Consider the system of differential equations:

$$x'=-x-y+4$$ $$y'=3-xy$$

a. Find the fixed points.







fixed points: (1,3), (3,1)

b. Determine the type of the linearized system at each fixed point.

calculating the Jacobian:

\begin{array}{cc} -1 & -1 \\ -y & -x \\ \end{array}

for the fixed point (1,3): \begin{array}{cc} -1 & -1 \\ -3 & -1 \\ \end{array}

calculating eigenvalues:

$λ_1=-1-\sqrt 3$ (could be positive or negative) $λ_2=\sqrt 3-1$ (negative)

So it is unstable (I think, because if we use the negative root of 3, then the first eigenvalue is positive, is this correct?

for the fixed point (3,1): \begin{array}{cc} -1 & -1 \\ -1 & -3 \\ \end{array}

calculating eigenvalues:

$λ_1=-2-\sqrt 2$ (negative) $λ_2=\sqrt 2-2$ (negative)

So it is stable

C. Determine the nullclines and the signs of $x'$ and $y'$ on the nullclines and in the various regions determined by them. (I'm not sure I am calculating the signs of $x'$ and $y'$ correctly)

y-nullcine: $y'=3/x$ x-nullcine: $y=4-x$


$x'<00$ $y'>0$


$x'<0$ $y'>0$


$x'>0$ $y'<0$


$x'<00$ $y'<0$


$x'<0$ $y'<0$


$x'0$ $y'<0$

enter image description here

d. Draw the phase plane portraitenter image description here

  • $\begingroup$ So is the general method to look at the phase portrait to determine the positive and negative roots, or the opposite, do I need to determine the roots to draw the phase portrait? $\endgroup$
    – Math Major
    Nov 14 '14 at 16:27
  • $\begingroup$ I get everything about the critical points and nullclines, but I guess the problem is I am still a little confused about how to draw the direction fields or how I can determine the direction lines if I don't have the phase portrait to work backwards from $\endgroup$
    – Math Major
    Nov 14 '14 at 16:33

Phase portrait:

$\qquad\qquad\qquad\qquad\qquad$enter image description here

Fixed points:

  • At $(3,1)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-1&-3\end{pmatrix}$, trace $-4$ (negative), determinant $+2$ (positive), discriminant $(-4)^2-4\cdot(+2)=8$ (positive), hence two real negative eigenvalues: the point $(3,1)$ is a stable node
  • At $(1,3)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-3&+1\end{pmatrix}$, trace $0$, determinant $-4$ (negative), hence two real eigenvalues of opposite signs: the point $(1,3)$ is a saddle point

...As explained there:

$\qquad$enter image description here


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