9. The general solution of x' = -6 -4 x is: 2 | ++(-:)| (a) x = Cịe-2t + Cze¬2t (b) x = Cye" () + Cac"(1) +*(:) 2t (e) x = Cw (;) +c=* [( -})(-^)] (). (-1) +*( -2t -2t -5/4 (d) x = C1e²t + C2 2t +t (e) None of the above.
9. The general solution of x' = -6 -4 x is: 2 | ++(-:)| (a) x = Cịe-2t + Cze¬2t (b) x = Cye" () + Cac"(1) +*(:) 2t (e) x = Cw (;) +c=* [( -})(-^)] (). (-1) +*( -2t -2t -5/4 (d) x = C1e²t + C2 2t +t (e) None of the above.
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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![The problem presents a differential equation solution task:
9. The general solution of the differential equation \( \mathbf{x}' = \begin{pmatrix} -6 & -4 \\ 4 & 2 \end{pmatrix} \mathbf{x} \) is given in different forms, and the task is to identify the correct one:
(a) \( \mathbf{x} = C_1 e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 e^{-2t} \left[ \begin{pmatrix} -5/4 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right] \)
(b) \( \mathbf{x} = C_1 e^{2t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{2t} \left[ \begin{pmatrix} 5 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right] \)
(c) \( \mathbf{x} = C_1 e^{-2t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-2t} \left[ \begin{pmatrix} 1 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5/4 \end{pmatrix} \right] \)
(d) \( \mathbf{x} = C_1 e^{2t} \begin{pmatrix} -1 \\ -1 \end{pmatrix} + C_2 \left[ e^{2t} \begin{pmatrix} -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right] \)
(e) None of the above.
The matrices and vector expressions involve exponential terms and linear functions with respect to \( t \), each associated with constants \( C_1 \) and \( C_2 \). The choices are formatted to show different combinations of these terms.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69d33b94-47ab-4124-8ded-71fa51390a3f%2F4e266e5a-1baf-44c4-afef-776486ee015a%2Foq0qm_processed.png&w=3840&q=75)
Transcribed Image Text:The problem presents a differential equation solution task:
9. The general solution of the differential equation \( \mathbf{x}' = \begin{pmatrix} -6 & -4 \\ 4 & 2 \end{pmatrix} \mathbf{x} \) is given in different forms, and the task is to identify the correct one:
(a) \( \mathbf{x} = C_1 e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 e^{-2t} \left[ \begin{pmatrix} -5/4 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right] \)
(b) \( \mathbf{x} = C_1 e^{2t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{2t} \left[ \begin{pmatrix} 5 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right] \)
(c) \( \mathbf{x} = C_1 e^{-2t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-2t} \left[ \begin{pmatrix} 1 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5/4 \end{pmatrix} \right] \)
(d) \( \mathbf{x} = C_1 e^{2t} \begin{pmatrix} -1 \\ -1 \end{pmatrix} + C_2 \left[ e^{2t} \begin{pmatrix} -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right] \)
(e) None of the above.
The matrices and vector expressions involve exponential terms and linear functions with respect to \( t \), each associated with constants \( C_1 \) and \( C_2 \). The choices are formatted to show different combinations of these terms.
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