ODE Cauchy problem Consider the Cauchy problem $ x'=f(t,x)  ,x(0)=0 $,
where 
$$ f(t,x)=\begin{cases}
\left(\frac{2}{\sqrt{\|x\|}}(x_{1}+x_{2}),\frac{2}{\sqrt{\|x\|}}(x_{2}-x_{1})\right)  &x\neq 0\\
(0,0) & x=0
\end{cases}$$
Solving the above system by converting from Cartesian coordinates $(x_{1},x_{2})$ to polar coordinates $(r,\theta)$.
After solving what conclusion can you draw about?
 A: If the initial condition is $x(0)=0$, there is nothing to do. The solution is $x(t)\equiv(0,0)$.
Otherwise:
We rewrite in polar coordinates $x(t)=(r(t)\cos(\theta(t)),r(t)\sin(\theta(t)))$, then
$$
x'(t)=(r'\cos(\theta)-r\sin(\theta)\theta', r'\sin(\theta)+r\cos(\theta)\theta')
$$
Therefore we have:
$$
x_1'=r'\cos(\theta)-r\sin(\theta)\theta'= 2(\,\cos(\theta(t))+\sin(\theta(t))\,) \\
x_2'=r'\sin(\theta)+r\cos(\theta)\theta' =2(\,-\cos(\theta(t))+\sin(\theta(t))\,) 
$$
Combining these relations we obtain: 
$$
x_1'\cos(\theta)+x_2'\sin (\theta)=r'=2
$$
and
$$
x_1'\sin (\theta) -x_2'\cos(\theta)=-r\theta'=2
$$
From the first equation we find that $r(t)=2t+a$, where $a$ a constant. From this and the second equation, we get
$$
\theta'(t)=\frac{-1}{t+\frac{a}{2}},\;\;\mbox{which implies } \theta(t)=-\log(t+\frac{a}{2})+b
$$
Example: Suppose $x(0)=(1,0)$, then $r(0)=1$ and $\theta(0)=0$. Then $r=2t+1$ and $\theta=-\log(t+\frac{1}{2})+\log(\frac{1}{2})=-\log(2t+1)=-\log(r)$. That is $r=e^{-\theta}$. 
