Differential Equations Quiz: Systems Real Distinct Eigenvalues

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.

Question 1

1. All solutions approach the origin along the direction of $\mathbf{v}_2$
2. All solutions grow unboundedly in the direction of $\mathbf{v}_1$ (correct answer)
3. Solutions oscillate between the directions of $\mathbf{v}_1$ and $\mathbf{v}_2$
4. Solutions approach a steady state determined by both eigenvalues

Explanation: The eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = 1$ (both positive), with $\lambda_1 > \lambda_2$. Since both eigenvalues are positive, the origin is an unstable node. As $t \to \infty$, the term $c_1 e^{2t} \mathbf{v}_1$ dominates because $e^{2t}$ grows faster than $e^t$. When $c_1 \neq 0$, solutions grow unboundedly in the direction of $\mathbf{v}_1$. 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

1. The origin is a stable node with trajectories approaching along the eigenvector for $\lambda_2$
2. The origin is a saddle point with stable manifold along the eigenvector for $\lambda_1$
3. The origin is a saddle point with unstable manifold along the eigenvector for $\lambda_1$ (correct answer)
4. The origin is an unstable node with trajectories moving away along the eigenvector for $\lambda_2$

Explanation: With eigenvalues $\lambda_1 = 1 > 0$ and $\lambda_2 = -6 < 0$, 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 $\lambda_1 = 1$, while the stable manifold (where trajectories approach the origin) corresponds to the negative eigenvalue $\lambda_2 = -6$. Choice A is wrong because this is not a node. Choice B is wrong because the stable manifold corresponds to $\lambda_2$, not $\lambda_1$. Choice D is wrong because this is not a node, and the behavior described doesn't match saddle point dynamics.

Question 3

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 $x_1(0) + x_2(0) = 0$
4. The initial condition must satisfy $2x_1(0) - 3x_2(0) = 0$

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

Question 4

1. The ratio approaches $\frac{1}{2}$ (correct answer)
2. The ratio approaches $\frac{3}{4}$
3. The ratio approaches $1$
4. The ratio approaches $\frac{2}{3}$

Explanation: The solutions are $\mathbf{x}_\mathbf{u}(t) = 2e^{3t}\mathbf{v}_1 + 3e^{-2t}\mathbf{v}_2$ and $\mathbf{x}_\mathbf{w}(t) = 4e^{3t}\mathbf{v}_1 - e^{-2t}\mathbf{v}_2$. As $t \to \infty$, the terms with $e^{-2t}$ become negligible compared to those with $e^{3t}$, so $\mathbf{x}_\mathbf{u}(t) \approx 2e^{3t}\mathbf{v}_1$ and $\mathbf{x}_\mathbf{w}(t) \approx 4e^{3t}\mathbf{v}_1$. Therefore, $\frac{\|\mathbf{x}_\mathbf{u}(t)\|}{\|\mathbf{x}_\mathbf{w}(t)\|} \approx \frac{\|2e^{3t}\mathbf{v}_1\|}{\|4e^{3t}\mathbf{v}_1\|} = \frac{2e^{3t}\|\mathbf{v}_1\|}{4e^{3t}\|\mathbf{v}_1\|} = \frac{1}{2}$.

Question 5

1. The trajectory approaches the origin along the eigenvector for $\lambda_2$
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 $\lambda_1$ (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 $\lambda_1 = 2 + \sqrt{2} \approx 3.41 > 0$ and $\lambda_2 = 2 - \sqrt{2} \approx 0.59 > 0$. Since both eigenvalues are positive, both corresponding solution components grow exponentially over time. The larger eigenvalue $\lambda_1$ grows faster and will dominate the long-term behavior. For any initial condition, the general solution is $\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2$, where $\mathbf{v}_1$ and $\mathbf{v}_2$ are the eigenvectors. As $t \to \infty$, the term with $\lambda_1$ dominates unless $c_1 = 0$. The trajectory will move away from the origin in the direction determined by $\lambda_1$'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

1. $\mathbf{x}(t) = c_1 e^{2t}\begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{3t}\begin{pmatrix} 1 \\ -3 \end{pmatrix}$
2. $\mathbf{x}(t) = c_1 e^{-3t}\begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{-2t}\begin{pmatrix} 1 \\ -3 \end{pmatrix}$
3. $\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{-3t}\begin{pmatrix} 1 \\ 3 \end{pmatrix}$
4. $\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{-3t}\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ (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 $\lambda^2 + 5\lambda + 6 = 0$, you first factor to get $(\lambda + 2)(\lambda + 3) = 0$, yielding eigenvalues $\lambda_1 = -2$ and $\lambda_2 = -3$. For each eigenvalue, you find eigenvectors by solving $(A - \lambda I)\mathbf{v} = \mathbf{0}$: For $\lambda_1 = -2$: $(A + 2I) = \begin{pmatrix} 2 & 1 \\ -6 & -3 \end{pmatrix}$. This gives $2v_1 + v_2 = 0$, so $v_2 = -2v_1$. With $v_1 = 1$, 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

1. $\|\mathbf{x}(t)\|$ approaches zero exponentially as $t \to \infty$
2. $\|\mathbf{x}(t)\|$ grows exponentially without bound as $t \to \infty$ (correct answer)
3. $\|\mathbf{x}(t)\|$ approaches a non-zero constant as $t \to \infty$
4. $\|\mathbf{x}(t)\|$ oscillates with increasing amplitude as $t \to \infty$

Explanation: To find the solution, we need to express the initial condition as a linear combination of eigenvectors: $\begin{pmatrix} 1 \\ 1 \end{pmatrix} = c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. This gives us $c_1 + c_2 = 1$ and $2c_1 - c_2 = 1$. Solving: $c_1 = \frac{2}{3}$ and $c_2 = \frac{1}{3}$. The solution is $\mathbf{x}(t) = \frac{2}{3}e^{5t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + \frac{1}{3}e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Since $\lambda_1 = 5 > 0$, the term $\frac{2}{3}e^{5t}\mathbf{v}_1$ dominates as $t \to \infty$, causing $\|\mathbf{x}(t)\|$ 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.