Differential Equations Quiz: Phase Portraits And Trajectories

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.

Question 1

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 $(0,0)$: $\frac{\partial f}{\partial x} = 2-2x-y$, $\frac{\partial f}{\partial y} = -x$, $\frac{\partial g}{\partial x} = -y$, $\frac{\partial g}{\partial y} = 1-x-4y$. At $(0,0)$, the Jacobian is $J = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ with eigenvalues $\lambda_1 = 2, \lambda_2 = 1$. 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 2

1. Trajectories spiral inward toward the origin since $\frac{dr}{dt} = -r^3 < 0$ for all $r > 0$
2. Trajectories spiral outward from the origin since $\frac{dr}{dt} = r^3 > 0$ for all $r > 0$ (correct answer)
3. Trajectories form closed circular orbits since $\frac{dr}{dt} = 0$ and $\frac{d\theta}{dt} = 1$
4. Trajectories approach the unit circle since $\frac{dr}{dt} = r(1-r^2)$ changes sign at $r = 1$

Explanation: Converting to polar coordinates: $x = r\cos\theta$, $y = r\sin\theta$. We have $\frac{dr}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{r} = \frac{x(-y + x(x^2+y^2)) + y(x + y(x^2+y^2))}{r} = \frac{-xy + x^2(x^2+y^2) + xy + y^2(x^2+y^2)}{r} = \frac{(x^2+y^2)^2}{r} = r^3$. Since $r^3 > 0$ for all $r > 0$, trajectories spiral outward. Choice A has the wrong sign. Choice C is wrong because $\frac{dr}{dt} \neq 0$. Choice D uses an incorrect formula for $\frac{dr}{dt}$.

Question 3

1. $(3, -1)$ corresponding to the eigenvector of the smaller eigenvalue
2. $(1, -1)$ corresponding to the eigenvector of the smaller eigenvalue
3. $(1, 3)$ corresponding to the eigenvector of the larger eigenvalue
4. $(3, 1)$ 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 $t \to \infty$, 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 $A = \begin{pmatrix} 2 & 3 \\ -1 & 6 \end{pmatrix}$. The characteristic equation is $\det(A - \lambda I) = (2-\lambda)(6-\lambda) - 3(-1) = \lambda^2 - 8\lambda + 15 = 0$. This gives eigenvalues $\lambda_1 = 3$ and $\lambda_2 = 5$. Since $\lambda_2 = 5 > \lambda_1 = 3$, the larger eigenvalue is 5. For $\lambda = 5$, solve $(A - 5I)\vec{v} = 0$: $\begin{pmatrix} -3 & 3 \\ -1 & 1 \end{pmatrix}\vec{v} = 0$. This yields the eigenvector $(3, 1)$. As $t \to \infty$, the $e^{5t}$ term grows faster than $e^{3t}$, so trajectories align with the eigenvector of the larger eigenvalue. Choice A incorrectly identifies $(3, -1)$ and references the smaller eigenvalue. Choice B gives the wrong eigenvector $(1, -1)$ for the smaller eigenvalue. Choice C correctly identifies the larger eigenvalue but gives the wrong eigenvector $(1, 3)$. Only choice D correctly pairs the eigenvector $(3, 1)$ with the larger eigenvalue $\lambda = 5$. Key strategy: For long-term behavior, always find the eigenvalue with the largest real part—its eigenvector determines the asymptotic direction of trajectories.

Question 4

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: $\frac{dx}{dt} = \frac{\partial H}{\partial y} = y^3 - y = y(y^2 - 1)$, $\frac{dy}{dt} = -\frac{\partial H}{\partial x} = -x$. The Jacobian at $(0,0)$ is $J = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$ with eigenvalues from $\lambda^2 - 1 = 0$, giving $\lambda = \pm 1$. 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 $(0,0)$ is still a saddle due to the specific form of $H$. Choice A is wrong because the eigenvalues are real, not purely imaginary. Choices C and D are wrong because the eigenvalues are $\pm 1$, both nonzero and real.

Question 5

1. All trajectories approach the equilibrium point $(3, 0)$ where species $x$ survives alone
2. All trajectories approach the equilibrium point $(0, 2)$ where species $y$ survives alone
3. All trajectories approach the coexistence equilibrium $(1, 1)$ where both species survive
4. Trajectory behavior depends on initial conditions, with some approaching $(3, 0)$ and others approaching $(0, 2)$ (correct answer)

Explanation: The equilibrium points are: $(0, 0)$, $(3, 0)$, $(0, 2)$, and $(1, 1)$. For the coexistence point $(1, 1)$: the Jacobian is $J = \begin{pmatrix} -1 & -2 \\ -1 & -1 \end{pmatrix}$ with eigenvalues from $\lambda^2 + 2\lambda + 1 - 2 = \lambda^2 + 2\lambda - 1 = 0$, giving $\lambda = -1 \pm \sqrt{2}$. Since one eigenvalue is positive and one negative, $(1, 1)$ is a saddle point. This means trajectories starting in the first quadrant will approach either $(3, 0)$ or $(0, 2)$ 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 6

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 $\det(A - \lambda I) = (-1-\lambda)^2 + 4 = \lambda^2 + 2\lambda + 5 = 0$, giving eigenvalues $\lambda = -1 \pm 2i$. Since the real part is negative (-1), trajectories spiral toward the origin (stable spiral). To determine rotation direction: the general solution has the form $\mathbf{x}(t) = e^{-t}[c_1\cos(2t)\mathbf{u} + c_2\sin(2t)\mathbf{v}]$ 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 7

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 $(-\pi, 0)$: $\frac{\partial f}{\partial x} = 0$, $\frac{\partial f}{\partial y} = 1$, $\frac{\partial g}{\partial x} = -\cos(x)$, $\frac{\partial g}{\partial y} = 0$. The Jacobian is $J = \begin{pmatrix} 0 & 1 \\ -\cos(-\pi) & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ since $\cos(-\pi) = -1$. The characteristic equation is $-\lambda^2 - 1 = 0$, giving $\lambda = \pm 1$. With eigenvalues $+1$ and $-1$, 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 $(-\pi, 0)$ corresponds to the pendulum balanced upside-down, which is indeed unstable (saddle).

Question 8

1. The trajectory spirals inward toward the origin as $t \to \infty$
2. The trajectory approaches a straight-line solution along the eigenvector direction as $t \to \infty$
3. The trajectory spirals outward away from the origin as $t \to \infty$ (correct answer)
4. The trajectory oscillates periodically around a closed orbit as $t \to \infty$

Explanation: The coefficient matrix is $A = \begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix}$. The characteristic equation is $(3-\lambda)(1-\lambda) + 2 = \lambda^2 - 4\lambda + 5 = 0$, giving eigenvalues $\lambda = 2 \pm i$. Since the real part is positive ($2 > 0$), 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.