I need help please #20 5.5 ProblemsFind general solutions of the systems in Problems 1 through22. In Problems 1 through 6, use a computer system or graph-ing calculator to construct a direction field and typical solutioncurves for the given system. 1-1-1]-1113. x' =1-4X1-3114. x' =-5-1-5X41-2-2-915. x' =1X11116. х'-2-2-3X234117. x' =1874X-27-9 -5118. х —131X-2-4-11-4-2119. х —6.X-6-12-1-4-1211120. x' =210 04+L

given x'=2101021000210002x we will find the eigenvalue A-λI=2101021000210002-λ1000010000100001…

Answered: 1 1 1 20. x' = %3D 2 1 2 N O O

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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I need help please #20

5.5 Problems
Find general solutions of the systems in Problems 1 through
22. In Problems 1 through 6, use a computer system or graph-
ing calculator to construct a direction field and typical solution
curves for the given system.

1
-1
-1]
-1
1
13. x' =
1
-4
X
1
-3
1
14. x' =
-5
-1
-5
X
4
1
-2
-2
-9
15. x' =
1
X
1
1
1
16. х'
-2
-2
-3
X
2
3
4
1
17. x' =
18
7
4
X
-27
-9 -5
1
18. х —
1
3
1
X
-2
-4
-1
1
-4
-2
1
19. х —
6.
X
-6
-12
-1
-4
-1
2
1
1
1
20. x' =
2
1
0 0
4+
L

Expert Solution

Step 1

given

$x^{'} = [\begin{matrix} 2 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix}] x$

we will find the eigenvalue

$A - λ I = [\begin{matrix} 2 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix}] - λ [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 - λ & 1 & 0 & 1 \\ 0 & 2 - λ & 1 & 0 \\ 0 & 0 & 2 - λ & 1 \\ 0 & 0 & 0 & 2 - λ \end{matrix}]$

$|A - λ I| = |\begin{matrix} 2 - λ & 1 & 0 & 1 \\ 0 & 2 - λ & 1 & 0 \\ 0 & 0 & 2 - λ & 1 \\ 0 & 0 & 0 & 2 - λ \end{matrix}| = (2 - λ) [(2 - λ) (2 - λ)] - 0 + 0 + 0 = {(2 - λ)}^{3}$

Step 2

${(2 - λ)}^{3} = 0 \Rightarrow λ = 2, 2, 2$

implies 2 is the only eigenvalue with multiplicity 3

now, we will find the eigenvector for $λ = 2$

$A - 2 I = [\begin{matrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = 0 \Rightarrow x_{2} + x_{4} = 0 \Rightarrow x_{3} = 0 \Rightarrow x_{4} = 0 \Rightarrow x_{2} = 0$

implies we have a eigenvector $v_{4} = [\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}]$

now, we will find other vectors

$[\begin{matrix} 2 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix}] [\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 0 \\ 0 \end{matrix}] = v_{3}$

$[\begin{matrix} 2 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix}] [\begin{matrix} 2 \\ 0 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 4 \\ 0 \\ 0 \\ 0 \end{matrix}] = v_{2} \begin{matrix} \end{matrix}$

$[\begin{matrix} 2 & 1 & 0 & 1 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix}] [\begin{matrix} 4 \\ 0 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 8 \\ 0 \\ 0 \\ 0 \end{matrix}] = v_{1}$

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