Solving non-linear second order differential equation: radius of curvature $= k \theta$ I'm trying to find any curve where the radius of curvature increases linearly with angular displacement. So in polar coordinates
radius of curvature $= k \theta$
$$ \frac{(r^2 + r'^2)^{3/2}}{r^2 + 2r'^2 - rr''} = k \theta$$
where k is a constant
I just need a single curve that fits this equation, does not matter if it is defined as an integral or power series. But, best to have a closed form solution. Thank you.

If $\psi $ is angle between radius vector and tangent and primes are taken with respect to arc length, then we need to solve for $ r,\theta $ via $ \psi$ for reciprocal radius of curvature:
$$ \dfrac{d \psi}{ds} + \dfrac{\sin \psi}{r} = k\, \int  \dfrac{\sin \psi}{r} ds, \, r^{'} = \cos \psi,\, \theta^{'}= \dfrac{\sin \psi}{r}.  $$
EDIT1:
As a first step numerical solution is given below for assumed boundary conditions near origin.

