Find locus of point (in the plane, polar coordinates $r,\theta$ ) if variable curvature $1/R$ is proportional to $ \sin^3\psi , \psi$ being variable angle between radius vector and tangent to the curve.
$$ R = c/ \sin ^3 \psi $$
EDIT1/2:
Tangent slope $\phi$ ccw angle convention
$$ \phi = \theta + \psi \tag1 $$
Differentiate w.r.t. arc
$$ \frac{1}{R} = \frac{\sin \psi}{r} +\psi^{'} \tag2 $$
$$ \theta ^{'}= {\sin \psi}/{r}, \quad r^{'}= \cos \psi; \tag3 $$
in differential geometric form when primed w.r.t. arc for $(r,\theta)$. Next after differentiating (2) w.r.t arc
$$ \frac{-R^{'}}{R^2} = \frac{ r \cos \psi \;\psi^{'}-\sin\psi \cos \psi} {r^2} +\psi^{''} \tag4 $$
to eliminate $r$ we substitute from (2)
Please help to find ODE involving $ \psi $ and its derivatives only.