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 $y(0)=\mu$.
Prove that there exists a $\tau = \tau(\mu)$ such that $0<x_{\mu}(\tau)\leq 1$ and $y_{\mu}(\tau)=1$.
Find $\tau(\mu)$ and show that $x(\tau(\mu))={\mu}^2$.
APPROACH
This is an exam question that I can't seem to figure out what they exactly are asking. The question actually consisted 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 is what you see above as my question.
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.)
You have found that $x_{\mu}(t) = e^{-2t}$ and $y_{\mu}(t) = \mu e^t$.
Let $\tau(\mu)= \frac 1 {\mu}$. Note that $\tau(\mu) > 0$ since $0 < \mu < 1$. Then $y_{\mu}(\tau)=1$ tand $x_{\mu}(\tau)=\mu^2$ so $0 < x_{\mu}(\tau) < 1$.