Systems: Real Distinct Eigenvalues - Differential Equations

1. For the system $\mathbf{x}' = \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}\mathbf{x}$, suppose the eigenvalues are $\lambda_1 = 2 + \sqrt{2}$ and $\lambda_2 = 2 - \sqrt{2}$. If a solution trajectory starts at $\mathbf{x}(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, which statement best describes the long-term behavior?
   Answer: The trajectory moves away from the origin along the eigenvector for $\lambda_1$ 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.
2. Consider the system $\mathbf{x}' = A\mathbf{x}$ where $A = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}$. If the general solution is written as $\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2$ where $\lambda_1 > \lambda_2$, what is the behavior of solutions as $t \to \infty$ when $c_1 \neq 0$?
   Answer: All solutions grow unboundedly in the direction of $\mathbf{v}_1$ 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.
3. For the system $\mathbf{x}' = \begin{pmatrix} -1 & 3 \\ 2 & -4 \end{pmatrix} \mathbf{x}$, the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = -6$. Which statement about the phase portrait is correct?
   Answer: The origin is a saddle point with unstable manifold along the eigenvector for $\lambda_1$ 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.
4. Consider the initial value problem $\mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\mathbf{x}$ with $\mathbf{x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. The eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = -1$ with corresponding eigenvectors $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. What happens to $\|\mathbf{x}(t)\|$ as $t$ increases?
   Answer: $\|\mathbf{x}(t)\|$ grows exponentially without bound as $t \to \infty$ 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.
5. The system $\mathbf{x}' = A\mathbf{x}$ has the general solution $\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{4t}\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. If we want the solution to remain bounded as $t \to \infty$, what constraint must be placed on the initial conditions?
   Answer: The initial condition must be a multiple of $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ 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 $c_2 = 0$.
6. Consider the linear system $\mathbf{x}' = A\mathbf{x}$ where $A$ has eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -2$ with eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$ respectively. If $\mathbf{u}$ and $\mathbf{w}$ are two different initial conditions such that $\mathbf{u} = 2\mathbf{v}_1 + 3\mathbf{v}_2$ and $\mathbf{w} = 4\mathbf{v}_1 - \mathbf{v}_2$, what is the ratio $\frac{\|\mathbf{x}_\mathbf{u}(t)\|}{\|\mathbf{x}_\mathbf{w}(t)\|}$ as $t \to \infty$?
   Answer: The ratio approaches $\frac{1}{2}$ 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, $\frac{\|\mathbf{x}_\mathbf{u}(t)\|}{\|\mathbf{x}_\mathbf{w}(t)\|} \approx \frac{\|2e^{3t}\mathbf{v}_1\|}{\|4e^{3t}\mathbf{v}_1\|} = \frac{2e^{3t}\|\mathbf{v}_1\|}{4e^{3t}\|\mathbf{v}_1\|} = \frac{1}{2}$.
7. The system $\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -6 & -5 \end{pmatrix}\mathbf{x}$ has characteristic polynomial $\lambda^2 + 5\lambda + 6 = 0$. After finding the eigenvalues and eigenvectors, what is the general solution?
   Answer: $\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{-3t}\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ 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.

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