One more classification
in Assignments
(5 points)
Again open this webpage and now refer to the second section on "One more classification" to get one more system.
For the given system, classify the origin as
- A saddle point,
- An unstable node,
- An unstable spiral,
- A stable spiral, or
- An unstable node.
Explain your answer using the full power of theorem 4.40.
Again be sure to include
- The system,
- The determinant and trace, and
- The classification
Here is another copy of the template that you can use to enter a 2D system into the forum:
$$
\begin{align}
x' &= ax + by \\\\
y' &= cx + dy
\end{align}
$$
Comments
$$
\begin{align}
x' &= 2x - 8y \\
y' &= 9x + 1y
\end{align}
$$
$$
\begin{align}
tr &= 3 \\
det &= 74
\end{align}
$$
It is an unstable spiral because tr>0, det>0, and 4det>tr^2
$$
\begin{align}
x' &= -9x + 4y \\
y' &= 5x + 6y
\end{align}
$$
Det= -74 so, the origin is an unstable saddle point.
$$
\begin{align}
x' &= 4x - 8y \\
y' &= 9x + 3y
\end{align}
$$
determinant=-59, Saddle Point
$$
\begin{align}
x' &= -2x - 8y \\
y' &= 0x - 10y
\end{align}
$$
Since the determinant (ad-cb) is -12 (negative), the origin is a saddle point.
$$
\begin{align}
x' &= -2x -7y \\
y' &= 3x -9y
\end{align}
$$
Det(A)=39
Tr(A)=-11
Stable Spiral because, $$4(39)>(-11)^2$$
$$
\begin{align}
x' &= -5x - 7y \\
y' &= 6x - y
\end{align}
$$
tr = -6
det = 48
tr(A)<4det(A). So the Matrix forms a stable spiral
My system of equations is
$$
\begin{align}
x' &= 8x + 0y \\
y' &= -6x + 9y
\end{align}
$$
The trace is 17 and the determinant is 72.
From theorem 4.41
4det(A) = 288
(trA)^2 = 289
Since 288<289 this is an unstable node.
$$
\begin{align}
x' &= 4x + 0y \\
y' &= -4x + 5y
\end{align}
$$
Since $tr(A)=9>0$, $det(A)=20>0$, and $tr(A)^2>4det(A)$ the origin is an unstable node.
$$
\begin{align}
x' &= -8x - 10y \\
y' &= 6x + 2y
\end{align}
$$
$det = 44$, $tr = -6$, and $4 * det = 178 > 36 = tr ^ 2$, so the origin is an aymptotically unstable spiral.
My system of equations is:
$$
\begin{align}
x' &= 5x - 10y \\
y' &= 3x - 5y
\end{align}
$$
Determinant = 5
Trace = 0
4(5) = 20
(0)^2 = 0
Since 20>0 this is a stable spiral
$$
\begin{align}
x' &= 2x + 5y \\
y' &= -3x + 4y
\end{align}
$$
Trace = 6
Determinant = 23
By theorem 4.40, 4det(A) = 144 and (trA)^2 = 36, so its a unstable spiral.
$$
\begin{align}
x' &= 8x - 2y \\
y' &= -9x + 9y
\end{align}
$$
determinant = $54$
trace = $17$
Here, $tr^2 = 17^2 = 289$ and $4det = 4 \cdot 54 = 216$. Since $tr, det > 0$ and $tr^2 > 4det$, it follows by Theorem 4.40 that the origin is an unstable node.
$$x'=-x+3y$$
$$y'=-6x+6y$$
determinate=12
trace=5
$4det=48$
$tr^2=25$
Since the trace and the determinate are positive that indicates it is unstable and since $4det>tr^2$ then that means that this is an unstable spiral.
$$
\begin{align}
x' &= 7x + 4y \\
y' &= -9x - 2y
\end{align}
$$
$det = 7*-2 - 4*-9 = 22$
$tr = 7 + -2 = 5$
Since $det$ is positive and $tr$ is positive, we have to determine if it's an unstable node or spiral.
$4det > (tr)^2$
$4(22) > 5^2$
$88 > 25$
Since $4det > (tr)^2$, this is an unstable spiral
$$
\begin{align}
x' &= -7x + 8y \\
y' &= -5x + y
\end{align}
$$
Det=33
Trace=-6
4(33)>36
the origin is an asymptomatically unstable spiral
$$
\begin{align}
x' &= -5x - 6y \\
y' &= 7x - 5y
\end{align}
$$
Determinant = 67
Trace = 10
Because the trace is greater than zero and 4*determinant is greater than the square of the trace, this is classified as an:
Unstable Spiral.
$$ x'=2x+4y$$ $$y'=-6x-8y$$
det = 8
tr = -6
Since $ 4det < tr^2$ then by theorem 4.40 this is an unstable node
$$
\begin{align}
x' &= -5x -6y \\
y' &= 3x + 3y
\end{align}
$$
Det=2
tr=-3
Tr^2> Det It is a saddle point
$$
\begin{align}
x' &= -10x + 3y \\
y' &= -4x - 10y
\end{align}
$$
det -> (100 - (-12)) -> 122
tr -> (-10 + (-10)) -> -20
since..
4(122) =488
(-20)^2 = 400
Since 488>400, the origin is an aymptotically unstable spiral.