How to solve the system of ODEs $f'(x)=-f(x)\cdot \|f(x)\|$ I am looking for continuously differentiable functions $$f:I\to\mathbb R^2, x\mapsto (f_1(x),f_2(x))$$ that satisfy $$f'(x)=-f(x)\cdot \|f(x)\|$$ for all $x\in I$, where $I\subset\mathbb R$ is some interval. (Here, $\|f(x)\|=\sqrt{f_1(x)^2+f_2(x)^2}$.)
I noted that this ODE is separable:
$$\frac{f'(x)}{f(x) \|f(x)\|}=-1,$$ but I don't think that integrating $\frac{f'(x)}{f(x) \|f(x)\|}$ is possible in general.
So what can I do to find the solutions here?
 A: It will help to convert to polar here. I am going to use $(x(t),y(t))$ instead of $(f_1(x),f_2(x))$. Let $x = r\cos(\theta)$ and $y = r\sin(\theta)$, then your ODE reads
$$\dot{x} = -x\sqrt{x^2+y^2}, \quad \dot{y} = -y\sqrt{x^2+y^2}$$
so that
$$\begin{bmatrix}\cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{bmatrix}\begin{bmatrix} \dot{r} \\ \dot{\theta} \end{bmatrix} = -r^2\begin{bmatrix} \cos(\theta) \\ \sin(\theta)\end{bmatrix}$$
inverting gives us
$$\begin{bmatrix} \dot{r} \\ \dot{\theta}\end{bmatrix} = \begin{bmatrix} -r^2 \\ 0 \end{bmatrix}$$
hence $r(t) = \frac{r(0)}{1 + r(0)t}$ and $\theta(t) = \theta(0)$. Then just convert back. Giving
$$f(x) = \frac{f(0)}{1 + ||f(0)||t}$$
A: Let $g=\|f\|^2$. Then
$$
g'=2\langle f,f'\rangle=-2\langle f,f\rangle \|f\|=-2g^{3/2},
$$
and hence $g'(t)g^{-3/2}=-2\,\,$ or $\,\,(g^{-½})'=1$, and hence 
$$
g(t)=(t+c)^{-2} \quad\Longrightarrow\quad \|f(t)\|=\frac{1}{t+c}=\frac{\|f(0)\|}{t+\|f(0)\|}
$$ 
Our system becomes
$$
f_1'=-\frac{\|f(0)\|}{t+\|f(0)\|} f_1,\quad
f_2'=-\frac{\|f(0)\|}{t+\|f(0)\|} f_2
$$
and hence
$$
f_i(t)=f_i(0)\exp\left(-\int_0^t\frac{\|f(0)\|\,ds}{s+\|f(0)\|}\right)
=f_i(0)\exp\big(\log(\|f(0)\|)-\log(t+\|f(0)\|)\big)=\frac{f_i(0)\|f(0)\|}{t+\|f(0)\|}
$$
for $i=1,2$.
