Solving a system of equations with proof: I can't figure out what exactly the question is asking me.
Question 
There is a system such as follows: 
$\dot{x}=-2x$, $\dot{y}=y$, given that $0<\eta<1$ and $(x_{\eta}(t),y_{\eta}(t))^T$ is the solution of the system with initial conditions $x(0)=1$ and $y(0)=\eta$. 
Prove that there exists a $\tau=\tau(\eta)$ such that $0<x_{\eta}\leq 1$ and $y_{\eta}(\tau)=1$. Find $\tau(\eta)$ and show that $x(\tau(\eta))={\eta}^2$.
Attempt at solving: 
I have found the general solution to be $x= c_1 (1,0)e^{-2t}+c_2(0,1)e^{t}$ by solving the system in matrix form and finding the eigenvalues and eigenvectors. I also tried solving it by turning it into a seperable equation through $\frac{\dot{y}}{\dot{x}}=\frac{dy}{dx}=-\frac{y}{2x}$ and finding the solution $y=\frac{C}{\sqrt{x}}$. 
However I can't connect any of these to what the question is asking and I don't know what else to try. This is a sample exam question and I can't find another example of it. Any help or tips would be much appreciated. 

Yes, the general solution is $x(t)=c_1e^{-2t},y(t)=c_2e^t$ where $c_1,c_2$ are constants.
You now apply the initial conditions to get the particular solution $x_0(t),y_0(t)$.
So we have $x_0(0)=1$ and hence $c_1=1$. Similarly, $y_0(0)=\eta$, so $c_2=\eta$.
We want to find $\tau$ so that $y_0(\tau)=1$. That implies that $\eta e^\tau=1$. So $e^\tau=1/\eta$ and hence $\tau=\ln(1/\eta)$. Note that $0<x_0(t)\le1$ for all $t>0$, so it is certainly true for $t=\tau$.
Finally, we want $x_0(\tau)$. It is $e^{-2\tau}=1/(e^\tau)^2=\eta^2$.
