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A fundamental set of solutions of x' = 3 2 x is: 1 (w) x. - “(;). x-() (F). (1). *--(;) (a) x1 = et X2 = e (b) x1 = e-4t X2 = e? (1)» (F)- (c) x1 = e-4t > X2 = e (). (d) x1 = eat • X2 = et (e) None of the above.

A fundamental set of solutions of x' = 3 2 x is: 1 (w) x. - “(;). x-() (F). (1). *--(;) (a) x1 = et X2 = e (b) x1 = e-4t X2 = e? (1)» (F)- (c) x1 = e-4t > X2 = e (). (d) x1 = eat • X2 = et (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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**7. A fundamental set of solutions of \(\mathbf{x'} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} \mathbf{x}\) is:** (a) \(\mathbf{x}_1 = e^{4t} \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (b) \(\mathbf{x}_1 = e^{-4t} \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) (c) \(\mathbf{x}_1 = e^{-4t} \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) (d) \(\mathbf{x}_1 = e^{4t} \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (e) None of the above. **8. The solution of the initial-value problem \(\mathbf{x'} = \begin{pmatrix} 1 & 1 \\ 4 & -2 \end{pmatrix} \mathbf{x}, \, \mathbf{x}(0) = \begin{pmatrix} 2 \\ 5 \end{pmatrix}\) is:** (a) \(\mathbf{x} = \frac{3}{5} e^{3t} \begin{pmatrix} -1 \\ 4 \end{pmatrix} - \frac{13}{5} e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (b) \(\mathbf{x} = -\frac{3}{5} e^{-3t} \begin{pmatrix} 1 \\ -4 \end{pmatrix} + \frac{13}{5} e^{2t} \begin{pmatrix} 1 \\ 1
Transcribed Image Text:**7. A fundamental set of solutions of \(\mathbf{x'} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} \mathbf{x}\) is:** (a) \(\mathbf{x}_1 = e^{4t} \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (b) \(\mathbf{x}_1 = e^{-4t} \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) (c) \(\mathbf{x}_1 = e^{-4t} \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) (d) \(\mathbf{x}_1 = e^{4t} \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad \mathbf{x}_2 = e^{t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (e) None of the above. **8. The solution of the initial-value problem \(\mathbf{x'} = \begin{pmatrix} 1 & 1 \\ 4 & -2 \end{pmatrix} \mathbf{x}, \, \mathbf{x}(0) = \begin{pmatrix} 2 \\ 5 \end{pmatrix}\) is:** (a) \(\mathbf{x} = \frac{3}{5} e^{3t} \begin{pmatrix} -1 \\ 4 \end{pmatrix} - \frac{13}{5} e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (b) \(\mathbf{x} = -\frac{3}{5} e^{-3t} \begin{pmatrix} 1 \\ -4 \end{pmatrix} + \frac{13}{5} e^{2t} \begin{pmatrix} 1 \\ 1
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