Let $f: \mathbb{R}^m \rightarrow \mathbb{R}^m$ is a locally Lipschitz function and homogeneous with a degree greater Let $f: \mathbb{R}^m \rightarrow \mathbb{R}^m$ is a locally Lipschitz function and homogeneous with a degree greater than one, that is, there exists $r> 1$ such that $f (sx) = s^rf (x)$ for all $x \in \mathbb{R}^m$ and all $s> 0$.
Show that if $f (x)$ is a positive multiple of $x$ for some nonzero $x \in \mathbb{R}^m$, then
some solution of the differential equation $x'=f(x)$ explodes in finite time, that is, not
is defined for all $t ≥ 0$.
Can anyone give a hint how to proceed? I have tried in many ways, but without success.
 A: Let $x_0$ be a point for which there is some positive constant $c$ with $f(x_0)=cx_0.$
We will show that there is a solution of the ODE $\gamma'(t)=f(\gamma (t))$ of the form $\gamma (t)=\lambda (t) x_0,$ where $\lambda$ is a positive valued function. Using this ‚ansatz‘ we get:
\begin{equation}\lambda'(t)x_0=\gamma'(t)=f(\gamma (t))=f(\lambda (t) x_0)=\lambda^r (t)f(x_0)=c\lambda^r(t) x_0
\end{equation}
So such a $\gamma$ is a solution if $\lambda$ solves the ODE
\begin{equation}\lambda'=c\lambda^r
\end{equation}
with an arbitrary initial value condition (let‘s take $\lambda(0)=1$). But this ODE can easily be solved, because
\begin{equation}c=\frac{\lambda'}{\lambda^r}=\left( \frac{1}{(1-r)\lambda^{r-1}}\right)',
\end{equation}
from which it follows by integration, using the initial value condition and multiplying by $1-r$ that
\begin{equation}c(1-r)t+1=\frac{1}{\lambda^{r-1}(t)}.
\end{equation}
So we get that 
\begin{equation}\lambda(t)=\left(\frac{1}{c(1-r)t+1}\right)^{\frac{1}{r-1}},
\end{equation}
which explodes in finite time. So $\gamma$ also explodes in finite time.
