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1 -1 4 x is: -2 10. A fundamental set of solutions of x' = 3 -3 3 -1 () (a) x1 = et 1 X2 = e-2t 1 (b) x1 = e' 1 X2 = e-2t 1 -2t , X3 = te- 1. (c) x1 = e , X2 = e-2t 1 , X3 = e-2t 1 + te-2t 1 (:) (d) x1 = et , X2 = e-2t X3 = e-2t + te-2t (e) None of the above.

1 -1 4 x is: -2 10. A fundamental set of solutions of x' = 3 -3 3 -1 () (a) x1 = et 1 X2 = e-2t 1 (b) x1 = e' 1 X2 = e-2t 1 -2t , X3 = te- 1. (c) x1 = e , X2 = e-2t 1 , X3 = e-2t 1 + te-2t 1 (:) (d) x1 = et , X2 = e-2t X3 = e-2t + te-2t (e) None of the above.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem Statement:** 10. A fundamental set of solutions of the differential equation \( \mathbf{x}' = \begin{pmatrix} -2 & 1 & -1 \\ 3 & -3 & 4 \\ 3 & -1 & 2 \end{pmatrix} \mathbf{x} \) is: **Options:** (a) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) (b) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_3 = te^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \) (c) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_3 = e^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + te^{-2t} \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} \) (d) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_3 = e^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + te^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \) (e) None of the above.
Transcribed Image Text:**Problem Statement:** 10. A fundamental set of solutions of the differential equation \( \mathbf{x}' = \begin{pmatrix} -2 & 1 & -1 \\ 3 & -3 & 4 \\ 3 & -1 & 2 \end{pmatrix} \mathbf{x} \) is: **Options:** (a) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) (b) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_3 = te^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \) (c) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_3 = e^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + te^{-2t} \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} \) (d) \( \mathbf{x}_1 = e^t \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-2t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{x}_3 = e^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + te^{-2t} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \) (e) None of the above.
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