Differential Equations Quiz: Classifying Critical Points

What this quiz covers

This quiz focuses on Classifying Critical Points, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $k > 2\sqrt{2}$ or $k < -2\sqrt{2}$ (correct answer)
2. $-2\sqrt{2} < k < 2\sqrt{2}$
3. Only for $k=0$
4. For all real values of $k$ except $k = \pm 2\sqrt{2}$

Explanation: Let the matrix be $A$. The classification of the critical point depends on the trace $\tau = \mathrm{tr}(A)$ and the determinant $\Delta = \det(A)$. For this system, $\tau = k + (-k) = 0$ and $\Delta = (k)(-k) - (2)(-4) = 8 - k^2$. A critical point is a saddle point if and only if $\Delta < 0$. Therefore, we must have $8 - k^2 < 0$, which simplifies to $k^2 > 8$. Taking the square root of both sides gives $|k| > \sqrt{8} = 2\sqrt{2}$. This corresponds to the range $k > 2\sqrt{2}$ or $k < -2\sqrt{2}$.

Question 2

1. $k > 9$
2. $k < 0$
3. $0 < k < 9$
4. $0 < k \le 9$ (correct answer)

Explanation: Let $y=x'$. The second-order equation can be written as the first-order system $\vec{z}' = \begin{pmatrix} 0 & 1 \\ -k & -6 \end{pmatrix} \vec{z}$, where $\vec{z} = \begin{pmatrix} x \\ y \end{pmatrix}$. The trace is $\tau = -6$ and the determinant is $\Delta = k$. For a stable node, we need real, negative eigenvalues. The characteristic equation is $\lambda^2 + 6\lambda + k = 0$. The eigenvalues are real if the discriminant $D = (6)^2 - 4k = 36 - 4k \ge 0$, which implies $k \le 9$. The eigenvalues have the same sign if $\Delta = k > 0$. The eigenvalues are negative if $\tau = -6 < 0$, which is always true. Combining the conditions, we need $0 < k \le 9$. The case $k=9$ corresponds to a repeated eigenvalue (a degenerate or improper stable node), which is still a type of stable node.

Question 3

1. Saddle point (correct answer)
2. Stable node
3. Unstable node
4. Stable spiral

Explanation: The trace of a matrix is the sum of its eigenvalues. Let the eigenvalues be $\lambda_1$ and $\lambda_2$. We are given $\tau = \lambda_1 + \lambda_2 = -5$. We are also given that one eigenvalue is $1$, so let $\lambda_1 = 1$. Substituting this into the trace equation gives $1 + \lambda_2 = -5$, which means $\lambda_2 = -6$. The eigenvalues are $1$ and $-6$. Since the eigenvalues are real, distinct, and have opposite signs, the critical point at the origin is a saddle point.

Question 4

1. Stable node
2. Saddle point (correct answer)
3. Unstable spiral
4. Stable spiral

Explanation: First, find the critical point by setting $x'=0$ and $y'=0$. For $x>0, y>0$, we must solve the system $2-x-y=0$ and $3-2x-y=0$. Subtracting the first equation from the second gives $(3-2x-y) - (2-x-y) = 0$, which simplifies to $1-x=0$, so $x=1$. Substituting $x=1$ into the first equation gives $2-1-y=0$, so $y=1$. The critical point is $(1,1)$. Next, find the Jacobian matrix: $J(x,y) = \begin{pmatrix} 2-2x-y & -x \\ -2y & 3-2x-2y \end{pmatrix}$. Evaluate the Jacobian at $(1,1)$: $J(1,1) = \begin{pmatrix} 2-2-1 & -1 \\ -2 & 3-2-2 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -2 & -1 \end{pmatrix}$. For this linearized system, the determinant is $\Delta = (-1)(-1) - (-1)(-2) = 1 - 2 = -1$. Since $\Delta < 0$, the critical point is a saddle point.

Question 5

1. $\vec{x}' = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \vec{x}$ (correct answer)
2. $\vec{x}' = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \vec{x}$
3. $\vec{x}' = \begin{pmatrix} 1 & 4 \\ 1 & 1 \end{pmatrix} \vec{x}$
4. $\vec{x}' = \begin{pmatrix} -2 & 0 \\ 0 & 0 \end{pmatrix} \vec{x}$

Explanation: A critical point is stable but not asymptotically stable if it is a center. A center occurs for a linear system when the eigenvalues are purely imaginary, which corresponds to having trace $\tau = 0$ and determinant $\Delta > 0$. Let's check the options:\nA) $\tau = 0+0=0$, $\Delta = (0)(0) - (2)(-2) = 4 > 0$. This is a center. Correct.\nB) $\tau = -1-1=-2$, $\Delta = 1 > 0$. This is a stable node, which is asymptotically stable.\nC) $\tau = 1+1=2$, $\Delta = 1-4=-3 < 0$. This is a saddle point, which is unstable.\nD) $\tau = -2+0=-2$, $\Delta = 0$. The eigenvalues are $-2$ and $0$. The system has a line of critical points ($x=0$). A trajectory starting at $(\epsilon, \delta)$ goes to $(0, \delta)$ as $t \to \infty$. Thus, the origin is unstable.

Question 6

1. $\begin{pmatrix} 4 & 10 \\ 1 & 1 \end{pmatrix}$
2. $\begin{pmatrix} 1 & 5 \\ -1 & 3 \end{pmatrix}$
3. $\begin{pmatrix} 3 & 1 \\ -1 & 1 \end{pmatrix}$ (correct answer)
4. $\begin{pmatrix} -2 & -1 \\ 1 & -4 \end{pmatrix}$

Explanation: When analyzing the stability of a linear system $\vec{x}' = A\vec{x}$, you need to examine the eigenvalues of matrix $A$. An unstable node occurs when both eigenvalues are real and positive, causing solutions to grow exponentially away from the origin along straight-line trajectories. For an unstable node, we need both eigenvalues to be real and positive. Let's check each matrix by finding their eigenvalues using the characteristic equation $\det(A - \lambda I) = 0$. Option C gives us the correct answer. For $\begin{pmatrix} 3 & 1 \\ -1 & 1 \end{pmatrix}$, the characteristic equation is $(3-\lambda)(1-\lambda) - (1)(-1) = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 = 0$. This yields a repeated eigenvalue $\lambda = 2 > 0$, creating an unstable node. Option A has eigenvalues that are real but include negative values (you can verify the trace and determinant suggest mixed signs). Option B produces complex eigenvalues since the discriminant of its characteristic equation is negative - this would create a spiral, not a node. Option D has negative eigenvalues (notice the negative diagonal entries and positive determinant), making it a stable node rather than unstable. Remember this key pattern: for a 2×2 system to have an unstable node, both eigenvalues must be real and positive. This happens when the determinant is positive (ensuring real eigenvalues of the same sign) and the trace is positive (ensuring that common sign is positive). Always check both conditions when classifying equilibrium points.

Question 7

1. Stable node, indicating stable coexistence.
2. Saddle point, indicating unstable coexistence. (correct answer)
3. Stable spiral, indicating oscillating coexistence.
4. Unstable node, indicating unstable coexistence.

Explanation: To find the coexistence equilibrium, we set the growth rates to zero for $x>0, y>0$: $3-x-2y=0$ and $2-x-y=0$. Solving this system yields $x=1, y=1$. To classify this point, we compute the Jacobian matrix $J(x,y) = \begin{pmatrix} 3-2x-2y & -2x \\ -y & 2-x-2y \end{pmatrix}$. At $(1,1)$, the Jacobian is $J(1,1) = \begin{pmatrix} -1 & -2 \\ -1 & -1 \end{pmatrix}$. The determinant of this matrix is $\Delta = (-1)(-1) - (-2)(-1) = 1 - 2 = -1$. Since the determinant is negative, the equilibrium point is a saddle point. In the context of population dynamics, a saddle point is unstable, meaning that small perturbations from this equilibrium will lead to one of the species outcompeting the other.

Question 8

1. Stable spiral, clockwise rotation
2. Center, counter-clockwise rotation
3. Center, clockwise rotation (correct answer)
4. Saddle point, direction not applicable

Explanation: When analyzing linear systems of differential equations, you need to examine the eigenvalues of the coefficient matrix to classify the critical point's behavior and stability. For the matrix $A = \begin{pmatrix} 3 & 5 \\ -2 & -3 \end{pmatrix}$, find the eigenvalues using the characteristic equation $\det(A - \lambda I) = 0$: $\det\begin{pmatrix} 3-\lambda & 5 \\ -2 & -3-\lambda \end{pmatrix} = (3-\lambda)(-3-\lambda) - (5)(-2) = \lambda^2 + 1 = 0$ This gives eigenvalues $\lambda = \pm i$, which are purely imaginary with zero real part. Since the eigenvalues are purely imaginary, the critical point is a center with closed, elliptical trajectories around the origin. The system is neutrally stable—trajectories neither spiral inward nor outward. To determine rotation direction, examine how the vector field moves a test point. At $(1,0)$, the system gives $\vec{x}' = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$. This vector points right and down, indicating clockwise rotation. Looking at the wrong answers: (A) incorrectly identifies this as a stable spiral—that would require eigenvalues with negative real parts, not purely imaginary ones. (B) correctly identifies the center but gets the rotation direction wrong. (D) confuses this with a saddle point, which requires one positive and one negative real eigenvalue. Study tip: Remember that purely imaginary eigenvalues always create centers with periodic orbits. Check rotation direction by evaluating the vector field at a simple test point like $(1,0)$ or $(0,1)$.

Question 9

1. The origin is a center for all non-zero $\alpha$ and a spiral for $\alpha=0$.
2. The origin is a stable spiral if $\alpha < 0$, a center if $\alpha = 0$, and an unstable spiral if $\alpha > 0$. (correct answer)
3. The origin is a saddle point for all $\alpha$ such that $|\alpha| > 4$.
4. The origin is a stable node if $\alpha < 0$ and an unstable node if $\alpha > 0$.

Explanation: Let $A$ be the matrix of the system. The trace is $\tau = \alpha + \alpha = 2\alpha$, and the determinant is $\Delta = (\alpha)(\alpha) - (-2)(8) = \alpha^2 + 16$. The discriminant of the characteristic equation is $D = \tau^2 - 4\Delta = (2\alpha)^2 - 4(\alpha^2 + 16) = 4\alpha^2 - 4\alpha^2 - 64 = -64$. Since $D < 0$ for all real $\alpha$, the eigenvalues are always complex conjugates: $\lambda = \frac{\tau \pm i\sqrt{|D|}}{2} = \frac{2\alpha \pm i\sqrt{64}}{2} = \alpha \pm 4i$. The classification depends on the real part of the eigenvalues, which is $\alpha$. If $\alpha < 0$, it is a stable spiral. If $\alpha > 0$, it is an unstable spiral. If $\alpha = 0$, it is a center.

Question 10

1. $c = 2.5$
2. $c = 2$
3. $c = -2 + 2\sqrt{5}$
4. $c = -2 - 2\sqrt{5}$ (correct answer)

Explanation: The transition between a spiral and a node occurs when the eigenvalues switch from complex to real, which is when the discriminant of the characteristic equation is zero. The trace is $\tau = c-2$ and the determinant is $\Delta = -2c+5$. The discriminant is $D = \tau^2 - 4\Delta = (c-2)^2 - 4(5-2c) = c^2 - 4c + 4 - 20 + 8c = c^2 + 4c - 16$. Setting $D=0$ and solving the quadratic equation $c^2 + 4c - 16 = 0$ gives $c = \frac{-4 \pm \sqrt{16 - 4(1)(-16)}}{2} = \frac{-4 \pm \sqrt{80}}{2} = -2 \pm 2\sqrt{5}$. For the critical point to be stable, we need $\tau = c-2 < 0$ (so $c<2$) and $\Delta = 5-2c > 0$ (so $c<2.5$). The root $c = -2 + 2\sqrt{5} \approx -2 + 2(2.236) = 2.472$ violates the condition $c<2$. The root $c = -2 - 2\sqrt{5} \approx -2 - 4.472 = -6.472$ satisfies both stability conditions. Thus, the transition occurs at $c = -2 - 2\sqrt{5}$.

Question 11

1. The sign of the trace, since $\lambda_1 = 1$ and $\lambda_2 = -1$ already determine this is a saddle point (correct answer)
2. The geometric multiplicity of the eigenvalues to determine if there are sufficient linearly independent eigenvectors
3. The determinant of the coefficient matrix to confirm whether the eigenvalues have the same or opposite signs
4. No additional information is needed since the eigenvalues $\lambda = \pm 1$ uniquely determine an unstable node

Explanation: From λ² - 1 = 0, we get λ₁ = 1 and λ₂ = -1. These are real eigenvalues with opposite signs (one positive, one negative), which uniquely determines that the critical point is a saddle point. No additional information is needed for classification - the eigenvalues alone are sufficient. The trace confirmation (τ = 1 + (-1) = 0) and determinant (Δ = 1×(-1) = -1 < 0) are consistent with a saddle. Choice B is incorrect because distinct eigenvalues automatically have geometric multiplicity 1. Choice C is unnecessary since we already know the signs differ. Choice D is wrong because real eigenvalues of opposite signs indicate a saddle, not a node.

Question 12

1. It's a stable spiral with complex eigenvalues having negative real parts, indicating the populations will converge to equilibrium values
2. It's a center with eigenvalues $\pm i\sqrt{ab}$, suggesting neutrally stable oscillations, but higher-order terms determine actual stability (correct answer)
3. It's a saddle point with real eigenvalues $\pm\sqrt{ab}$, suggesting one population grows while the other declines
4. It's an unstable node with positive real eigenvalues, indicating both populations will grow without bound from the equilibrium

Explanation: When analyzing linearized predator-prey systems, you need to find the eigenvalues of the Jacobian matrix to classify the equilibrium point and understand the system's local behavior. To find the eigenvalues of $J = \begin{pmatrix} 0 & -a \\ b & 0 \end{pmatrix}$, solve the characteristic equation $\det(J - \lambda I) = 0$. This gives us $\lambda^2 + ab = 0$, so $\lambda = \pm i\sqrt{ab}$. These are purely imaginary eigenvalues, which classify the equilibrium as a center. In the linearized system, this indicates neutrally stable oscillations around the equilibrium. However, since linearization only captures first-order behavior, the actual stability of the nonlinear system depends on higher-order terms that were neglected during linearization. Looking at the wrong answers: Choice A incorrectly suggests complex eigenvalues with negative real parts, but our eigenvalues have zero real parts. Choice C mistakes the eigenvalues as real ($\pm\sqrt{ab}$), but they're purely imaginary ($\pm i\sqrt{ab}$). Choice D claims positive real eigenvalues, which would occur if we had $\lambda^2 - ab = 0$, but our characteristic equation has $\lambda^2 + ab = 0$. Study tip: For predator-prey Jacobians with this zero-diagonal, opposite-sign off-diagonal structure, expect purely imaginary eigenvalues and a center classification. Remember that linearization around centers doesn't determine true stability—you'd need to analyze the full nonlinear system or use methods like Lyapunov functions to determine whether oscillations grow, decay, or remain constant.

Question 13

1. The linearization predicts a stable spiral, and this classification is reliable for the nonlinear system (correct answer)
2. The linearization predicts an unstable spiral, but the classification may not hold for the nonlinear system
3. The linearization predicts a center, so the nonlinear behavior requires higher-order analysis to determine stability
4. The linearization predicts a stable node, and this classification transfers directly to the nonlinear system

Explanation: The eigenvalues λ₁ = -1 + 2i and λ₂ = -1 - 2i are complex conjugates with negative real part (-1) and nonzero imaginary parts (±2), indicating a stable spiral in the linearization. Since the real parts are nonzero (hyperbolic case), the linearization theorem guarantees that the nonlinear system has the same qualitative behavior near the critical point. Choice B incorrectly identifies it as unstable. Choice C incorrectly calls it a center (which would require purely imaginary eigenvalues). Choice D incorrectly identifies it as a node rather than a spiral.

Question 14

1. Stable proper node, with trajectories approaching along multiple straight-line directions
2. Stable improper node, with trajectories approaching the origin tangentially to the eigenspace direction (correct answer)
3. Stable spiral node, requiring Jordan form analysis due to the deficient eigenspace
4. Stable degenerate node, with characteristic curved approach to the equilibrium point

Explanation: When analyzing critical points in linear systems, the key is understanding how eigenvalue properties determine the behavior of solution trajectories near equilibrium. The algebraic multiplicity (how many times an eigenvalue appears) versus geometric multiplicity (number of linearly independent eigenvectors) tells you everything about trajectory structure. With repeated eigenvalue $\lambda = -3$ having geometric multiplicity 1, you have a deficient eigenspace—fewer eigenvectors than the algebraic multiplicity would suggest. Since $\lambda < 0$, the critical point is stable (trajectories approach the origin). The deficient eigenspace creates an improper node, where most trajectories approach tangentially to the single eigenvector direction rather than along multiple straight-line paths. Choice A incorrectly describes a proper node, which would require geometric multiplicity equal to algebraic multiplicity, giving multiple linearly independent eigenvectors and thus multiple straight-line approach directions. Choice C misidentifies this as a spiral node—spirals require complex eigenvalues, not repeated real ones. Choice D uses the non-standard term "degenerate node" and incorrectly emphasizes curved approaches, when the key feature is tangential approach to the eigenspace direction. The Jordan canonical form is indeed needed to analyze this system completely, but that mathematical requirement doesn't change the classification—it's still an improper node. Study tip: Remember the classification hierarchy: negative real eigenvalues mean stability, while geometric multiplicity deficiency (fewer eigenvectors than expected) always creates "improper" nodes with tangential approach patterns, never spirals or multiple straight-line trajectories.

Question 15

1. $\tau = 0$ and $\Delta > 0$, yielding purely imaginary eigenvalues that create the observed straight-line approaches
2. $\Delta > 0$, $\tau < 0$, and $\tau^2 - 4\Delta > 0$, creating two negative real eigenvalues with different magnitudes
3. $\Delta > 0$, $\tau > 0$, and $\tau^2 - 4\Delta = 0$, producing a repeated positive eigenvalue with geometric multiplicity 1
4. $\Delta < 0$ and $\tau^2 - 4\Delta > 0$, ensuring real eigenvalues of opposite signs with distinct eigendirections (correct answer)

Explanation: When analyzing phase portraits near critical points, you need to connect the visual behavior to the trace-determinant conditions that determine eigenvalue types. The key insight is recognizing what "trajectories approaching along two distinct straight-line directions while others curve away" tells you about the eigenvalues. This description matches a saddle point, which occurs when you have real eigenvalues of opposite signs. The straight-line approaches happen along the eigendirections—one eigenvalue is negative (creating the stable direction where trajectories approach), while the other is positive (creating the unstable direction where trajectories move away). The curved trajectories result from the competing influences of both eigenvalues. Answer D is correct because $\Delta < 0$ guarantees eigenvalues of opposite signs (since $\Delta = \lambda_1 \lambda_2$), and $\tau^2 - 4\Delta > 0$ ensures they're real and distinct, giving you the two separate eigendirections described. Answer A fails because $\tau = 0, \Delta > 0$ gives purely imaginary eigenvalues, creating a center with circular or elliptical trajectories—no straight-line approaches. Answer B produces a stable node since both eigenvalues are negative ($\Delta > 0, \tau < 0$). While trajectories approach along eigendirections, they don't curve away from the origin. Answer C creates an unstable node with a repeated eigenvalue. All trajectories move away from the origin, contradicting the "approach" behavior described. Study tip: Remember the saddle signature: $\Delta < 0$ always means opposite-signed eigenvalues, which creates the mixed approach/repel behavior characteristic of saddle points.

Question 16

1. Stable node → stable spiral → center, with the transition occurring when the discriminant changes sign
2. Stable spiral → stable node → unstable node, with transitions based on the trace becoming positive
3. The critical point remains a stable spiral for all $k > 0$, since the discriminant $\tau^2 - 4\Delta = -4k^2 < 0$ (correct answer)
4. Center → stable spiral → stable node, with stability determined by the sign of the determinant

Explanation: The coefficient matrix is A = [[-1, k], [-k, -1]], so trace τ = -2 < 0 and determinant Δ = 1 + k² > 0 for all k > 0. The discriminant is τ² - 4Δ = 4 - 4(1 + k²) = -4k² < 0 for all k > 0. Since the discriminant is always negative (complex eigenvalues) and τ < 0 (negative real parts), the critical point is always a stable spiral for any k > 0. At k = 0, we get eigenvalue λ = -1 with multiplicity 2 (stable proper node), but for any k > 0, it's a stable spiral.

Question 17

1. Unstable node because the discriminant is positive and both eigenvalues are positive real numbers
2. Stable node because the discriminant is positive and both eigenvalues are negative real numbers
3. Stable spiral because the discriminant is negative and the trace indicates complex eigenvalues with negative real parts (correct answer)
4. Saddle point because the positive determinant with negative trace creates eigenvalues of opposite signs

Explanation: The characteristic equation is λ² - τλ + Δ = λ² + 2λ + 8 = 0. The discriminant is τ² - 4Δ = 4 - 32 = -28 < 0, indicating complex eigenvalues. Using the quadratic formula: λ = (-2 ± i√28)/2 = -1 ± i√7. Since the real part is -1 < 0, this is a stable spiral. Choice A is wrong because the eigenvalues are complex, not real. Choice B is wrong for the same reason. Choice D is incorrect because saddle points require real eigenvalues of opposite signs, which would need Δ < 0.

Question 18

1. The critical point will become a stable spiral since perturbations typically introduce complex eigenvalues with negative real parts
2. The critical point could become a saddle point if the perturbation is large enough to make one eigenvalue positive
3. The classification will depend on whether the perturbation preserves the reality and negativity of both eigenvalue real parts
4. The critical point will remain a stable node since small perturbations cannot change the fundamental character of hyperbolic critical points (correct answer)

Explanation: When analyzing how perturbations affect linear systems, the key concept is hyperbolic critical points and their structural stability. A critical point is hyperbolic when all eigenvalues have nonzero real parts, and such points are remarkably robust to small changes. Your original system has eigenvalues $\lambda_1 = -2$ and $\lambda_2 = -\frac{1}{2}$, both negative and real, making the origin a stable node and the critical point hyperbolic. The fundamental theorem of structural stability tells us that hyperbolic critical points maintain their essential character under small perturbations - the eigenvalues will shift slightly, but they cannot cross the imaginary axis or change the overall stability picture when $\epsilon$ is sufficiently small. Answer (A) incorrectly assumes perturbations typically create complex eigenvalues. While possible, this isn't guaranteed, and the stability would be preserved regardless. Answer (B) suggests a saddle point could form, but this would require an eigenvalue to become positive, which contradicts the "small perturbation" condition - you'd need the perturbation to overcome the substantial negative real parts. Answer (C) overcomplicates the situation by suggesting the classification depends on preserving specific eigenvalue properties, when structural stability theory already guarantees this for hyperbolic points. Answer (D) correctly identifies that small perturbations cannot fundamentally alter hyperbolic critical points - the stable node character persists. Study tip: Remember that hyperbolic critical points are "structurally stable" - when all eigenvalues have nonzero real parts, small perturbations preserve the qualitative behavior. Focus on whether critical points are hyperbolic when analyzing perturbation effects.