This is the fourth part of a four-part problem. If the given solutions 7₁ (t) = [ 26² +6e-1]. 2e³t 3e3t +15e-t' 9 ÿ' = [15 form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem ÿ(t) = (0)[, -4 -7 - 3e³t ₂ (t) = = ÿ, ÿ(0) 2e³t+6e7 impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. +15e-t [4e³t +2e- 6e³t +5e-t + = [28] " [4e³t +2e-t 6est + 5e7
This is the fourth part of a four-part problem. If the given solutions 7₁ (t) = [ 26² +6e-1]. 2e³t 3e3t +15e-t' 9 ÿ' = [15 form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem ÿ(t) = (0)[, -4 -7 - 3e³t ₂ (t) = = ÿ, ÿ(0) 2e³t+6e7 impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. +15e-t [4e³t +2e- 6e³t +5e-t + = [28] " [4e³t +2e-t 6est + 5e7
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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![This is the fourth part of a four-part problem.
If the given solutions
\[
\vec{y}_1(t) = \begin{bmatrix} 2e^{3t} + 6e^{-t} \\ 3e^{3t} + 15e^{-t} \end{bmatrix}, \quad \vec{y}_2(t) = \begin{bmatrix} 4e^{3t} + 2e^{-t} \\ 6e^{3t} + 5e^{-t} \end{bmatrix}.
\]
form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem
\[
\vec{y}' = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{y}, \quad \vec{y}(0) = \begin{bmatrix} 8 \\ 28 \end{bmatrix},
\]
impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks.
\[
\vec{y}(t) = \begin{bmatrix} (\text{\_\_\_}) \begin{bmatrix} 2e^{3t} + 6e^{-t} \\ 3e^{3t} + 15e^{-t} \end{bmatrix} + (\text{\_\_\_}) \begin{bmatrix} 4e^{3t} + 2e^{-t} \\ 6e^{3t} + 5e^{-t} \end{bmatrix} \end{bmatrix}.
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Transcribed Image Text:This is the fourth part of a four-part problem.
If the given solutions
\[
\vec{y}_1(t) = \begin{bmatrix} 2e^{3t} + 6e^{-t} \\ 3e^{3t} + 15e^{-t} \end{bmatrix}, \quad \vec{y}_2(t) = \begin{bmatrix} 4e^{3t} + 2e^{-t} \\ 6e^{3t} + 5e^{-t} \end{bmatrix}.
\]
form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem
\[
\vec{y}' = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{y}, \quad \vec{y}(0) = \begin{bmatrix} 8 \\ 28 \end{bmatrix},
\]
impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks.
\[
\vec{y}(t) = \begin{bmatrix} (\text{\_\_\_}) \begin{bmatrix} 2e^{3t} + 6e^{-t} \\ 3e^{3t} + 15e^{-t} \end{bmatrix} + (\text{\_\_\_}) \begin{bmatrix} 4e^{3t} + 2e^{-t} \\ 6e^{3t} + 5e^{-t} \end{bmatrix} \end{bmatrix}.
\]
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