1 -3 1. 10. The general solution of x' = -1 1 -1 |х is 3 -3 -1 3 (a) x = Ce-t -2 + C2e*| 3+ C3e* | 1 -3 3 (b) x = Cet 2 + C2e-2t| 0 + C3e-t -3 1 3 3 + C2e-2t| 1 + C3e-t 3 (c) x = C, et -2 3 1 -3 (d) x = C1e¬t 2 + C2e-2t -3 |+ C3 e -3 (e) None of the above.
1 -3 1. 10. The general solution of x' = -1 1 -1 |х is 3 -3 -1 3 (a) x = Ce-t -2 + C2e*| 3+ C3e* | 1 -3 3 (b) x = Cet 2 + C2e-2t| 0 + C3e-t -3 1 3 3 + C2e-2t| 1 + C3e-t 3 (c) x = C, et -2 3 1 -3 (d) x = C1e¬t 2 + C2e-2t -3 |+ C3 e -3 (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 presented is to find the general solution of the differential equation given by:
\[ x' = \begin{pmatrix} 1 & -3 & 1 \\ -1 & 1 & -1 \\ 3 & -3 & -1 \end{pmatrix} x \]
The options provided for the general solution are:
(a) \[ x = C_1 e^{-4t} \begin{pmatrix} 3 \\ -2 \\ -3 \end{pmatrix} + C_2 e^{2t} \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} + C_3 e^t \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
(b) \[ x = C_1 e^{4t} \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + C_3 e^t \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix} \]
(c) \[ x = C_1 e^{4t} \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} + C_3 e^{-t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
(d) \[ x = C_1 e^{-4t} \begin{pmatrix} -3 \\ 2 \\ -3 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} + C_3 e^t \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \]
(e) None of the above.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69d33b94-47ab-4124-8ded-71fa51390a3f%2F2f5a624f-bd34-434f-b7b7-86de8c7edfd4%2Fomorv7a_processed.png&w=3840&q=75)
Transcribed Image Text:The problem presented is to find the general solution of the differential equation given by:
\[ x' = \begin{pmatrix} 1 & -3 & 1 \\ -1 & 1 & -1 \\ 3 & -3 & -1 \end{pmatrix} x \]
The options provided for the general solution are:
(a) \[ x = C_1 e^{-4t} \begin{pmatrix} 3 \\ -2 \\ -3 \end{pmatrix} + C_2 e^{2t} \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} + C_3 e^t \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
(b) \[ x = C_1 e^{4t} \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + C_3 e^t \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix} \]
(c) \[ x = C_1 e^{4t} \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} + C_3 e^{-t} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
(d) \[ x = C_1 e^{-4t} \begin{pmatrix} -3 \\ 2 \\ -3 \end{pmatrix} + C_2 e^{-2t} \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} + C_3 e^t \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \]
(e) None of the above.
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