Q:

Charged particle moving in magnetic field using cylindrical coordinate

It's knowen that a charged particle take a helix trajectory in a uniform magnetic field $B =B e_z$ 
I tried to study this problem using cylindrical coordinate and i get that 
$$F=q v ×B = m a$$
in cylindrical coordinate (noting $\varphi$  by $f$)  we have 
$$ r= r e_r+ z e_z$$
$$ v = \frac{dr} {dt} = r' e_r +r f'e_f +z'e_z $$
$$  a=   \frac{dv} {dt} = (r''- r f'^2) e_r + (2 r'f' +r f'') e_f +z''e_z  $$
By equalling terms i get
Three equations  one is simple, and others are  a system  of nonlinear differential equations 
,
$$
\left\{ 
\begin{array}{c}
k r f'=r''- r f'^2 \\
-k r'=2 r'f' +r f'' \\
0=z''
\end{array}
\right. 
$$
Where $k=q B/m$
I tried to solve the two first equations but I can't, so I need help 

A:

You are very near the end, you just need to rewrite the equations as:
$$\begin{cases}
\dfrac{\ddot{r}}{r} = \dot{f} \left(k + \dot{f} \right) \\
\ddot{f} = -\dfrac{\dot{r}}{r} \left(k + 2 \dot{f} \right)
\end{cases}$$
with $r \neq 0$.
In the first equation the function are completely separated and this means that they have to be separately equal to zero:
$$\begin{cases}
\ddot{r} = 0 \\
\dot{f} \left(k + \dot{f} \right) = 0 \\
\ddot{f} = -\dfrac{\dot{r}}{r} \left(k + 2 \dot{f} \right)
\end{cases}$$
From this second equation we can see that either $\dot{f} = 0$ (trivial case) or $\dot{f} = -k$. In both cases $\ddot{f} = 0$, so the third equation becomes $\dot{r} = 0$.
So to summarize the radius doesn't change with time ($\dot{r} = 0$) and the angular velocity is constant ($\dot{f} = -k$), which is the definition of uniform circular motion. Add $\ddot{z} = 0$ to the picture and you get a helix.
Edit: I need to clarify a bit what I did when I separated the two sides of the first equation. Let's assume that both parts are equal to a function $C(t)$.
$$\begin{cases}
\dfrac{\ddot{r}}{r} = C \\
\dot{f} \left(k + \dot{f} \right) = C
\end{cases}$$
If $C$ is constant, then it's still easy to calculate $\dot{f}$:
$$\dot{f} = \frac{-k \pm \sqrt{k^2+4C}}{2}$$
with the condition that $C > -\frac{k^2}{4}$. This still means that $\ddot{f} = 0$ and thus $\dot{r} = 0$ from the second equation. The problem is, this now contradicts the first equation, where $\ddot{r} \neq 0$, except when $C = 0$.
The case where $C$ isn't constant is a bit more complicated. While $\dot{f}$ remains the same, $\ddot{f} \neq 0$. Explicitly:
$$\ddot{f} = \pm \frac{\dot{C}}{\sqrt{k^2 + 4C}}$$
This, together with the fact that $C = \frac{\ddot{r}}{r}$, transforms the second equation into:
$$\dddot{r} = - \dot{r} \left( k^2 + 3 \frac{\ddot{r}}{r} \right)$$
Wolfram Alpha says that this equation is solvable, but I don't know at the moment how the solution can be found, other than reading it there. But again the problem that arises from all solutions is that they are all complex in general. 
I have to admit at this point that I can't demonstrate mathematically that the only real solution emerges when $C=0$, but it's surely a solution and the easiest one to find. I'm not even sure this is the best way to demonstrate there aren't other solutions other than the uniform circular motion.

