Determine the stability of the critical point $(0,0)$ for the system of differential equations $dot{x} = x^2 - y$, $dot{y} = y^2 - x$.

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∻Solution by Steps∻ ‖ step 1 ⋮ We start with the system of differential equations: $dot{x} = x^2 - y$, $dot{y} = y^2 - x$. We need to find the critical points by setting $dot{x} = 0$ and $dot{y} = 0$. This gives us the equations: $x^2 - y = 0$ and $y^2 - x = 0$. ‖ ‖ step 2 ⋮ Solving these equations simultaneously at the critical point $(0,0)$, we substitute $x = 0$ and $y = 0$: $0^2 - 0 = 0$ and $0^2 - 0 = 0$. Thus, $(0,0)$ is indeed a critical point. ‖ ‖ step 3 ⋮ To determine the stability of the critical point, we analyze the Jacobian matrix of the system. The Jacobian matrix $J$ is given by: $J = \begin{pmatrix} \frac{\partial dot{x}}{\partial x} & \frac{\partial dot{x}}{\partial y} \ \frac{\partial dot{y}}{\partial x} & \frac{\partial dot{y}}{\partial y} \end{pmatrix}$. Calculating the partial derivatives at $(0,0)$, we find: $J(0,0) = \begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}$. ‖ ‖ step 4 ⋮ The eigenvalues of the Jacobian matrix are found by solving the characteristic equation: $det(J - \lambda I) = 0$. This leads to the equation: $\lambda^2 + 1 = 0$, giving us eigenvalues $\lambda = i$ and $\lambda = -i$. Since the eigenvalues are purely imaginary, the critical point $(0,0)$ is a center and is stable. ‖ ∻Answer∻ ⚹ The critical point $(0,0)$ is stable and behaves like a center. ⚹ ∻Key Concept∻ ⚹ Stability of critical points in dynamical systems is determined by the eigenvalues of the Jacobian matrix. ⚹ ∻Explanation∻ ⚹ The eigenvalues being purely imaginary indicates that trajectories near the critical point will oscillate around it, confirming its stability. ⚹◊What method can you use to analyze the stability of the critical point $(0,0)$ for the given system of differential equations?⍭ Generate me a similar question◊

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