Answered: Matrix System -4 -2 1 X; Xі — е t -12…

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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Question

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system

Matrix System
-4
-2
1
X; Xі — е t
-12
-1
-6
-4
0 -1
X2 = e=t
et
X4 = e
-2

Expert Solution

Step 1

Given : $x' = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] x; x_{1} = e^{- t} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]; x_{2} = e^{- t} [\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}]; x_{3} = e^{t} [\begin{matrix} 0 \\ 1 \\ 0 \\ - 2 \end{matrix}]; x_{4} = e^{t} [\begin{matrix} 1 \\ 0 \\ 3 \\ 0 \end{matrix}]$

(a) To verify: these vectors are solutions of given system

(b) To Show: $W r o n s k i a n$ is nonzero and Solution of the system

Step 2

$x' = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] x = P x x_{1} = e^{- t} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]; x_{2} = e^{- t} [\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}]; x_{3} = e^{t} [\begin{matrix} 0 \\ 1 \\ 0 \\ - 2 \end{matrix}]; x_{4} = e^{t} [\begin{matrix} 1 \\ 0 \\ 3 \\ 0 \end{matrix}]$

To verify that the solution the vector function $x_{1}, x_{2}, x_{3} & x_{4}$

$x_{1} = e^{- t} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]; x_{2} = [\begin{matrix} 0 \\ 0 \\ e^{- t} \\ 0 \end{matrix}]; x_{3} = [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}]; x_{4} = e^{t} [\begin{matrix} 1 \\ 0 \\ 3 \\ 0 \end{matrix}] x_{1}^{'} = e^{- t} [\begin{matrix} - e^{- t} \\ 0 \\ 0 \\ - e^{- t} \end{matrix}]; x_{2}^{'} = [\begin{matrix} 0 \\ 0 \\ - e^{- t} \\ 0 \end{matrix}]; x_{3}^{'} = [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}]; x_{4}^{'} = [\begin{matrix} e^{t} \\ 0 \\ 3 e^{t} \\ 0 \end{matrix}]$

are both solution of matrix differential equation with coefficient matrix P . we need only Calculate .

$P x_{1} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} e^{- t} \\ 0 \\ 0 \\ e^{- t} \end{matrix}] = [\begin{matrix} - e^{- t} \\ 0 \\ 0 \\ - e^{- t} \end{matrix}] = x_{1}^{'} P x_{2} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} 0 \\ 0 \\ e^{- t} \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ - e^{- t} \\ 0 \end{matrix}] = x_{2}^{'} P x_{3} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}] = [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}] = x_{3}^{'} P x_{4} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} e^{t} \\ 0 \\ 3 e^{t} \\ 0 \end{matrix}] = [\begin{matrix} e^{t} \\ 0 \\ 3 e^{t} \\ 0 \end{matrix}] = x_{4}^{'}$

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