Differential Equations Quiz: Systems Real Distinct Eigenvalues
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Systems Real Distinct EigenvaluesQuestion 1 of 7

Consider the system x′=Ax\mathbf{x}' = A\mathbf{x} where A=(3−210)A = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix} . If the general solution is written as x(t)=c1eλ1tv1+c2eλ2tv2\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 where λ1>λ2\lambda_1 > \lambda_2, what is the behavior of solutions as t→∞t \to \infty when c1≠0c_1 \neq 0?

All solutions approach the origin along the direction of v2\mathbf{v}_2
All solutions grow unboundedly in the direction of v1\mathbf{v}_1
Solutions oscillate between the directions of v1\mathbf{v}_1 and v2\mathbf{v}_2
Solutions approach a steady state determined by both eigenvalues
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Differential Equations Quiz

Differential Equations Quiz: Systems Real Distinct Eigenvalues

Practice Systems Real Distinct Eigenvalues 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 Systems Real Distinct Eigenvalues, 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 x′=Ax\mathbf{x}' = A\mathbf{x} where A=(3−210)A = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix} . If the general solution is written as x(t)=c1eλ1tv1+c2eλ2tv2\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 where λ1>λ2\lambda_1 > \lambda_2, what is the behavior of solutions as t→∞t \to \infty when c1≠0c_1 \neq 0?

  1. All solutions approach the origin along the direction of v2\mathbf{v}_2
  2. All solutions grow unboundedly in the direction of v1\mathbf{v}_1 (correct answer)
  3. Solutions oscillate between the directions of v1\mathbf{v}_1 and v2\mathbf{v}_2
  4. Solutions approach a steady state determined by both eigenvalues
Explanation: The eigenvalues are λ1=2\lambda_1 = 2 and λ2=1\lambda_2 = 1 (both positive), with λ1>λ2\lambda_1 > \lambda_2. Since both eigenvalues are positive, the origin is an unstable node. As t→∞t \to \infty, the term c1e2tv1c_1 e^{2t} \mathbf{v}_1 dominates because e2te^{2t} grows faster than ete^t. When c1≠0c_1 \neq 0, solutions grow unboundedly in the direction of v1\mathbf{v}_1. Choice A is wrong because solutions don't approach the origin when eigenvalues are positive. Choice C is wrong because real eigenvalues don't produce oscillations. Choice D is wrong because there's no steady state when both eigenvalues are positive.

Question 2

For the system x′=(−132−4)x\mathbf{x}' = \begin{pmatrix} -1 & 3 \\ 2 & -4 \end{pmatrix} \mathbf{x}, the eigenvalues are λ1=1\lambda_1 = 1 and λ2=−6\lambda_2 = -6. Which statement about the phase portrait is correct?

  1. The origin is a stable node with trajectories approaching along the eigenvector for λ2\lambda_2
  2. The origin is a saddle point with stable manifold along the eigenvector for λ1\lambda_1
  3. The origin is a saddle point with unstable manifold along the eigenvector for λ1\lambda_1 (correct answer)
  4. The origin is an unstable node with trajectories moving away along the eigenvector for λ2\lambda_2
Explanation: With eigenvalues λ1=1>0\lambda_1 = 1 > 0 and λ2=−6<0\lambda_2 = -6 < 0, we have one positive and one negative eigenvalue, making the origin a saddle point. The unstable manifold (where trajectories move away from the origin) corresponds to the positive eigenvalue λ1=1\lambda_1 = 1, while the stable manifold (where trajectories approach the origin) corresponds to the negative eigenvalue λ2=−6\lambda_2 = -6. Choice A is wrong because this is not a node. Choice B is wrong because the stable manifold corresponds to λ2\lambda_2, not λ1\lambda_1. Choice D is wrong because this is not a node, and the behavior described doesn't match saddle point dynamics.

Question 3

The system x′=Ax\mathbf{x}' = A\mathbf{x} has the general solution x(t)=c1e−2t(13)+c2e4t(21)\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{4t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} . If we want the solution to remain bounded as t→∞t \to \infty, what constraint must be placed on the initial conditions?

  1. The initial condition must be a multiple of $$ \begin{pmatrix} 1 \ 3 \end{pmatrix} (correct answer)
  2. The initial condition must be a multiple of $$ \begin{pmatrix} 2 \ 1 \end{pmatrix}
  3. The initial condition must satisfy x1(0)+x2(0)=0x_1(0) + x_2(0) = 0
  4. The initial condition must satisfy 2x1(0)−3x2(0)=02x_1(0) - 3x_2(0) = 0
Explanation: For the solution to remain bounded as t→∞t \to \infty, we need the coefficient of the growing term e4te^{4t} to be zero, i.e., c2=0c_2 = 0. This means x(t)=c1e−2t(13)\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ 3 \end{pmatrix} , so x(0)=c1(13)\mathbf{x}(0) = c_1\begin{pmatrix} 1 \\ 3 \end{pmatrix} . Therefore, the initial condition must be a multiple of the eigenvector corresponding to the negative eigenvalue λ=−2\lambda = -2. Choice B is wrong because (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix} corresponds to the positive eigenvalue, leading to unbounded growth. Choice C gives x1(0)+x2(0)=c1+2c2+3c1+c2=4c1+3c2=0x_1(0) + x_2(0) = c_1 + 2c_2 + 3c_1 + c_2 = 4c_1 + 3c_2 = 0, which doesn't force c2=0c_2 = 0. Choice D gives 2x1(0)−3x2(0)=2(c1+2c2)−3(3c1+c2)=2c1+4c2−9c1−3c2=−7c1+c2=02x_1(0) - 3x_2(0) = 2(c_1 + 2c_2) - 3(3c_1 + c_2) = 2c_1 + 4c_2 - 9c_1 - 3c_2 = -7c_1 + c_2 = 0, which also doesn't force c2=0c_2 = 0.

Question 4

Consider the linear system x′=Ax\mathbf{x}' = A\mathbf{x} where AA has eigenvalues λ1=3\lambda_1 = 3 and λ2=−2\lambda_2 = -2 with eigenvectors v1\mathbf{v}_1 and v2\mathbf{v}_2 respectively. If u\mathbf{u} and w\mathbf{w} are two different initial conditions such that u=2v1+3v2\mathbf{u} = 2\mathbf{v}_1 + 3\mathbf{v}_2 and w=4v1−v2\mathbf{w} = 4\mathbf{v}_1 - \mathbf{v}_2, what is the ratio ∥xu(t)∥∥xw(t)∥\frac{\|\mathbf{x}_\mathbf{u}(t)\|}{\|\mathbf{x}_\mathbf{w}(t)\|} as t→∞t \to \infty?

  1. The ratio approaches 12\frac{1}{2} (correct answer)
  2. The ratio approaches 34\frac{3}{4}
  3. The ratio approaches 11
  4. The ratio approaches 23\frac{2}{3}
Explanation: The solutions are xu(t)=2e3tv1+3e−2tv2\mathbf{x}_\mathbf{u}(t) = 2e^{3t}\mathbf{v}_1 + 3e^{-2t}\mathbf{v}_2 and xw(t)=4e3tv1−e−2tv2\mathbf{x}_\mathbf{w}(t) = 4e^{3t}\mathbf{v}_1 - e^{-2t}\mathbf{v}_2. As t→∞t \to \infty, the terms with e−2te^{-2t} become negligible compared to those with e3te^{3t}, so xu(t)≈2e3tv1\mathbf{x}_\mathbf{u}(t) \approx 2e^{3t}\mathbf{v}_1 and xw(t)≈4e3tv1\mathbf{x}_\mathbf{w}(t) \approx 4e^{3t}\mathbf{v}_1. Therefore, ∥xu(t)∥∥xw(t)∥≈∥2e3tv1∥∥4e3tv1∥=2e3t∥v1∥4e3t∥v1∥=12\frac{\|\mathbf{x}_\mathbf{u}(t)\|}{\|\mathbf{x}_\mathbf{w}(t)\|} \approx \frac{\|2e^{3t}\mathbf{v}_1\|}{\|4e^{3t}\mathbf{v}_1\|} = \frac{2e^{3t}\|\mathbf{v}_1\|}{4e^{3t}\|\mathbf{v}_1\|} = \frac{1}{2}.

Question 5

For the system x′=(3−111)x\mathbf{x}' = \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}\mathbf{x}, suppose the eigenvalues are λ1=2+2\lambda_1 = 2 + \sqrt{2} and λ2=2−2\lambda_2 = 2 - \sqrt{2}. If a solution trajectory starts at $$\mathbf{x}(0) = \begin{pmatrix} 1 \ 0 \end{pmatrix}

  1. The trajectory approaches the origin along the eigenvector for λ2\lambda_2
  2. The trajectory approaches the origin, but the approach direction depends on the eigenvector coefficients
  3. The trajectory exhibits bounded oscillatory behavior around the origin
  4. The trajectory moves away from the origin along the eigenvector for λ1\lambda_1 (correct answer)
Explanation: When analyzing linear systems of differential equations, the eigenvalues determine the fundamental behavior of solution trajectories. Since both eigenvalues here are real and distinct, you'll get solutions that are combinations of exponential functions along their corresponding eigenvectors. The key insight is examining the signs and magnitudes of the eigenvalues. We have λ1=2+2≈3.41>0\lambda_1 = 2 + \sqrt{2} \approx 3.41 > 0 and λ2=2−2≈0.59>0\lambda_2 = 2 - \sqrt{2} \approx 0.59 > 0. Since both eigenvalues are positive, both corresponding solution components grow exponentially over time. The larger eigenvalue λ1\lambda_1 grows faster and will dominate the long-term behavior. For any initial condition, the general solution is x(t)=c1eλ1tv1+c2eλ2tv2\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2, where v1\mathbf{v}_1 and v2\mathbf{v}_2 are the eigenvectors. As t→∞t \to \infty, the term with λ1\lambda_1 dominates unless c1=0c_1 = 0. The trajectory will move away from the origin in the direction determined by λ1\lambda_1's eigenvector. Answer A is wrong because positive eigenvalues mean trajectories move away from, not toward, the origin. Answer B incorrectly suggests approach to the origin, which only happens when eigenvalues are negative. Answer C describes oscillatory behavior, which occurs with complex eigenvalues, not real ones. Therefore, D correctly identifies that the trajectory moves away from the origin along the dominant eigenvector direction. Study tip: Remember that positive real eigenvalues always indicate instability (growth away from equilibrium), while the largest eigenvalue determines the dominant long-term direction.

Question 6

The system x′=(01−6−5)x\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -6 & -5 \end{pmatrix}\mathbf{x} has characteristic polynomial λ2+5λ+6=0\lambda^2 + 5\lambda + 6 = 0. After finding the eigenvalues and eigenvectors, what is the general solution?

  4. (correct answer)
Explanation: When you encounter a linear system of differential equations, you're looking to find eigenvalues and eigenvectors to construct the general solution. The key is systematically working through each eigenvalue to find its corresponding eigenvector. Given the characteristic polynomial λ2+5λ+6=0\lambda^2 + 5\lambda + 6 = 0, you first factor to get (λ+2)(λ+3)=0(\lambda + 2)(\lambda + 3) = 0, yielding eigenvalues λ1=−2\lambda_1 = -2 and λ2=−3\lambda_2 = -3. For each eigenvalue, you find eigenvectors by solving (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0}: For λ1=−2\lambda_1 = -2: (A+2I)=(21−6−3)(A + 2I) = \begin{pmatrix} 2 & 1 \\ -6 & -3 \end{pmatrix} . This gives 2v1+v2=02v_1 + v_2 = 0, so v2=−2v1v_2 = -2v_1. With v1=1v_1 = 1, the eigenvector is $$ \begin{pmatrix} 1 \ -2 \end{pmatrix} For $$\lambda_2 = -3$$: $$(A + 3I) = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} $$. This gives $$3v_1 + v_2 = 0$$, so $$v_2 = -3v_1$$. With $$v_1 = 1$$, the eigenvector is $$ \begin{pmatrix} 1 \\ -3 \end{pmatrix} $$. Option A has positive eigenvalues (should be negative). Option B switches the eigenvector assignments between eigenvalues. Option C has incorrect signs in the eigenvector components (positive instead of negative). Only option D correctly pairs $$e^{-2t}$$ with $$ \begin{pmatrix} 1 \\ -2 \end{pmatrix} $$ and $$e^{-3t}$$ with $$ \begin{pmatrix} 1 \\ -3 \end{pmatrix} $$. Always double-check your eigenvector calculations by substituting back into the original equation—small arithmetic errors in finding eigenvectors are extremely common on exams.

Question 7

Consider the initial value problem x′=(1243)x\mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\mathbf{x} with x(0)=(11)\mathbf{x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix} . The eigenvalues are λ1=5\lambda_1 = 5 and λ2=−1\lambda_2 = -1 with corresponding eigenvectors v1=(12)\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} and v2=(1−1)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} . What happens to ∥x(t)∥\|\mathbf{x}(t)\| as tt increases?

  1. ∥x(t)∥\|\mathbf{x}(t)\| approaches zero exponentially as t→∞t \to \infty
  2. ∥x(t)∥\|\mathbf{x}(t)\| grows exponentially without bound as t→∞t \to \infty (correct answer)
  3. ∥x(t)∥\|\mathbf{x}(t)\| approaches a non-zero constant as t→∞t \to \infty
  4. ∥x(t)∥\|\mathbf{x}(t)\| oscillates with increasing amplitude as t→∞t \to \infty
Explanation: To find the solution, we need to express the initial condition as a linear combination of eigenvectors: (11)=c1(12)+c2(1−1)\begin{pmatrix} 1 \\ 1 \end{pmatrix} = c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}. This gives us c1+c2=1c_1 + c_2 = 1 and 2c1−c2=12c_1 - c_2 = 1. Solving: c1=23c_1 = \frac{2}{3} and c2=13c_2 = \frac{1}{3}. The solution is x(t)=23e5t(12)+13e−t(1−1)\mathbf{x}(t) = \frac{2}{3}e^{5t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + \frac{1}{3}e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} . Since λ1=5>0\lambda_1 = 5 > 0, the term 23e5tv1\frac{2}{3}e^{5t}\mathbf{v}_1 dominates as t→∞t \to \infty, causing ∥x(t)∥\|\mathbf{x}(t)\| to grow exponentially. Choice A is wrong because the positive eigenvalue causes growth, not decay. Choice C is wrong because there's no equilibrium with a positive eigenvalue. Choice D is wrong because real eigenvalues don't produce oscillations.