Flows of the system $\dot{x}=y$ and $\dot{y}=x$ are area-preserving. Show by computation that the area of the unit disk is the same after time is moved forward 2 units by the flow.
My solution to the system is $x=\frac{1}{2}(e^{t}+e^{-t})$ and $y=\frac{1}{2}(e^{t}-e^{-t})$. I can easily check that these are solutions. But I don't know whether they are flows, because when t=0, both solutions equal constants.
More importantly, I don't understand why they make a disk. $x^{2}+y^{2}=(e^{2t}+e^{-2t})/2$. So a circle means $(e^{2t}+e^{-2t})/2\leq1$?
Since the original area is 1, I guess I need to show that the new integral is equal to 1, too. After the time moves forward, $x=\frac{1}{2}(e^{t+2}+e^{-t+2})$ and $y=\frac{1}{2}(e^{t+2}-e^{-t+2})$. I notice that we can let $u=x+y=e^{t+2}$ and $v=x-y=e^{-t+2}$. But how to set up the integration? Thanks.