Graphing solutions of a system of differential equations. I have a system
$$\dot{x} = y, \dot{y} =-4x-5y$$
I have found is general solution to be
$$\textbf{z} = B\textbf{u}e^{-t}+C\textbf{v}e^{-4t}, \textbf{u} = (1,-1),\textbf{v}=(1,-4)$$
Now, I have to sketch typical trajectories.
I find that we have an equilibrium point at the origin and x nullcline $y = 0$ and y nullcline $y = \frac{-4}{5}x$
How do interpret this? 

 A: While you can use the nullclines, you actually already have the eigenvectors, which are usually more useful. Basically, if you start on the line $y=-x$, then you decay to the origin along that line. If you start on the line $y=-4x$, then you decay to the origin along that line. The decay rate along the second line is much faster than the decay along the first. That means that for points in between the two lines, you will tend to approach the first line and then decay near it. For instance consider the initial condition $(1,-1)+(1,-4)=(2,-5)$. You get $z=(e^{-t}+e^{-4t},-e^{-t}-4e^{-4t})$. The $e^{-4t}$ part will quickly decay, so that $z$ will be nearly a multiple of $(1,-1)$ and decay close to that line.
This effect will occur even for points quite close to the line $y=-4x$, but you may have to run the trajectory for a long time and zoom in very close on the origin to actually see it.
Some MATLAB code to help visualize my example:
f=@(t,y) [y(2);-4*y(1)-5*y(2)];
[t,y]=ode45(f,[0 10],[2;-5]);
figure
hold on
scatter(y(:,1),y(:,2))
plot(0:0.01:2,0:(-0.01):-2)

