# Direction Field for System of Equations

I have a question that asks to draw the direction field for the set of linear systems.

\begin{align} \frac{d\,x}{dt} &= -x + y + 1 \\ \frac{d\,y}{dt}&=x+y+3\end{align}

My attempt:

What I did first was set the system in the form of $d(Q)/dt = KQ + b$, then I found the critical points which was at $[2,1]$ and then I chose an arbitrary point in the plane, lets pick $[1,2]$, and set it as my Q and then solved and found the vector $[1,3]$ to be associated to that point, but this is where I have a problem. How will this vector be plotted? I am on point $[1,2]$ and the tangent vector is $[1,3]$ but how can I draw this? Will it go up 1 unit and right 3?

-

## 1 Answer

See here for an explanation of what a direction field is.

At the point $(a,b)$, the tangent vector is $v=[-a+b+1,a+b+1]^T$, from the ODE system that you give.

Specifically at the point $(1,2)$, the tangent vector is $v=[0,4]^T$, which is a vector parallel to the $x$-axis ($0$ up and $4$ to the right). This process is then repeated for all points in the plane.

-
Apparently the vector <1,3> is wrong though because in my book it shows that the direction field is flat, almost like a 0 slope at point (1,2). –  Q.matin Oct 1 '12 at 6:53
I will edit my post to update. I assumed you were correct. –  Daryl Oct 1 '12 at 7:49