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

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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
Transcribed Image Text: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'=1-40-201006-12-1-60-40-1x  ; x1=e-t1001  ; x2=e-t0010  ; x3=et010-2 ; x4=et 1030

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

 (b) To Show: Wronskian is nonzero and Solution of the system

Step 2

x'=1-40-201006-12-1-60-40-1x  =Px x1=e-t1001  ; x2=e-t0010  ; x3=et010-2 ; x4=et 1030

To verify that the solution the vector function x1 , x2 , x3 & x4

x1=e-t1001  ; x2=00e-t0  ; x3=0et0-2et ; x4=et 1030x1'=e-t-e-t00-e-t  ; x2'=00-e-t0  ; x3'=0et0-2et ; x4'=et03et0

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

Px1=1-40-201006-12-1-60-40-1e-t00e-t=-e-t00-e-t=x1'Px2=1-40-201006-12-1-60-40-1 00e-t0=00-e-t0=x2'Px3=1-40-201006-12-1-60-40-10et0-2et=0et0-2et=x3'Px4=1-40-201006-12-1-60-40-1et03et0=et03et0=x4'

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