0 -2 has characteristic equation The matrix A det [0 -> a11 a12 = = a21 a22 1 -2 -2 = 1 -2-X (0-1)(-2-A) — (1)(−2) = - 12+2)+(2) = 0. The eigenvalue with positive imaginary part is 1 = -1+1 i For eigenvalue 1, the eigenvector has components a and b that satisfy (0-1)a+(-2)6 = 0, (1)a+(-21)b = 0 which is the system 1+-1 ia +-2b=0, 1a+1+i−1 ]b = 0. Choosing the first component of the eigenvector as a = 1, for 1 the eigenvector is a v = = b where b = +i The system of differential equations x' = 1 -2 - -2 has a general solution x(t) = = c1x1(t) + c2x2(t) where [Re (eta)] x1(t) = = [Re (etb)] [Im (e¹¹t a)] x2(t) = = Im (etb)] et cos(t) et sin(t) =
0 -2 has characteristic equation The matrix A det [0 -> a11 a12 = = a21 a22 1 -2 -2 = 1 -2-X (0-1)(-2-A) — (1)(−2) = - 12+2)+(2) = 0. The eigenvalue with positive imaginary part is 1 = -1+1 i For eigenvalue 1, the eigenvector has components a and b that satisfy (0-1)a+(-2)6 = 0, (1)a+(-21)b = 0 which is the system 1+-1 ia +-2b=0, 1a+1+i−1 ]b = 0. Choosing the first component of the eigenvector as a = 1, for 1 the eigenvector is a v = = b where b = +i The system of differential equations x' = 1 -2 - -2 has a general solution x(t) = = c1x1(t) + c2x2(t) where [Re (eta)] x1(t) = = [Re (etb)] [Im (e¹¹t a)] x2(t) = = Im (etb)] et cos(t) et sin(t) =
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![0 -2
has characteristic equation
The matrix A
det
[0 ->
a11
a12
=
=
a21 a22
1 -2
-2
=
1 -2-X
(0-1)(-2-A) — (1)(−2) =
-
12+2)+(2) = 0.
The eigenvalue with positive imaginary part is 1
= -1+1 i
For eigenvalue 1, the eigenvector has components a and b that satisfy
(0-1)a+(-2)6 = 0,
(1)a+(-21)b = 0
which is the system
1+-1 ia
+-2b=0,
1a+1+i−1 ]b = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F164d6224-1ff2-4b80-979f-023e65564011%2F3e6337df-2aa7-42e3-b2b6-6312539d2975%2F7vg4cuh_processed.png&w=3840&q=75)
Transcribed Image Text:0 -2
has characteristic equation
The matrix A
det
[0 ->
a11
a12
=
=
a21 a22
1 -2
-2
=
1 -2-X
(0-1)(-2-A) — (1)(−2) =
-
12+2)+(2) = 0.
The eigenvalue with positive imaginary part is 1
= -1+1 i
For eigenvalue 1, the eigenvector has components a and b that satisfy
(0-1)a+(-2)6 = 0,
(1)a+(-21)b = 0
which is the system
1+-1 ia
+-2b=0,
1a+1+i−1 ]b = 0.
![Choosing the first component of the eigenvector as a = 1, for 1 the eigenvector is
a
v =
=
b
where b = +i
The system of differential equations
x'
=
1
-2
-
-2
has a general solution
x(t) =
= c1x1(t) + c2x2(t)
where
[Re (eta)]
x1(t)
=
=
[Re (etb)]
[Im (e¹¹t a)]
x2(t) =
=
Im (etb)]
et cos(t)
et sin(t)
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F164d6224-1ff2-4b80-979f-023e65564011%2F3e6337df-2aa7-42e3-b2b6-6312539d2975%2Ffc7wxy_processed.png&w=3840&q=75)
Transcribed Image Text:Choosing the first component of the eigenvector as a = 1, for 1 the eigenvector is
a
v =
=
b
where b = +i
The system of differential equations
x'
=
1
-2
-
-2
has a general solution
x(t) =
= c1x1(t) + c2x2(t)
where
[Re (eta)]
x1(t)
=
=
[Re (etb)]
[Im (e¹¹t a)]
x2(t) =
=
Im (etb)]
et cos(t)
et sin(t)
=
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