Question

classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. (order your answers from smallest to largest \\(x\\), then from smallest to largest \\(y\\).)

\\
\

$$\begin{aligned} x &= 2x - y^2 \\\\ y &= -y + xy \\end{aligned}$$

\\

Explanation:

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<reasoning>

Find the critical points

Using the Nonlinear Systems of ODEs knowledge point
\[

$$\begin{cases} 2x - y^2 = 0 \\ -y + xy = 0 \implies y(x - 1) = 0 \end{cases}$$

\]
If \(y = 0\):
\[
2x - 0^2 = 0 \implies x = 0 \implies (0, 0)
\]
If \(x = 1\):
\[
2(1) - y^2 = 0 \implies y^2 = 2 \implies y = \pm\sqrt{2} \implies (1, -\sqrt{2}),\ (1, \sqrt{2})
\]
Ordered from smallest \(x\), then smallest \(y\):
\[
(0, 0),\ (1, -\sqrt{2}),\ (1, \sqrt{2})
\]

Find the Jacobian matrix

Using the Jacobian Matrix Linearization knowledge point
\[
J(x, y) =

$$\begin{pmatrix} \frac{\partial}{\partial x}(2x - y^2) & \frac{\partial}{\partial y}(2x - y^2) \\ \frac{\partial}{\partial x}(-y + xy) & \frac{\partial}{\partial y}(-y + xy) \end{pmatrix}$$

=

$$\begin{pmatrix} 2 & -2y \\ y & x - 1 \end{pmatrix}$$

\]

Classify the critical point (0, 0)

Using the Phase Portrait Analysis knowledge point
\[
J(0, 0) =

$$\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$$

\]
\[
\lambda_1 = 2 > 0,\quad \lambda_2 = -1 < 0 \implies \text{saddle point}
\]

Classify the critical point (1, -sqrt(2))

Using the Phase Portrait Analysis knowledge point
\[
J(1, -\sqrt{2}) =

$$\begin{pmatrix} 2 & 2\sqrt{2} \\ -\sqrt{2} & 0 \end{pmatrix}$$

\]
\[
\text{tr}(J) = 2,\quad \det(J) = 0 - (2\sqrt{2})(-\sqrt{2}) = 4
\]
\[
\lambda^2 - 2\lambda + 4 = 0 \implies \lambda = 1 \pm i\sqrt{3}
\]
\[
\text{Re}(\lambda) = 1 > 0 \implies \text{unstable spiral point}
\]

Classify the critical point (1, sqrt(2))

Using the Phase Portrait Analysis knowledge point
\[
J(1, \sqrt{2}) =

$$\begin{pmatrix} 2 & -2\sqrt{2} \\ \sqrt{2} & 0 \end{pmatrix}$$

\]
\[
\text{tr}(J) = 2,\quad \det(J) = 0 - (-2\sqrt{2})(\sqrt{2}) = 4
\]
\[
\lambda^2 - 2\lambda + 4 = 0 \implies \lambda = 1 \pm i\sqrt{3}
\]
\[
\text{Re}(\lambda) = 1 > 0 \implies \text{unstable spiral point}
\]
</reasoning>

<answer>

\((x, y)\)	Conclusion
\((1, -\sqrt{2})\)	unstable spiral point
\((1, \sqrt{2})\)	unstable spiral point

</answer>

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</post_analysis>

Answer:

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<reasoning>

Find the critical points

Using the Nonlinear Systems of ODEs knowledge point
\[

$$\begin{cases} 2x - y^2 = 0 \\ -y + xy = 0 \implies y(x - 1) = 0 \end{cases}$$

\]
If \(y = 0\):
\[
2x - 0^2 = 0 \implies x = 0 \implies (0, 0)
\]
If \(x = 1\):
\[
2(1) - y^2 = 0 \implies y^2 = 2 \implies y = \pm\sqrt{2} \implies (1, -\sqrt{2}),\ (1, \sqrt{2})
\]
Ordered from smallest \(x\), then smallest \(y\):
\[
(0, 0),\ (1, -\sqrt{2}),\ (1, \sqrt{2})
\]

Find the Jacobian matrix

Using the Jacobian Matrix Linearization knowledge point
\[
J(x, y) =

$$\begin{pmatrix} \frac{\partial}{\partial x}(2x - y^2) & \frac{\partial}{\partial y}(2x - y^2) \\ \frac{\partial}{\partial x}(-y + xy) & \frac{\partial}{\partial y}(-y + xy) \end{pmatrix}$$

=

$$\begin{pmatrix} 2 & -2y \\ y & x - 1 \end{pmatrix}$$

\]

Classify the critical point (0, 0)

Using the Phase Portrait Analysis knowledge point
\[
J(0, 0) =

$$\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$$

\]
\[
\lambda_1 = 2 > 0,\quad \lambda_2 = -1 < 0 \implies \text{saddle point}
\]

Classify the critical point (1, -sqrt(2))

Using the Phase Portrait Analysis knowledge point
\[
J(1, -\sqrt{2}) =

$$\begin{pmatrix} 2 & 2\sqrt{2} \\ -\sqrt{2} & 0 \end{pmatrix}$$

\]
\[
\text{tr}(J) = 2,\quad \det(J) = 0 - (2\sqrt{2})(-\sqrt{2}) = 4
\]
\[
\lambda^2 - 2\lambda + 4 = 0 \implies \lambda = 1 \pm i\sqrt{3}
\]
\[
\text{Re}(\lambda) = 1 > 0 \implies \text{unstable spiral point}
\]

Classify the critical point (1, sqrt(2))

Using the Phase Portrait Analysis knowledge point
\[
J(1, \sqrt{2}) =

$$\begin{pmatrix} 2 & -2\sqrt{2} \\ \sqrt{2} & 0 \end{pmatrix}$$

\]
\[
\text{tr}(J) = 2,\quad \det(J) = 0 - (-2\sqrt{2})(\sqrt{2}) = 4
\]
\[
\lambda^2 - 2\lambda + 4 = 0 \implies \lambda = 1 \pm i\sqrt{3}
\]
\[
\text{Re}(\lambda) = 1 > 0 \implies \text{unstable spiral point}
\]
</reasoning>

<answer>

\((x, y)\)	Conclusion
\((1, -\sqrt{2})\)	unstable spiral point
\((1, \sqrt{2})\)	unstable spiral point

</answer>

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