Given the system of linear differential equations: $$ \dot{x}=-2x $$
$$\dot{y}=y $$ given that $0 < \mu<1$, show that $(x_{\mu}(t),y_{\mu}(t))^T$ is the solution of the given system with the initial values $x(0)=1$ and $y(0)=\mu$.
There exists a $\tau = \tau(\mu)$ such that $0<x_{\mu}(\tau)\leq 1$ and $y_{\mu}(\tau)=1$.
Also given $x(\tau(\mu))={\mu}^2$.
Define $f:[0,1] \rightarrow [0,1]$ with
\begin{equation} f(\mu) = \begin{cases} x(\tau(\mu)) & \text{if $0<\mu\leq1$}\\ 0 & \text{if $\mu=0$} \end{cases} \end{equation}
Show that there exists a $0\leq k <1$ such that
$$|f(x)-f(y)|\leq k|x-y| $$
for $x,y \in [0,\epsilon]$ where $\epsilon >0$ but is small enough.
APPROACH
This is part of an exam question that I can't seem to figure out what exactly is being asked. The question actually consists of three parts, the first part being asking for a phase plane and the character of the origin which is easy to deduce that it's an unstable node. The second part also got me puzzled and I posted that as a separate question earlier.
I solved for the given systems and got $x(t)=c_1e^{-2t}$ and $y(t)=c_2e^t$. Using the given initial values I get $c_1=1$ and $c_2=\mu$. However this is how far I get, I don't understand how I can prove the existence of the mentioned $\tau$ and also don't see any relevance between the first part of the question (not that it has to but I thought maybe I am missing something.)
Regarding the solution for $\tau$: since $y_\mu(\tau) = 1$ we know that $\mu e^\tau = 1$, so $\tau = \ln\frac1\mu$ (which is permitted as $\mu > 0$ by definition). As $\mu < 1$, we thus know that $\tau > 0$ (if you're confused, think it through - what do we know about the graph of $\ln$?). Hence $x(\tau) = e^{-2\tau} < 1$ by properties of the exponential.
I assume the "also given $x(\tau(\mu)) = \mu^2$" is a subsequent part? Because if this is given, then the above is moot - we know $0 < \mu < 1$ hence $0 < \mu^2 < 1$...
Will think more on the second part of the question - this is only a partial solution at the moment.