I am being asked to "sketch the solution curve in the phase plane, and the x(t) and y(t) graphs" \begin{align} \frac{dx}{dt} &= -3x \\ \frac{dy}{dt} &= -x + 2y \\ Y(0) &=\binom{-2}{1} \end{align} I found the eigenvalues of $-3$ and $2$, as well as the (corresponding) eigenvectors $(5, 1)$ and $(0, 1)$.
From what I understand about phase plans, I should draw lines $y=5x$ and $x=0$ (these are the lines that the eigenvectors lie on). I should draw $y=5x$ with arrows pointing towards the origin (since its eigenvalue $-3$ is negative), and the line $x=0$ with arrows pointing away from the origin (since its eigenvalue $2$ is positive. Then, I should trace some lines that asymptotically follow the direction of what's already on the graph.
What confuses me is that on this phase plane plotter (http://www.bluffton.edu/homepages/facstaff/nesterd/java/slopefields.html), the trajectories seems like they're going in the wrong direction —— it seems like when I plotted my straight line solutions, I should have drawn arrows the other way. I'd like to know if I'm doing something wrong.
Next, if I am to draw the solution with the initial condition $Y(0)=(-2,1)$, should I just follow the trajectory from that point on the graph?
Lastly, when the question asks me to graph $x$ and $y$, I don't know what they mean. Do they mean the solutions for $x$ and $y$, separately graphed?