd appreciate any help.
 A: Take the second curve ( the first one is quite different) taking $\phi$ as slope and integrate
$$ \kappa = \frac{d \phi}{ds} = 1/\cosh s ; \, \phi = \tan^{-1}(\sinh s) + c ; \, \tan \phi =  \sinh s +c_1 \tag 1 $$
where the constant vanished when we started curve along x-axis as initial condition $\phi_0 = 0 ,\tag 2 $ then from 1)
$$ \sin \phi = \tanh s\, ; \cos \phi = sech s \, \tag 3 $$
$$ \frac{dy}{dx} = \frac{dy/ds}{dx/ds} = \frac{\sin \phi}{\cos \phi} = \frac{\tanh s}{sech \,s} \tag4 $$
Integrate numerator and denominator separately after setting an arbitrary coefficient $a$
$$ x = \int  a sech s ds\, ; y= a \,\int  \tanh s ds \, \tag 5 $$
$$ x = a \tan^{-1} ( \sinh s ) ,\, y =a  \log(\cosh s) \tag  6 $$
where I took initial value translations as zero. This is the required parametrization.
The curve is visualized, it has vertical asymptote at $ x= \pi a/2. $
Eliminating arc length $s$ between $x,y$ we get a special gravity hanging chain whose formulation may be recognized as a chain of uniform strength, 
$$ y = log \,sec (x/a) \tag 7 $$
a sort of "deep" catenary, where $a$ is specific strength of material.

Proceeding along similar lines parametrization of first curve can be likewise established... in the following way:
EDIT1:
For first curve we integrate the natural equation. Wlog we can set $b=0, a=1$; it only means we starting at a more convenent point elsewhere, to another $1$ parameter, geometrically similar figure or curve. By integrating with proper boundary conditions, setting $c=0$, we get
$$ \kappa = \frac{d \phi}{ds} = 1/s ; \, \phi = \log  \frac{s}{c}; \tag 8 $$
$$ \frac {dx}{ds} = \cos \log \frac{s}{c};\, \frac {dx}{ds} = \sin\log \frac{s}{c} \tag 9 $$
$$ (x,y) = \frac {s}2 (  \cos \log(s/c) + \sin \log(s/c) ,\,- \cos \log(s/c) + \sin \log(s/c) ) \tag {10} $$

This curve gets flatter when you let $s \rightarrow \infty$ as expected.
A: If a plane curve $$\gamma:\quad s\mapsto\bigl(x(s),y(s)\bigr)$$
is parametrized with respect to arc length then
$$\dot x(s)=\cos\theta(s),\quad \dot y(s)=\sin\theta(s)\ ,\tag{1}$$
whereby $s\mapsto\theta(s)$ denotes the changing polar angle of the unit tangent vector ${\bf t}(s)=\bigl(\dot x(s), \dot y(s)\bigr)$. The curvature $\kappa(s)$ then is the speed with which ${\bf t}(s)$ is turning counterclockwise, i.e.,
$$\kappa(s)=\dot\theta(s)\ .$$
In your case $\kappa$ is given as a function of $s$, and the indefinite integral 
$\int \kappa(s)\>ds$ gives $s\mapsto\theta(s)$ up to an additive constant. Changing this constant will just rotate $\gamma$ as a rigid object in the plane. When $\theta(s)$ has been determined a second integration using $(1)$ will then give $s\mapsto x(s)$ and $s\mapsto y(s)$, whereby these functions are of course only determined up to a translation.
A: If you use this curvature, well, it will be a messy. So we can translate the curve $\alpha$ assuming that $\alpha (1) = (1,1)$, $\alpha'(1)=(0,1)$ and take $\kappa(s) = \frac{1}{s}$, $s>0$.
Take
$$\theta(s) = \int_1^s \frac{1}{t} \mathrm{d}t = \ln(|s|)$$
and $\alpha(1) = (1,1) \Rightarrow s_0=x_0=y_0=1$.
So
$$x(s) = 1 \int_1^s \cos (\ln(|t|)+\lambda) \mathrm{d}t \Rightarrow x'(s) = \cos(\ln(|s|) + \lambda).$$
But $\alpha(1) = (0,1) \Rightarrow x'(1)=0$ and $y'(1) = 1$.
So
$$x'(1) = 0 \Rightarrow \cos(\lambda)=0 \Rightarrow \lambda = \frac{\pi}{2} + k \pi; \quad k \in \mathbb{Z}.$$
Taking $k=0$ we have $\lambda = \frac{\pi}{2}$;
So
$$x(s) = 1 + \int_1^s \cos \left( \ln (|t|) + \frac{\pi}{2} \right) dt = 1 - \int_1^s \sin ( \ln (|t|)) dt$$
$$= 1- \left( t \sin (\ln(|t|))|_1^s - \int_1^s \cos(\ln(|t|)) \mathrm{d}t\right) = 1 - s \cdot \sin (\ln(|s|)) + (t \cdot \cos (\ln(|s|))|_1^s - \int_1^s - \sin (\ln(|t|))\mathrm{d}t,$$
ie
$$1 - \int_1^s \sin (\ln(|t|))\mathrm{d}t = 1 - s \cdot \sin (\ln(|s|)) + (t \cdot \cos (\ln(|s|))|_1^s - \int_1^s - \sin (\ln(|t|))\mathrm{d}t \Rightarrow$$
$$\int_1^s \sin (\ln(|t|))\mathrm{d}t = \frac{1}{2} + \frac{s}{2} (\sin(\ln(|s|)) - \cos(\ln(|s|))).$$
Therefore,
$$x(s) = \frac{1}{2} - \frac{s}{2}(\sin(\ln(|s|)) - \cos(\ln(|s|))).$$
In the same way
$$y(s) = 1 + \int_1^s \sin \left( \ln(|t|) + \frac{\pi}{2}\right) \mathrm{d}t$$
where $y'(1) = 1$ and $y(s) = 1 + \int_1^s \cos(\ln(|t|))\mathrm{d}t$.
Integrating two times by parts we have
$$y(s) = \frac{1}{2} + \frac{s}{2} (\sin(\ln(|s|)) + \cos (\ln(|s|))).$$
But
$$\begin{cases}
 \cos(\ln(|s|)) + \sin (\ln(|s|)) = \sqrt{2} \sin \left(\ln(|s|) + \frac{\pi}{4} \right)\\
 \sin(\ln(|s|)) - \cos (\ln(|s|)) = -\sqrt{2} \cos \left(\ln(|s|) + \frac{\pi}{4} \right)\\
\end{cases}$$
So
$$x(s) = \frac{1}{2}+\frac{\sqrt{2}}{2} s \cos \left(\ln(|s|) + \frac{\pi}{4}\right);$$
$$y(s) = \frac{1}{2}+\frac{\sqrt{2}}{2} s \sin \left(\ln(|s|) + \frac{\pi}{4}\right)$$
Let $u = \frac{\pi}{4}$ implies $|s| = e^{u-\frac{\pi}{4}}$ and $s =e^{u-\frac{\pi}{4}}$ by $s>0$.
Taking the reparametrization $\beta(u) = \alpha \circ \psi(u)$ where $\psi (u) = e^{u-\frac{\pi}{4}}$. So
$$\beta(u) = (x(\psi(u)),y(\psi(u))) = \left( \frac{1}{2}+\frac{\sqrt{2}}{2} e^{u-\frac{\pi}{4}} \cdot \cos (u), \frac{1}{2}+\frac{\sqrt{2}}{2} e^{u-\frac{\pi}{4}} \cdot \sin (u)\right)$$
$$\beta(u) = \left( \frac{1}{2}, \frac{1}{2} \right) + \frac{\sqrt{2}}{2} e^{-\frac{\pi}{4}} (e^u \cos (u), e^u \sin (u)),$$
which is a logarithmic spiral.
