Consider the system of differential equations:
$$x'=-x-y+4$$
$$y'=3-xy$$
a. Find the fixed points.
$x'=-x-y+4$
$x+y=4$
$x+3/x=4$
x=3,x=1
$y'=3-xy$
$y=3/x$
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: $y=4-x$
R1
$x'<00$
$y'>0$
R2
$x'<0$
$y'>0$
R3
$x'>0$
$y'<0$
R4
$x'<00$
$y'<0$
R5
$x'<0$
$y'<0$
R6
$x'0$
$y'<0$
d. Draw the phase plane portrait
Best Answer
Phase portrait:
$\qquad\qquad\qquad\qquad\qquad$
Fixed points:
...As explained there:
$\qquad$