Differential Equations Quiz: Phase Portraits And Trajectories
8 questions · exam conditions
0:00
Phase Portraits And TrajectoriesQuestion 1 of 8

Consider the system dxdt=2x+3y\frac{dx}{dt} = 2x + 3y, dydt=−x+6y\frac{dy}{dt} = -x + 6y. A trajectory starts at (1,0)(1, 0) at t=0t = 0. As t→∞t \to \infty, the trajectory will become approximately parallel to which vector?

(3,−1)(3, -1) corresponding to the eigenvector of the smaller eigenvalue
(1,−1)(1, -1) corresponding to the eigenvector of the smaller eigenvalue
(1,3)(1, 3) corresponding to the eigenvector of the larger eigenvalue
(3,1)(3, 1) corresponding to the eigenvector of the larger eigenvalue
← Back to quizzes

Differential Equations Quiz

Differential Equations Quiz: Phase Portraits And Trajectories

Practice Phase Portraits And Trajectories in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Phase Portraits And Trajectories, 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.

All questions

Question 1

Consider the system dxdt=2x+3y\frac{dx}{dt} = 2x + 3y, dydt=−x+6y\frac{dy}{dt} = -x + 6y. A trajectory starts at (1,0)(1, 0) at t=0t = 0. As t→∞t \to \infty, the trajectory will become approximately parallel to which vector?

  1. (3,−1)(3, -1) corresponding to the eigenvector of the smaller eigenvalue
  2. (1,−1)(1, -1) corresponding to the eigenvector of the smaller eigenvalue
  3. (1,3)(1, 3) corresponding to the eigenvector of the larger eigenvalue
  4. (3,1)(3, 1) corresponding to the eigenvector of the larger eigenvalue (correct answer)
Explanation: When analyzing the long-term behavior of a linear system of differential equations, you need to understand how eigenvalues and eigenvectors determine trajectory directions. As t→∞t \to \infty, trajectories become parallel to the eigenvector corresponding to the eigenvalue with the largest real part, since that component dominates the solution. First, find the eigenvalues of the coefficient matrix A=(23−16)A = \begin{pmatrix} 2 & 3 \\ -1 & 6 \end{pmatrix} . The characteristic equation is det⁡(A−λI)=(2−λ)(6−λ)−3(−1)=λ2−8λ+15=0\det(A - \lambda I) = (2-\lambda)(6-\lambda) - 3(-1) = \lambda^2 - 8\lambda + 15 = 0. This gives eigenvalues λ1=3\lambda_1 = 3 and λ2=5\lambda_2 = 5. Since λ2=5>λ1=3\lambda_2 = 5 > \lambda_1 = 3, the larger eigenvalue is 5. For λ=5\lambda = 5, solve (A−5I)v⃗=0(A - 5I)\vec{v} = 0: (−33−11)v⃗=0 \begin{pmatrix} -3 & 3 \\ -1 & 1 \end{pmatrix}\vec{v} = 0. This yields the eigenvector (3,1)(3, 1). As t→∞t \to \infty, the e5te^{5t} term grows faster than e3te^{3t}, so trajectories align with the eigenvector of the larger eigenvalue. Choice A incorrectly identifies (3,−1)(3, -1) and references the smaller eigenvalue. Choice B gives the wrong eigenvector (1,−1)(1, -1) for the smaller eigenvalue. Choice C correctly identifies the larger eigenvalue but gives the wrong eigenvector (1,3)(1, 3). Only choice D correctly pairs the eigenvector (3,1)(3, 1) with the larger eigenvalue λ=5\lambda = 5. Key strategy: For long-term behavior, always find the eigenvalue with the largest real part—its eigenvector determines the asymptotic direction of trajectories.

Question 2

Consider the Hamiltonian system dxdt=∂H∂y\frac{dx}{dt} = \frac{\partial H}{\partial y}, dydt=−∂H∂x\frac{dy}{dt} = -\frac{\partial H}{\partial x} where H(x,y)=12x2+14y4−12y2H(x,y) = \frac{1}{2}x^2 + \frac{1}{4}y^4 - \frac{1}{2}y^2. What characterizes the trajectories near the equilibrium point (0,0)(0, 0)?

  1. Closed elliptical orbits since the linearization gives a center with purely imaginary eigenvalues
  2. Hyperbolic trajectories since the linearization gives a saddle point with real eigenvalues (correct answer)
  3. Spiral trajectories since the linearization gives complex eigenvalues with nonzero real parts
  4. Parabolic trajectories since one eigenvalue is zero and the system is degenerate
Explanation: The system is: dxdt=∂H∂y=y3−y=y(y2−1)\frac{dx}{dt} = \frac{\partial H}{\partial y} = y^3 - y = y(y^2 - 1), dydt=−∂H∂x=−x\frac{dy}{dt} = -\frac{\partial H}{\partial x} = -x. The Jacobian at (0,0)(0,0) is J=(0−1−10)J = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} with eigenvalues from λ2−1=0\lambda^2 - 1 = 0, giving λ=±1\lambda = \pm 1. With real eigenvalues of opposite signs, this is a saddle point. Note that even though this is a Hamiltonian system (which typically preserves area), the equilibrium at (0,0)(0,0) is still a saddle due to the specific form of HH. Choice A is wrong because the eigenvalues are real, not purely imaginary. Choices C and D are wrong because the eigenvalues are ±1\pm 1, both nonzero and real.

Question 3

Consider the linear system dxdt=3x−2y\frac{dx}{dt} = 3x - 2y, dydt=x+y\frac{dy}{dt} = x + y. If a trajectory passes through the point (2,−1)(2, -1) at t=0t = 0, which statement best describes the long-term behavior of this trajectory?

  1. The trajectory spirals inward toward the origin as t→∞t \to \infty
  2. The trajectory approaches a straight-line solution along the eigenvector direction as t→∞t \to \infty
  3. The trajectory spirals outward away from the origin as t→∞t \to \infty (correct answer)
  4. The trajectory oscillates periodically around a closed orbit as t→∞t \to \infty
Explanation: The coefficient matrix is A=(3−211)A = \begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix} . The characteristic equation is (3−λ)(1−λ)+2=λ2−4λ+5=0(3-\lambda)(1-\lambda) + 2 = \lambda^2 - 4\lambda + 5 = 0, giving eigenvalues λ=2±i\lambda = 2 \pm i. Since the real part is positive (2>02 > 0), the origin is an unstable spiral, so trajectories spiral outward. Choice A is wrong because positive real parts cause outward spiraling. Choice B is wrong because complex eigenvalues produce spirals, not straight-line approaches. Choice D is wrong because the positive real part prevents periodic behavior.

Question 4

For the nonlinear system dxdt=x(2−x−y)\frac{dx}{dt} = x(2-x-y), dydt=y(1−x−2y)\frac{dy}{dt} = y(1-x-2y), the equilibrium point (0,0)(0, 0) has what type of behavior in its phase portrait?

  1. Unstable node with trajectories moving away from the origin along both coordinate axes (correct answer)
  2. Saddle point with trajectories approaching along one axis and departing along the other axis
  3. Stable node with trajectories approaching the origin along both coordinate axes from all directions
  4. Center point with trajectories forming closed elliptical orbits around the origin
Explanation: Linearizing around (0,0)(0,0): ∂f∂x=2−2x−y\frac{\partial f}{\partial x} = 2-2x-y, ∂f∂y=−x\frac{\partial f}{\partial y} = -x, ∂g∂x=−y\frac{\partial g}{\partial x} = -y, ∂g∂y=1−x−4y\frac{\partial g}{\partial y} = 1-x-4y. At (0,0)(0,0), the Jacobian is J=(2001)J = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} with eigenvalues λ1=2,λ2=1\lambda_1 = 2, \lambda_2 = 1. Both positive eigenvalues make this an unstable node. The eigenvectors are along the coordinate axes, so trajectories depart along both axes. Choice B is wrong because both eigenvalues have the same sign. Choice C is wrong because positive eigenvalues indicate instability. Choice D is wrong because real eigenvalues don't produce centers.

Question 5

Consider the system dxdt=−y+x(x2+y2)\frac{dx}{dt} = -y + x(x^2 + y^2), dydt=x+y(x2+y2)\frac{dy}{dt} = x + y(x^2 + y^2). In polar coordinates (r,θ)(r, \theta), what is the correct description of the trajectory behavior?

  1. Trajectories spiral inward toward the origin since drdt=−r3<0\frac{dr}{dt} = -r^3 < 0 for all r>0r > 0
  2. Trajectories spiral outward from the origin since drdt=r3>0\frac{dr}{dt} = r^3 > 0 for all r>0r > 0 (correct answer)
  3. Trajectories form closed circular orbits since drdt=0\frac{dr}{dt} = 0 and dθdt=1\frac{d\theta}{dt} = 1
  4. Trajectories approach the unit circle since drdt=r(1−r2)\frac{dr}{dt} = r(1-r^2) changes sign at r=1r = 1
Explanation: Converting to polar coordinates: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta. We have drdt=xdxdt+ydydtr=x(−y+x(x2+y2))+y(x+y(x2+y2))r=−xy+x2(x2+y2)+xy+y2(x2+y2)r=(x2+y2)2r=r3\frac{dr}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{r} = \frac{x(-y + x(x^2+y^2)) + y(x + y(x^2+y^2))}{r} = \frac{-xy + x^2(x^2+y^2) + xy + y^2(x^2+y^2)}{r} = \frac{(x^2+y^2)^2}{r} = r^3. Since r3>0r^3 > 0 for all r>0r > 0, trajectories spiral outward. Choice A has the wrong sign. Choice C is wrong because drdt≠0\frac{dr}{dt} \neq 0. Choice D uses an incorrect formula for drdt\frac{dr}{dt}.

Question 6

For the competitive species system dxdt=x(3−x−2y)\frac{dx}{dt} = x(3 - x - 2y), dydt=y(2−x−y)\frac{dy}{dt} = y(2 - x - y), what happens to trajectories that start in the region where both x>0x > 0 and y>0y > 0?

  1. All trajectories approach the equilibrium point (3,0)(3, 0) where species xx survives alone
  2. All trajectories approach the equilibrium point (0,2)(0, 2) where species yy survives alone
  3. All trajectories approach the coexistence equilibrium (1,1)(1, 1) where both species survive
  4. Trajectory behavior depends on initial conditions, with some approaching (3,0)(3, 0) and others approaching (0,2)(0, 2) (correct answer)
Explanation: The equilibrium points are: (0,0)(0, 0), (3,0)(3, 0), (0,2)(0, 2), and (1,1)(1, 1). For the coexistence point (1,1)(1, 1): the Jacobian is J=(−1−2−1−1)J = \begin{pmatrix} -1 & -2 \\ -1 & -1 \end{pmatrix} with eigenvalues from λ2+2λ+1−2=λ2+2λ−1=0\lambda^2 + 2\lambda + 1 - 2 = \lambda^2 + 2\lambda - 1 = 0, giving λ=−1±2\lambda = -1 \pm \sqrt{2}. Since one eigenvalue is positive and one negative, (1,1)(1, 1) is a saddle point. This means trajectories starting in the first quadrant will approach either (3,0)(3, 0) or (0,2)(0, 2) depending on which side of the stable manifold they start. The separatrix divides the first quadrant into regions of attraction for each boundary equilibrium.

Question 7

For the linear system with coefficient matrix $$A = \begin{pmatrix} -1 & 4 \ -1 & -1 \end{pmatrix}

  1. Trajectories spiral clockwise toward the origin with decreasing amplitude over time (correct answer)
  2. Trajectories spiral counterclockwise toward the origin with decreasing amplitude over time
  3. Trajectories spiral clockwise away from the origin with increasing amplitude over time
  4. Trajectories spiral counterclockwise away from the origin with increasing amplitude over time
Explanation: The characteristic equation is det⁡(A−λI)=(−1−λ)2+4=λ2+2λ+5=0\det(A - \lambda I) = (-1-\lambda)^2 + 4 = \lambda^2 + 2\lambda + 5 = 0, giving eigenvalues λ=−1±2i\lambda = -1 \pm 2i. Since the real part is negative (-1), trajectories spiral toward the origin (stable spiral). To determine rotation direction: the general solution has the form x(t)=e−t[c1cos⁡(2t)u+c2sin⁡(2t)v]\mathbf{x}(t) = e^{-t}[c_1\cos(2t)\mathbf{u} + c_2\sin(2t)\mathbf{v}] where the eigenvectors and matrix structure determine clockwise rotation. Since both choices C and D involve outward spiraling, they are eliminated by the negative real part. The specific form of matrix A with positive off-diagonal element in position (1,2) and negative in position (2,1) produces clockwise rotation.

Question 8

The system dxdt=y\frac{dx}{dt} = y, dydt=−sin⁡(x)\frac{dy}{dt} = -\sin(x) represents a nonlinear pendulum. Near the equilibrium point (−π,0)(-\pi, 0), what type of phase portrait behavior occurs?

  1. Stable spiral due to the linearization having complex eigenvalues with negative real parts
  2. Unstable spiral due to the linearization having complex eigenvalues with positive real parts
  3. Saddle point due to the linearization having real eigenvalues of opposite signs (correct answer)
  4. Center point due to the linearization having purely imaginary eigenvalues
Explanation: At (−π,0)(-\pi, 0): ∂f∂x=0\frac{\partial f}{\partial x} = 0, ∂f∂y=1\frac{\partial f}{\partial y} = 1, ∂g∂x=−cos⁡(x)\frac{\partial g}{\partial x} = -\cos(x), ∂g∂y=0\frac{\partial g}{\partial y} = 0. The Jacobian is J=(01−cos⁡(−π)0)=(0110)J = \begin{pmatrix} 0 & 1 \\ -\cos(-\pi) & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} since cos⁡(−π)=−1\cos(-\pi) = -1. The characteristic equation is −λ2−1=0-\lambda^2 - 1 = 0, giving λ=±1\lambda = \pm 1. With eigenvalues +1+1 and −1-1, this is a saddle point. Choices A, B, and D are wrong because the eigenvalues are real, not complex or purely imaginary. The equilibrium at (−π,0)(-\pi, 0) corresponds to the pendulum balanced upside-down, which is indeed unstable (saddle).