Motion of charged particle in uniform magnetic field and a radially symmetric electric field This question posted by me on MSE talks about a physics problem. This is what it was: (I hope someone can help me with this)
Consider a region of 2-dimensional space with a uniform magnetic field of magnitude $B$ directed perpendicular to the plane. With respect to some fixed point (origin), at a point with position vector $\vec {r}$, there exists an electric field $\vec{E} = E_0\vec {r}$ where $E_0$ is some positive constant. Both the magnetic and electric fields are constant with respect to time and exist independently of each other. We have a particle of mass $m$ that has charge $q$ (where $q > 0$) which we release in this region (from rest) at a distance $a$ from the origin.
The aim is to realise (and prove) that the subsequent path described by the particle is an epicycloid
However, this is true only if $E_0 < \frac {qB^2}{4m}$
Apparently, for larger magnitudes of electric field the motion of particle becomes unbounded (its distance from the origin keeps increasing with time).
Is this a known result? I can't seem to find this anywhere on the internet.
How do I prove this?
(Edit: This is the progress I had already made with the problem. I should have added this when I asked the question, can only offer an apology)
I used polar coordinates $(r, \theta)$ with the same origin as above.
I write $\omega$ for $\frac {d \theta}{d t}$. The equations would be: (assuming $\vec{B}$ is along $\vec{\theta} \times \vec {r}$)
$$m (\ddot{r} - r \omega^2) =qE_0 r - qr\omega B  \dots(1)$$
$$m(2 \dot{r} \omega + r \dot{\omega}) = q\dot{r} B \dots (2)$$
From $(2)$, multiply both sides with $r$ and integrate w.r.t. time: (this is equivalent to writing torque equation about origin)
$$r^2 \omega = \frac {qB}{2m} (r^2 - a^2) \dots (3)$$
Substitute $\omega$ from $(3)$ into $(1)$ to get:
$$\ddot {r} = \left( \frac {qE_0}{m} - \left( \frac {qB}{2m} \right)^2 \right) r + \left( \frac {qB}{2m} \right)^2 \frac {a^4}{r^3}$$
Integrating both sides with respect to $r$:
$$(\dot {r})^2 = \left( \frac {qE_0}{m} - \left( \frac {qB}{2m} \right)^2 \right) (r^2 - a^2) + \left( \frac {qB}{2m} \right)^2 \frac {a^2(r^2 - a^2)}{r^2}$$
In an attempt to eliminate the variable time from these equations, I tried to compute $\frac{dr}{rd\theta}$ as $\frac {\dot {r}}{r \omega}$ :
$$\left(\frac {dr}{r d \theta} \right)^2 = \frac {a^2 - r^2 \left(1- \frac {4mE_0}{qB^2} \right)}{r^2 - a^2} = \frac {a^2}{b^2} \frac {(b^2 - r^2)}{(r^2 - a^2)}$$
where $1- \frac {4mE_0}{qB^2} = \frac {a^2}{b^2}$ for some $b$. ($b$ happens to be the maximum distance from the origin during motion)
Now, although this does let me describe $\theta$ as a function of $r$, I am having trouble concluding from there that the path is an epicycloid. As an example, this should represent a part of the epicycloid:
$$2 \theta = \frac {b}{a} \cos^{-1} \left(\frac {a^2 + b^2 - 2 r^2}{b^2 - a^2} \right) - \cos ^{-1} \left(\frac {\frac {2a^2b^2}{r^2} - a^2 -b^2}{b^2 - a^2} \right) $$
This is where I am stuck
Edit: Thanks to Hint II by Frobenius, specifically the substitution $\xi = x + iy$, I am now able to conclude that $\xi$ is of the form:
$$\frac {b+a}{b-a} e^{i \phi} - e^{i \phi \frac {b + a}{b - a}}$$
Where $\phi$ is a real parameter (happens to be proportional to time).
This I am able to realise and prove as an epicycloid. All that remains to figure out is how do my equations lead to an epicycloid as well..
 A: 
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
For the equation of motion we have
\begin{align}
& q\left(\mathbf E\boldsymbol{+}\mathbf v\boldsymbol{\times}\mathbf B\right)\boldsymbol{=}m\mathbf a \quad \boldsymbol{\Longrightarrow} \quad
q E_0\,\mathbf r\boldsymbol{+}q\,\mathbf{\dot{r}}\boldsymbol{\times}\mathbf B\boldsymbol{=}m\mathbf{\ddot{r}} \quad \stackrel{\mathbf B\boldsymbol{=}B_0 \mathbf e_z}{\boldsymbol{=\!=\!=\!\Longrightarrow}} 
\nonumber\\
& q E_0\left(x,y\right)\boldsymbol{+}q B_0 \left(\dot{y},\boldsymbol{-}\dot{x}\right)\boldsymbol{=}m\left(\ddot{x},\ddot{y}\right) \quad \boldsymbol{\Longrightarrow} \quad
\tag{01}\label{01}   
\end{align}
so the following system of two scalar linear differential equations of 2nd order
\begin{align}
\ddot{x}\boldsymbol{-}b\dot{y}\boldsymbol{-}\epsilon x & \boldsymbol{=}0
\tag{02a}\label{02a}\\
\ddot{y}\boldsymbol{+}b\dot{x}\boldsymbol{-}\epsilon  y & \boldsymbol{=}0
\tag{02b}\label{02b}\\
\epsilon \boldsymbol{=}\dfrac{qE_0}{m}\,,\quad b&\boldsymbol{=}\dfrac{qB_0}{m}
\tag{02c}\label{02c}
\end{align}
Multiplying \eqref{02b} by the imaginary unit $\,i\,$ and adding/subtracting the result to/from \eqref{02a}  we have respectively
\begin{align}
\left(\ddot{x}\boldsymbol{+}i\ddot{y}\right)\boldsymbol{+}i b\left(\dot{x}\boldsymbol{+}i\dot{y}\right)\boldsymbol{-}\epsilon \left(x\boldsymbol{+}iy\right) & \boldsymbol{=}0
\tag{03a}\label{03a}\\
\left(\ddot{x}\boldsymbol{-}i\ddot{y}\right)\boldsymbol{-}i b\left(\dot{x}\boldsymbol{-}i\dot{y}\right)\boldsymbol{-}\epsilon \left(x\boldsymbol{-}iy\right) & \boldsymbol{=}0
\tag{03b}\label{03b}
\end{align}
Defining the complex variable
\begin{equation}
\xi\boldsymbol{=}x\boldsymbol{+}iy
\tag{04}\label{04}  
\end{equation}
equations \eqref{03a},\eqref{03b} yield respectively
\begin{align}
\ddot{\xi}\boldsymbol{+}i b\,\dot{\xi}\boldsymbol{-}\epsilon \,\xi & \boldsymbol{=}0
\tag{05a}\label{05a}\\
\ddot{\overline{\xi}}\boldsymbol{-}i b\,\dot{\overline{\xi}}\boldsymbol{-}\epsilon \,\overline{\xi} & \boldsymbol{=}0
\tag{05b}\label{05b}
\end{align}
Note that equation \eqref{05b} is the complex conjugate of \eqref{05a}, so in order to find explicitly the orbit of the particle it's sufficient to solve one of them, let it be equation \eqref{05a}.
Beyond the possibility to use an integral transform, for example Laplace or Fourier or etc, it's convenient to make the following trick that converts the second order equation \eqref{05a} to a first order. Suppose that we split the coefficient $\,ib\,$ to two coefficients
\begin{equation}
ib\boldsymbol{=}\rho_{1}\boldsymbol{+}\rho_{2}
\tag{06}\label{06}  
\end{equation}
Then equation \eqref{05a} could be expressed as
\begin{equation}
\left(\ddot{\xi}\boldsymbol{+}\rho_{1}\dot{\xi}\right)\boldsymbol{+}\rho_{2}\left(\dot{\xi}\boldsymbol{-}\dfrac{\epsilon}{\rho_{2}} \,\xi \right) \boldsymbol{=}0
\tag{07}\label{07}  
\end{equation}
If it would be possible the two coefficients $\,\rho_{1},\rho_{2}\,$  of equation \eqref{06} to satisfy also the condition
\begin{equation}
\boldsymbol{-}\dfrac{\epsilon}{\rho_{2}} \boldsymbol{=}\rho_{1}
\tag{08}\label{08}  
\end{equation}
then \eqref{06} yields
\begin{equation}
\left(\ddot{\xi}\boldsymbol{+}\rho_{1}\dot{\xi}\right)\boldsymbol{+}\rho_{2}\left(\dot{\xi}\boldsymbol{+}\rho_{1} \,\xi \right) \boldsymbol{=}0
\tag{09}\label{09}  
\end{equation}
and defining the new complex variable
\begin{equation}
\eta \boldsymbol{=}\dot{\xi}\boldsymbol{+}\rho_{1} \,\xi 
\tag{10}\label{10}  
\end{equation}
we have the first order equation
\begin{equation}
\dot{\eta}\boldsymbol{+}\rho_{2} \,\eta \boldsymbol{=}0
\tag{11}\label{11}  
\end{equation}
From equations \eqref{06}, \eqref{08} the unknown coefficients $\,\rho_{1},\rho_{2}\,$ must be the roots of the quadratic equation
\begin{equation}
\rho^2\boldsymbol{-}ib\rho-\epsilon\boldsymbol{=}0
\tag{12}\label{12}  
\end{equation}
Looking carefully in \eqref{12} we note that its  discriminant is
\begin{equation}
\Delta\boldsymbol{=}4\epsilon\boldsymbol{+}\left(\boldsymbol{-}ib\right)^2\boldsymbol{=}4\epsilon\boldsymbol{-}b^2\stackrel{\eqref{02c}}{\boldsymbol{=\!=\!=}}4\left(\dfrac{qE_0}{m}\right)\boldsymbol{-}\left(\dfrac{qB_0}{m}\right)^2
\tag{13}\label{13}  
\end{equation}
that is
\begin{equation}
\Delta\boldsymbol{=}\dfrac{4q}{m}\left(E_0\boldsymbol{-}\dfrac{qB^2_0}{4m}\right)\boldsymbol{\longrightarrow}
\begin{cases}
\boldsymbol{<}0 \quad \text{if} \quad E_0\boldsymbol{<}\dfrac{qB^2_0}{4m}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\
\boldsymbol{=}0 \quad \text{if} \quad E_0\boldsymbol{=}\dfrac{qB^2_0}{4m}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\
\boldsymbol{>}0 \quad \text{if} \quad E_0\boldsymbol{>}\dfrac{qB^2_0}{4m}\vphantom{\dfrac{a}{\dfrac{a}{b}}}
\end{cases}
\tag{14}\label{14}  
\end{equation}
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

A: Your charged particle is subject to an electric force, q(Eo)r, directed radially outward from the origin where E is zero. As it gains speed, it will experience a magnetic force, qvB, at a right angle to its velocity.  The particle will follow a curved path until it reaches a point where the radial velocity is zero. From that point, the curve back down will be symmetrical with the one going up.  If you set the difference of the two forces equal to the required centripetal force at the transition point, use conservation of energy to get the velocity in terms of r, and assume r is much larger than a, you get a close approximation to their limit on Eo.
