Consider the system dx · (1) dt 3 (a) Let x= ge" be the solution of the system (1) then, 5-r -1 (2) 3 1-r5. The eigenvalues are the roots of the equation. 5-r | 3 1-r| p2 - 6r +8 =0 (r-2)(r-4)=0 r= 2,4 5 -1 are r = 2 and r, = 4 Therefore, the eigenvalues of Commont
Consider the system dx · (1) dt 3 (a) Let x= ge" be the solution of the system (1) then, 5-r -1 (2) 3 1-r5. The eigenvalues are the roots of the equation. 5-r | 3 1-r| p2 - 6r +8 =0 (r-2)(r-4)=0 r= 2,4 5 -1 are r = 2 and r, = 4 Therefore, the eigenvalues of Commont
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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![Consider the system
dx
· (1)
dt
3
(a)
Let x= ge" be the solution of the system (1) then,
200
5-r -1
(2)
3
1-r5.
The eigenvalues are the roots of the equation.
5-r
| 3-0
1-r
p? - 6r +8 =0
(r-2)(r-4)=0
r= 2,4
(5 -1
Therefore, the eigenvalues of
are r = 2 and r, = 4
3 1
Comment](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e18929d-c1af-48ad-8fab-37aadb4f52ee%2F6f512d52-2acd-49d3-833f-49012a2dcaac%2Fr54sf7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the system
dx
· (1)
dt
3
(a)
Let x= ge" be the solution of the system (1) then,
200
5-r -1
(2)
3
1-r5.
The eigenvalues are the roots of the equation.
5-r
| 3-0
1-r
p? - 6r +8 =0
(r-2)(r-4)=0
r= 2,4
(5 -1
Therefore, the eigenvalues of
are r = 2 and r, = 4
3 1
Comment
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