Advice on how to draw phase portrait Let's say we already have two curves (from solving a differential equation):
$$x(t) = 2C e^{-t}+De^{-4t}\\
y(t) = Ce^{-t} - De^{-4t}\\$$
I can draw it for $C = 0$ or $D = 0$, that was not hard. However I get into trouble when the other variants need to be drawn (so like $D \neq 0 $ and $C \neq 0$). What is the best systematic approach for drawing phase portraits like these?
EDIT:
post of the system, where the solution came from:
$$\frac{dx}{dt} = -2x+2y\\
\frac{dy}{dt} = x-3y\\$$
 A: From your solution we can see that the eigenvalues are $-1$ and $-4$ with corresponding eigenvectors $(2,1)$ and $(1,-1)$.  As $t$ goes to $-\infty$, $x$ and $y$ get very large, but $e^{-4t}$ gets larger much faster than $e^{-t}$, the trajectory must start heading in the direction of $(1,-1)$ as it leaves the critical point $(0,0)$.  (It's leaving backwards as $t$ heads toward $-\infty.$)
When $t$ approaches $+\infty$, $x$ and $y$ get smaller, but the $e^{-4t}$ gets smaller much quicker, so the other eigenvector becomes dominant.  The trajectory become tangent to the line through the origin in the direction of $(2,1)$.
So draw your two straight trajectories corresponding to when $C=0$ and when $D=0$.  Put your pencil on $(0,0)$ and draw a curve tangent to the $(2,1)$ trajectory, then have it slowly turn until it's going in the direction of the $(1,-1)$ trajectory.  Then put arrows on your curve to show that the trajectory is towards the origin.  Repeat a few times for each "eigenquadrant" and you'll have a pretty good picture.
If the eigenvalues are both positive, then reverse the reasoning and the arrows.  If they're opposite sign, then similar reasoning tells you when a trajectory should be parallel to which eigenvector.
