Differential Equations Quiz: Stability In Systems

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Stability In SystemsQuestion 1 of 19

For the system $\mathbf{x}' = \begin{pmatrix} 1 & k \\ k & -3 \end{pmatrix} \mathbf{x}$ , the origin is a saddle point for which values of the real parameter $k$ ?

For all real

k

For no real

k

Only for

k=0

For

|k| > \sqrt{3}

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Differential Equations Quiz

Differential Equations Quiz: Stability In Systems

Practice Stability In Systems in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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

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Question 1

For the system $\mathbf{x}' = \begin{pmatrix} 1 & k \\ k & -3 \end{pmatrix} \mathbf{x}$ , the origin is a saddle point for which values of the real parameter $k$ ?

For all real $k$ (correct answer)
For no real $k$
Only for $k=0$
For $|k| > \sqrt{3}$

Explanation: The classification of the equilibrium point at the origin for a linear system

\mathbf{x}' = A\mathbf{x}

can be determined by the trace,

T = \mathrm{tr}(A)

, and the determinant,

D = \det(A)

. For the given matrix,

T = 1 + (-3) = -2

and

D = (1)(-3) - (k)(k) = -3 - k^2

. An equilibrium point is a saddle point if and only if

D < 0

. In this case, we need

-3 - k^2 < 0

. This inequality can be written as

k^2 > -3

. Since

k

is a real parameter,

k^2

is always non-negative (

k^2 \ge 0

). Therefore, the condition

k^2 > -3

is satisfied for all real values of

k

. The trace is irrelevant for identifying a saddle point, as the condition

D<0

is sufficient.

Question 2

Consider the nonlinear system $\dot{x} = -x^3 - y$ and $\dot{y} = x - y^3$ . Using the Liapunov function $V(x,y) = \frac{1}{2}(x^2+y^2)$ , what can be concluded about the stability of the origin $(0,0)$ ?

The origin is unstable.
The origin is stable, but not asymptotically stable.
The origin is asymptotically stable. (correct answer)
Stability cannot be determined with the given function.

Explanation: To use the Liapunov function

V(x,y) = \frac{1}{2}(x^2+y^2)

, we first note that it is positive definite (

V(0,0)=0

and

V(x,y) > 0

otherwise). Next, we compute its time derivative along the trajectories of the system,

\dot{V} = \frac{\partial V}{\partial x}\dot{x} + \frac{\partial V}{\partial y}\dot{y}

. We have

\frac{\partial V}{\partial x} = x

and

\frac{\partial V}{\partial y} = y

. Substituting the expressions for

\dot{x}

and

\dot{y}

\dot{V} = x(-x^3 - y) + y(x - y^3) = -x^4 - xy + xy - y^4 = -(x^4 + y^4)

. Since

\dot{V}(0,0)=0

and

\dot{V}(x,y) < 0

for all

(x,y) \neq (0,0)

\dot{V}

is negative definite. According to Liapunov's stability theorem, if there exists a positive definite function

V

whose derivative

\dot{V}

is negative definite, then the equilibrium point at the origin is asymptotically stable.

Question 3

The motion of a damped pendulum can be modeled by the equation $\ddot{\theta} + c\dot{\theta} + \sin(\theta) = 0$ with a damping constant $c > 0$ . The equilibrium point $(\theta, \dot{\theta}) = (\pi, 0)$ corresponds to the pendulum balanced in the vertical, inverted position. What is the nature of this equilibrium point?

Asymptotically stable spiral
Saddle point (unstable) (correct answer)
Stable center
Unstable node

Explanation: First, we convert the second-order equation into a first-order system. Let

x = \theta

and

y = \dot{\theta}

. The system is

\dot{x} = y

and

\dot{y} = -\sin(x) - cy

. We analyze the stability at the equilibrium point

(\pi, 0)

by linearizing the system. The Jacobian matrix is

J(x,y) = \begin{pmatrix} 0 & 1 \\ -\cos(x) & -c \end{pmatrix}

. Evaluating at

(\pi, 0)

, we get

J(\pi, 0) = \begin{pmatrix} 0 & 1 \\ -\cos(\pi) & -c \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & -c \end{pmatrix}

. The determinant of this matrix is

\det(J) = (0)(-c) - (1)(1) = -1

. Since the determinant is negative, the equilibrium point is a saddle point, which is always unstable. This matches the physical intuition that it is impossible to balance a damped pendulum in the inverted position; any small perturbation will cause it to fall.

Question 4

The linearization of the system $\dot{x} = y + x(x^2+y^2)$ , $\dot{y} = -x + y(x^2+y^2)$ at the origin $(0,0)$ is $\dot{x} = y$ , $\dot{y} = -x$ , which corresponds to a center. What is the actual stability of the origin for the full nonlinear system?

The origin is a stable center.
The origin is unstable. (correct answer)
The origin is asymptotically stable.
The stability cannot be determined from the given information.

Explanation: When the linearization of a nonlinear system at an equilibrium point results in a center (eigenvalues are purely imaginary), the linearization is inconclusive and the stability is determined by the nonlinear terms. To analyze the full system, we can convert to polar coordinates. Let

x = r\cos\theta

and

y = r\sin\theta

. The time derivative of

r^2 = x^2+y^2

2r\dot{r} = 2x\dot{x} + 2y\dot{y}

. Substituting for

\dot{x}

and

\dot{y}

r\dot{r} = x(y + x(x^2+y^2)) + y(-x + y(x^2+y^2)) = xy + x^2(r^2) - xy + y^2(r^2) = (x^2+y^2)r^2 = r^4

. For

r \neq 0

, we have

\dot{r} = r^3

. Since

\dot{r} > 0

for all

r>0

, the radius

r(t)

is strictly increasing. Trajectories move away from the origin, so the origin is an unstable equilibrium point.

Question 5

Consider the 3D linear system $\mathbf{x}' = A\mathbf{x}$ where $A = \begin{pmatrix} -2 & 0 & 0 \\ 1 & -1 & 1 \\ 0 & -1 & -1 \end{pmatrix}$ . Which statement best describes the long-term behavior of solutions to this system?

All non-zero solutions approach the origin as $t \to \infty$ . (correct answer)
Solutions oscillate in a stable manner but do not approach the origin.
Some solutions approach the origin, while others grow without bound.
All non-zero solutions grow without bound as $t \to \infty$ .

Explanation: The long-term behavior of a linear system is determined by the eigenvalues of the matrix

A

. Since

A

is block lower triangular, its eigenvalues are the eigenvalues of its diagonal blocks. The first block is the

1 \times 1

matrix

(-2)

, so one eigenvalue is

\lambda_1 = -2

. The second block is the

2 \times 2

matrix

M = \begin{pmatrix} -1 & 1 \\ -1 & -1 \end{pmatrix}

. The characteristic equation for

M

(-1-\lambda)^2 - (1)(-1) = 0

, which is

(\lambda+1)^2 + 1 = 0

. This gives

\lambda+1 = \pm i

, so the other two eigenvalues are

\lambda_{2,3} = -1 \pm i

. The eigenvalues of

A

are therefore

\{-2, -1+i, -1-i\}

. For a system to be asymptotically stable, all eigenvalues must have negative real parts. The real parts are

-2

-1

, and

-1

. Since all are negative, the origin is an asymptotically stable equilibrium point. This means that all solutions, regardless of their initial condition, will approach the origin as

t \to \infty

Question 6

For the system $\mathbf{x}' = \begin{pmatrix} -3 & 1 \\ 1 & -3 \end{pmatrix} \mathbf{x}$ , trajectories approach the origin as $t \to \infty$ . As they approach the origin, they become tangent to which line?

$y = -x$
$y = x$ (correct answer)
$y = -3x$
$y = x/3$

Explanation: The asymptotic behavior of trajectories near a stable node is determined by the eigenvectors. First, we find the eigenvalues. The characteristic equation is

(-3-\lambda)^2 - 1 = 0

, which is

\lambda^2 + 6\lambda + 9 - 1 = 0

, or

\lambda^2 + 6\lambda + 8 = 0

. Factoring gives

(\lambda+4)(\lambda+2) = 0

, so the eigenvalues are

\lambda_1 = -2

and

\lambda_2 = -4

. The general solution is

\mathbf{x}(t) = c_1 e^{-2t} \mathbf{v}_1 + c_2 e^{-4t} \mathbf{v}_2

. As

t \to \infty

, the term

e^{-4t}

decays to zero much faster than

e^{-2t}

. Thus, for large

t

, the solution is dominated by the term with the less negative eigenvalue,

\lambda_1 = -2

. The trajectory will approach the origin tangent to the eigenvector

\mathbf{v}_1

. We find

\mathbf{v}_1

by solving

(A - (-2)I)\mathbf{v}_1 = \mathbf{0}

\begin{pmatrix} -3 - (-2) & 1 \\ 1 & -3 - (-2) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

This gives the equation

-x+y=0

, or

y=x

. So, trajectories approach the origin tangent to the line

y=x

Question 7

A model for two competing species is given by the nonlinear system:

\begin{cases} \dot{x} = x(1 - x - y) \\ \dot{y} = y(0.75 - y - 0.5x) \end{cases}

This system has a non-zero equilibrium point representing the coexistence of both species. What is the nature of the stability at this coexistence equilibrium point?

Asymptotically stable node (correct answer)
Asymptotically stable spiral
Saddle point (unstable)
Unstable node

Explanation: First, find the coexistence equilibrium point where

x > 0

and

y > 0

. This requires solving

\dot{x}=0

and

\dot{y}=0

1 - x - y = 0

and

0.75 - y - 0.5x = 0

. From the first equation,

y = 1 - x

. Substituting into the second gives

0.75 - (1 - x) - 0.5x = 0

, which simplifies to

-0.25 + 0.5x = 0

, so

x=0.5

. Then

y = 1 - 0.5 = 0.5

. The equilibrium point is

(0.5, 0.5)

. Next, linearize the system by finding the Jacobian matrix:

J(x,y) = \begin{pmatrix} 1-2x-y & -x \\ -0.5y & 0.75-2y-0.5x \end{pmatrix}

. Evaluate the Jacobian at

(0.5, 0.5)

J(0.5, 0.5) = \begin{pmatrix} 1-1-0.5 & -0.5 \\ -0.25 & 0.75-1-0.25 \end{pmatrix} = \begin{pmatrix} -0.5 & -0.5 \\ -0.25 & -0.5 \end{pmatrix}

. The trace is

T = -0.5 - 0.5 = -1

. The determinant is

D = (-0.5)(-0.5) - (-0.5)(-0.25) = 0.25 - 0.125 = 0.125

. Since

T < 0

and

D > 0

, the equilibrium is asymptotically stable. To determine if it is a node or spiral, we check the discriminant

T^2 - 4D = (-1)^2 - 4(0.125) = 1 - 0.5 = 0.5 > 0

. Since the discriminant is positive, the eigenvalues are real and distinct, meaning the point is an asymptotically stable node.

Question 8

A system is described in polar coordinates by the equations $\dot{r} = r(4 - r^2)$ and $\dot{\theta} = -1$ . What is the long-term behavior of a trajectory that starts at the Cartesian point $(x_0, y_0) = (3, 0)$ ?

The trajectory approaches the origin.
The trajectory spirals outwards towards infinity.
The trajectory is a circle of radius 3.
The trajectory approaches a limit cycle of radius 2. (correct answer)

Explanation: The initial point

(x_0, y_0) = (3, 0)

corresponds to a radius of

r_0 = \sqrt{3^2 + 0^2} = 3

. The equation for the radius is

\dot{r} = r(4 - r^2) = r(2-r)(2+r)

. For

r>0

, the radial velocity

\dot{r}

is zero when

r=2

. If

0 < r < 2

\dot{r} > 0

, so

r

increases. If

r > 2

\dot{r} < 0

, so

r

decreases. This means that

r=2

is a stable limit cycle. Since the trajectory starts at

r_0=3

, which is greater than 2, the radius

r(t)

will decrease and approach 2 as

t \to \infty

. The equation

\dot{\theta}=-1

indicates that the trajectory spirals in a clockwise direction. Therefore, the trajectory starting at

r=3

approaches the stable limit cycle at

r=2

Question 9

If the eigenvalues of the real matrix $A$ for the linear system $\mathbf{x}' = A\mathbf{x}$ are $\lambda_1 = 3i$ and $\lambda_2 = -3i$ , which of the following best describes the long-term behavior of non-zero solutions?

Trajectories are ellipses or circles, representing periodic solutions. (correct answer)
Trajectories spiral into the origin, representing damped oscillations.
Trajectories spiral away from the origin, representing growing oscillations.
Trajectories are hyperbolas approaching specific lines through the origin.

Explanation: The eigenvalues are a purely imaginary conjugate pair,

\lambda = \pm bi

with

b=3

. For a linear system, this corresponds to a stable center at the origin. The solutions are linear combinations of

\cos(3t)

and

\sin(3t)

, which are periodic. The trajectories are closed orbits around the origin, specifically ellipses (or circles in special cases). These solutions are stable because they remain in a bounded neighborhood of the origin, but they are not asymptotically stable because they do not approach the origin as

t \to \infty

Question 10

Consider the linear system $\mathbf{x}' = \begin{pmatrix} \alpha & -1 \\ 5 & -1 \end{pmatrix} \mathbf{x}$ . For which range of the parameter $\alpha$ is the origin an asymptotically stable spiral point?

$\alpha < 1$
$-1 - 2\sqrt{5} < \alpha < -1 + 2\sqrt{5}$
$-1 - 2\sqrt{5} < \alpha < 1$ (correct answer)
$\alpha < -1 - 2\sqrt{5}$

Explanation: The stability is determined by the eigenvalues of the matrix

A = \begin{pmatrix} \alpha & -1 \\ 5 & -1 \end{pmatrix}

. The characteristic equation is

\lambda^2 - \mathrm{tr}(A)\lambda + \det(A) = 0

, which is

\lambda^2 - (\alpha-1)\lambda + (5-\alpha) = 0

. For a spiral point, the eigenvalues must be complex, which means the discriminant of the characteristic equation must be negative:

(\alpha-1)^2 - 4(5-\alpha) < 0

. This simplifies to

\alpha^2 - 2\alpha + 1 - 20 + 4\alpha < 0

, or

\alpha^2 + 2\alpha - 19 < 0

. The roots of

\alpha^2 + 2\alpha - 19 = 0

are

\alpha = -1 \pm \sqrt{1 - (-19)}} = -1 \pm \sqrt{20} = -1 \pm 2\sqrt{5}

. Thus, we need

-1 - 2\sqrt{5} < \alpha < -1 + 2\sqrt{5}

. For the spiral to be asymptotically stable, the real part of the eigenvalues must be negative. The real part is

\frac{\mathrm{tr}(A)}{2} = \frac{\alpha-1}{2}

. We need

\frac{\alpha-1}{2} < 0

, which implies

\alpha < 1

. Combining both conditions, we need

-1 - 2\sqrt{5} < \alpha < 1

Question 11

Consider the system $\frac{dx}{dt} = y$ , $\frac{dy}{dt} = -\sin(x) - \delta y$ where $\delta > 0$ . This represents a damped pendulum. For small oscillations about the stable equilibrium, what determines the transition between overdamped and underdamped motion?

The transition occurs when $\delta = 1$ , separating exponential decay from oscillatory decay
Overdamped motion occurs for all $\delta > 0$ because the system is inherently dissipative
The transition depends on the initial amplitude and occurs when $\delta = \sqrt{2}$
The transition occurs when $\delta = 2$ , where the discriminant of the characteristic equation changes sign (correct answer)

Explanation: When analyzing damped oscillatory systems like the pendulum, you need to linearize around equilibrium and examine the characteristic equation to determine the nature of the motion. For small oscillations about

x = 0

, you can approximate

\sin(x) \approx x

, giving the linearized system:

\frac{dx}{dt} = y

and

\frac{dy}{dt} = -x - \delta y

. Converting to a second-order equation:

\frac{d^2x}{dt^2} + \delta \frac{dx}{dt} + x = 0

. The characteristic equation is

r^2 + \delta r + 1 = 0

with discriminant

\Delta = \delta^2 - 4

. The nature of the roots determines the motion type:

If $\Delta > 0$ ( $\delta > 2$ ): two real roots → overdamped motion (exponential decay)
If $\Delta = 0$ ( $\delta = 2$ ): repeated real root → critically damped
If $\Delta < 0$ ( $\delta < 2$ ): complex roots → underdamped motion (oscillatory decay)

The transition occurs precisely when

\delta = 2

, where the discriminant equals zero. Answer A incorrectly suggests

\delta = 1

as the transition point, confusing this with other damping criteria. Answer B is wrong because damped systems can still be underdamped—dissipation doesn't eliminate oscillations entirely, just causes them to decay. Answer C incorrectly involves initial amplitude (irrelevant for the overdamped/underdamped classification) and gives

\delta = \sqrt{2}

. Study tip: For any damped harmonic oscillator

\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0

, the transition occurs when the damping coefficient equals the natural frequency. Always check the discriminant of the characteristic equation to classify the motion.

Question 12

For the Van der Pol oscillator $\frac{d^2x}{dt^2} - \mu(1-x^2)\frac{dx}{dt} + x = 0$ with $\mu > 0$ , converted to the system $\frac{dx}{dt} = y$ , $\frac{dy}{dt} = \mu(1-x^2)y - x$ , what happens to the stability of the origin as $\mu$ increases from 0?

The origin remains stable for all $\mu > 0$ but becomes a weak focus
The origin undergoes a transcritical bifurcation at $\mu = 1$ where it changes stability
The origin is unstable only for $\mu > 1$ , with stability determined by the nonlinear terms
The origin becomes unstable for any $\mu > 0$ and a stable limit cycle emerges (correct answer)

Explanation: When analyzing the Van der Pol oscillator, you're dealing with a classic example of how nonlinear systems can exhibit bifurcations and limit cycle behavior. The key is to examine how the linear stability of the equilibrium point changes with the parameter

\mu

. To determine stability of the origin, you linearize the system around

(0,0)

. The Jacobian matrix is:

The eigenvalues satisfy

\lambda^2 - \mu\lambda + 1 = 0

, giving

\lambda = \frac{\mu \pm \sqrt{\mu^2 - 4}}{2}

. For any

\mu > 0

, the real parts are positive (since the sum of eigenvalues is

\mu > 0

and the product is 1), making the origin unstable. The Van der Pol oscillator is famous because when

\mu > 0

, trajectories spiral outward from the unstable origin but are eventually pulled back by the nonlinear damping term

\mu(1-x^2)y

, creating a stable limit cycle. Answer A is wrong because the origin is unstable, not stable, for

\mu > 0

. Answer B incorrectly identifies a transcritical bifurcation at

\mu = 1

- the stability actually changes at

\mu = 0

, and this is a Hopf bifurcation. Answer C is incorrect because the origin is unstable for any

\mu > 0

, not just

\mu > 1

. Remember: For the Van der Pol oscillator, any positive

\mu

creates an unstable spiral at the origin with an attracting limit cycle - this is the hallmark of a supercritical Hopf bifurcation at

\mu = 0

Question 13

For the nonlinear system $\frac{dx}{dt} = x(2-x-y)$ , $\frac{dy}{dt} = y(1-x-2y)$ , the equilibrium point $(1, \frac{1}{2})$ has been identified. To determine its stability, what is the trace of the Jacobian matrix evaluated at this equilibrium?

$-\frac{3}{2}$ (correct answer)
$-\frac{1}{2}$
$\frac{1}{2}$
$\frac{3}{2}$

Explanation: The Jacobian matrix J has entries: ∂f₁/∂x = 2-2x-y, ∂f₁/∂y = -x, ∂f₂/∂x = -y, ∂f₂/∂y = 1-x-4y. At (1, 1/2): ∂f₁/∂x = 2-2(1)-1/2 = -1/2, ∂f₁/∂y = -1, ∂f₂/∂x = -1/2, ∂f₂/∂y = 1-1-4(1/2) = -2. So J = [[-1/2, -1], [-1/2, -2]]. The trace is -1/2 + (-2) = -3/2. Choice B gives only the first diagonal entry. Choice C gives the negative of the first diagonal entry. Choice D gives the negative of the correct answer.

Question 14

For the system $\frac{dx}{dt} = \mu x - y - x^3$ , $\frac{dy}{dt} = x + \mu y - y^3$ where $\mu$ is a parameter, at what value of $\mu$ does the origin undergo a bifurcation, and what type occurs?

$\mu = 0$ ; transcritical bifurcation due to the cubic terms
$\mu = 0$ ; Hopf bifurcation as complex eigenvalues cross the imaginary axis (correct answer)
$\mu = 1$ ; pitchfork bifurcation due to the symmetric nonlinear terms
$\mu = -1$ ; saddle-node bifurcation as eigenvalues become equal

Explanation: The Jacobian at the origin is J = [[μ, -1], [1, μ]], with eigenvalues λ = μ ± i. At μ = 0, the eigenvalues are ±i (purely imaginary), and as μ increases through 0, they cross the imaginary axis with nonzero speed (Re(λ) = μ). This is the condition for a Hopf bifurcation. The cubic terms provide the nonlinearity needed to determine supercritical vs subcritical nature. Choice A is wrong because transcritical bifurcations involve real eigenvalues passing through zero. Choice C is wrong about both the parameter value and bifurcation type. Choice D is wrong because saddle-node bifurcations involve repeated real eigenvalues.

Question 15

A competitive species model has the form $\frac{dx}{dt} = x(1-x-ay)$ , $\frac{dy}{dt} = y(1-bx-y)$ where $a, b > 0$ . For the coexistence equilibrium $\left(\frac{1-a}{1-ab}, \frac{1-b}{1-ab}\right)$ to be stable when $ab > 1$ , which condition must hold?

The coexistence equilibrium cannot be stable when $ab > 1$ due to competitive exclusion (correct answer)
The condition $ab > 1$ is sufficient to guarantee stability of the coexistence equilibrium
Stability requires $ab < 1$ in addition to the equilibrium existing in the positive quadrant
The system exhibits bistability when $ab > 1$ with both boundary and interior equilibria stable

Explanation: When analyzing competitive species models, you're examining how two populations interact when competing for the same resources. The key insight is understanding when species can coexist versus when one will drive the other to extinction. To determine stability, you need to examine the coexistence equilibrium's existence and stability conditions. For the equilibrium

\left(\frac{1-a}{1-ab}, \frac{1-b}{1-ab}\right)

to exist in the positive quadrant (both species present), you need both coordinates positive. This requires

1-a > 0

and

1-b > 0

(so

a < 1

and

b < 1

), plus

1-ab > 0

(so

ab < 1

). However, the question states

ab > 1

. When

ab > 1

, the denominator

1-ab

becomes negative, making the equilibrium coordinates negative—meaning it doesn't exist in the biologically meaningful positive quadrant. Without a valid coexistence equilibrium, stability is impossible. Choice A correctly identifies that coexistence cannot be stable when

ab > 1

due to competitive exclusion—one species will always eliminate the other. Choice B is wrong because

ab > 1

actually prevents stability rather than guaranteeing it. Choice C incorrectly suggests the equilibrium could exist when

ab > 1

if additional conditions were met. Choice D mischaracterizes the dynamics, as there's no stable interior equilibrium when

ab > 1

. Study tip: In competitive models, always check equilibrium existence before analyzing stability. When competition coefficients satisfy

ab > 1

, expect competitive exclusion where only one species survives.

Question 16

A predator-prey system has the form $\frac{dx}{dt} = x(a-by)$ , $\frac{dy}{dt} = y(-c+dx)$ where $a, b, c, d > 0$ . The equilibrium point $\left(\frac{c}{d}, \frac{a}{b}\right)$ is analyzed for stability. Which statement correctly describes the nature of this equilibrium?

It is asymptotically stable because the determinant of the Jacobian is positive and the trace is negative
It is unstable because the determinant of the Jacobian is negative, indicating a saddle point
It is neutrally stable with periodic orbits because the eigenvalues are purely imaginary (correct answer)
It is asymptotically stable because both eigenvalues have negative real parts due to the predator-prey interaction

Explanation: The Jacobian at the equilibrium is J = [[0, -bc/d], [ad/b, 0]]. The characteristic equation is λ² + ac = 0, giving eigenvalues λ = ±i√(ac). Since a,c > 0, these are purely imaginary, indicating a center with periodic orbits (neutrally stable). Choice A is wrong because the trace is 0, not negative. Choice B is wrong because det(J) = ac > 0, not negative. Choice D is wrong because the eigenvalues don't have negative real parts; they have zero real parts.

Question 17

Consider the linear system $\frac{dx}{dt} = 3x + 2y$ , $\frac{dy}{dt} = -x + 4y$ . After determining the eigenvalues of the coefficient matrix, what can be concluded about the stability of the equilibrium point at the origin?

The equilibrium is asymptotically stable because both eigenvalues have negative real parts
The equilibrium is unstable because both eigenvalues have positive real parts (correct answer)
The equilibrium is a saddle point because the eigenvalues have opposite signs
The equilibrium is neutrally stable because the eigenvalues are purely imaginary

Explanation: The coefficient matrix is A = [[3, 2], [-1, 4]]. The characteristic equation is det(A - λI) = (3-λ)(4-λ) - (2)(-1) = λ² - 7λ + 14 = 0. Using the quadratic formula: λ = (7 ± √(49-56))/2 = (7 ± √(-7))/2 = 7/2 ± i√7/2. Both eigenvalues have positive real part 7/2 > 0, making the equilibrium unstable. Choice A is wrong because the real parts are positive, not negative. Choice C is wrong because both eigenvalues have the same sign (positive real part). Choice D is wrong because the eigenvalues have nonzero real parts.

Question 18

A biochemical reaction network has the linearized system $$\frac{d}{dt}\begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -2 & 1 & 0 \ 0 & -1 & 2 \ 1 & 0 & -3 \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix}

The equilibrium is stable because the trace of the matrix is negative
The equilibrium is unstable because at least one eigenvalue has positive real part
The equilibrium is stable because all eigenvalues have negative real parts (correct answer)
Stability cannot be determined without computing the eigenvalues explicitly

Explanation: The matrix A = [[-2,1,0],[0,-1,2],[1,0,-3]] has characteristic polynomial det(A-λI). We can use the Routh-Hurwitz criteria: the trace is -6 < 0, and we need to check that all principal minors alternate in sign appropriately. Computing: the characteristic polynomial is -λ³ - 6λ² - 9λ - 4. By Routh-Hurwitz, since all coefficients are negative (when written as λ³ + 6λ² + 9λ + 4), we need: 6 > 0 ✓, 6·9 - 1·4 = 50 > 0 ✓, and 4 > 0 ✓. All conditions are satisfied, so all eigenvalues have negative real parts. Choice A is incorrect because negative trace is necessary but not sufficient for stability in 3D. Choice B is wrong because all eigenvalues actually have negative real parts. Choice D is wrong because the Routh-Hurwitz criteria allow us to determine stability without explicit eigenvalue computation.

Question 19

Which of the following statements about the stability of equilibrium points for 2D autonomous systems is true?

If the linearization of a nonlinear system at an equilibrium point has purely imaginary eigenvalues, the equilibrium is a stable center.
If all trajectories of a system approach a point $\mathbf{x}_0$ as $t \to \infty$ , then $\mathbf{x}_0$ must be an asymptotically stable node.
For a linear system $\mathbf{x}' = A\mathbf{x}$ , if $\det(A) < 0$ , then some solutions will move away from the origin as $t \to \infty$ . (correct answer)
Any equilibrium point that is stable must also be asymptotically stable.

Explanation: Let's analyze each statement: A. False. If the linearization has purely imaginary eigenvalues, it is a borderline case. The stability of the nonlinear system's equilibrium point depends on the nonlinear terms and cannot be determined from the linearization alone; it could be a stable center, an unstable spiral, or a stable spiral. B. False. If all trajectories approach

\mathbf{x}_0

, it is an asymptotically stable equilibrium. However, it could be either a stable node (real negative eigenvalues) or a stable spiral (complex eigenvalues with negative real part). C. True. For a 2D linear system,

\det(A) = \lambda_1 \lambda_2

. If

\det(A) < 0

, the eigenvalues must be real and have opposite signs (e.g., one positive, one negative). This corresponds to a saddle point, which is an unstable equilibrium. Trajectories along the eigenvector corresponding to the positive eigenvalue will move away from the origin. D. False. A stable equilibrium point is not necessarily asymptotically stable. A center, for example, is stable (trajectories stay nearby) but not asymptotically stable (they do not approach the equilibrium point).