Using eigenvalues and eigenvectors to solve system of ODEs

  • #1

Homework Statement


Use eigenvalues and eigenvectors to find the general solution of the system of ODEs..

x1 = 3x1 - x2
x2 = -x1 + 2x2 - x3
x3 = -x2 + 3x3

Homework Equations





The Attempt at a Solution


I converted that into the matrix
(3-λ -1 0)
(-1 2-λ -1) using 0=[itex]|A-λI[/itex][itex]|[/itex]
(0 -1 3-λ)

Sorry, I am new and don't know how else to write out the matrix... but I hope you get the gist..


I then solved through to get (3-λ)(λ-3)(λ-2)

λ=2 got me an eigenvector (5,1,5)T

λ=3 got me an eigenvecotr (6,1,6)T


Not really sure if I have done this correctly, and if it's correct I'm not sure how I would get this into a general solution eλt(eigenvector)T seeing as there is a repeated root.


Thanks
 

Answers and Replies

  • #2
33,981
5,639

Homework Statement


Use eigenvalues and eigenvectors to find the general solution of the system of ODEs..

x1 = 3x1 - x2
x2 = -x1 + 2x2 - x3
x3 = -x2 + 3x3
You're missing the primes in the column vector at the left. These are supposed to be derivatives.

Homework Equations





The Attempt at a Solution


I converted that into the matrix
(3-λ -1 0)
(-1 2-λ -1) using 0=[itex]|A-λI[/itex][itex]|[/itex]
(0 -1 3-λ)

Sorry, I am new and don't know how else to write out the matrix... but I hope you get the gist..


I then solved through to get (3-λ)(λ-3)(λ-2)
I'm pretty sure you made a mistake. I get a different characteristic polynomial with three distinct roots. Show us what you did to get the char. polynomial.
λ=2 got me an eigenvector (5,1,5)T
When you reach the point that you have found an eigenvector, a sanity test is a good idea. For your matrix (call it A) and the eigenvector you found (call it x), verify that Ax = 2x.
λ=3 got me an eigenvecotr (6,1,6)T


Not really sure if I have done this correctly, and if it's correct I'm not sure how I would get this into a general solution eλt(eigenvector)T seeing as there is a repeated root.


Thanks
 
  • #3
Yeah, sorry they were meant to have primes on them, my bad.

Okay, from

(3-λ -1 0)
(-1 2-λ -1)
(0 -1 3-λ)

I did Column 1 - C3 to get

(3-λ -1 0)
(0 2-λ -1)
(-(3-λ) -1 3-λ)

I then took out (3-λ) as it's a common factor of C1 to get: (This is the step i wasn't too confident on)

(3-λ) (1 -1 0)
(0 2-λ -1)
(-1 -1 3-λ)

I then did C2 - C1:

(3-λ) (1 0 0)
(0 2-λ -1)
(-1 0 3-λ)

Sorry if this is quite hard to read! The second brackets is the matrix and the (3-λ) is the common factor i have extracted.

From here i simplified to:

(3-λ) (2-λ -1)
( 0 3-λ) which gave this answer:

(3-λ)(6 - 5λ + λ2)

=

(3-λ)(λ-3)(λ-2) = 0

therefore λ = 3,3,2

I think you can check to see if you're on the right step by seeing if the product of the eigenvalues is the same as the determinant of the original matrix, and they both come out to be 18... But i'm not too sure if this is correct.


''When you reach the point that you have found an eigenvector, a sanity test is a good idea. For your matrix (call it A) and the eigenvector you found (call it x), verify that Ax = 2x.''
For Ax i ended up with (14,8,14)T and 2x was (10,2,10)T, i presume this means I have made a mistake somewhere..

Thanks Mark.
 
  • #4
33,981
5,639
Yeah, sorry they were meant to have primes on them, my bad.

Okay, from

(3-λ -1 0)
(-1 2-λ -1)
(0 -1 3-λ)
I expanded the determinant across the first row to get
(3-λ)[(2 - λ)(3 - λ) - 1] + 1(λ - 3 - 0). I am confident of this work.
I did Column 1 - C3 to get

(3-λ -1 0)
(0 2-λ -1)
(-(3-λ) -1 3-λ)

I then took out (3-λ) as it's a common factor of C1 to get: (This is the step i wasn't too confident on)

(3-λ) (1 -1 0)
(0 2-λ -1)
(-1 -1 3-λ)

I then did C2 - C1:

(3-λ) (1 0 0)
(0 2-λ -1)
(-1 0 3-λ)

Sorry if this is quite hard to read! The second brackets is the matrix and the (3-λ) is the common factor i have extracted.

From here i simplified to:

(3-λ) (2-λ -1)
( 0 3-λ) which gave this answer:

(3-λ)(6 - 5λ + λ2)

=

(3-λ)(λ-3)(λ-2) = 0

therefore λ = 3,3,2

I think you can check to see if you're on the right step by seeing if the product of the eigenvalues is the same as the determinant of the original matrix, and they both come out to be 18... But i'm not too sure if this is correct.




For Ax i ended up with (14,8,14)T and 2x was (10,2,10)T, i presume this means I have made a mistake somewhere..

Thanks Mark.
 
  • #5
I expanded the determinant across the first row to get
(3-λ)[(2 - λ)(3 - λ) - 1] + 1(λ - 3 - 0). I am confident of this work.
Does this mean I have the correct roots?

If so, how should I approach the eigenvectors with a repeated root?

Thanks Mark
 
  • #6
33,981
5,639
No, you don't have the correct roots. There are three distinct roots.

What I had in my post was part of the work needed to find the roots. You can start from where I left off to find these roots.
 
  • #7
multiplying out gave me

λ3 + 2λ2 -19λ -12

which factorised (using grouping) to:

(λ+2)(λ2-19)


Am I doing it right?

Thanks again.
 
  • #8
33,981
5,639
multiplying out gave me

λ3 + 2λ2 -19λ -12

which factorised (using grouping) to:

(λ+2)(λ2-19)


Am I doing it right?

Thanks again.
No. Show me how you got this: λ3 + 2λ2 -19λ -12

The 3 roots are positive integers.
 
  • #9
I multiplied out the first brackets to get:

(3-λ)(5-5λ-λ2)+(-3) + λ

=

15-15λ-3λ2-5λ+5λ23-3+λ

=

λ3+2λ2-19λ-12...

Thanks
 
  • #10
33,981
5,639
I multiplied out the first brackets to get:

(3-λ)(5-5λ-λ2)+(-3) + λ
That's still not quite right.
Starting from here -- (3-λ)[(2 - λ)(3 - λ) - 1] + 1(λ - 3 - 0) --
you should have (3-λ)[5 -5λ + λ2] + λ - 3.

Instead of multiplying things out, it's quicker to notice that 3 - λ is a factor in the first product and the second expression. This gives you
(3-λ)[5 -5λ + λ2 - 1]
Can you work it out from here?
=

15-15λ-3λ2-5λ+5λ23-3+λ

=

λ3+2λ2-19λ-12...

Thanks
 
  • #11
33,981
5,639
I multiplied out the first brackets to get:

(3-λ)(5-5λ-λ2)+(-3) + λ
There's a sign error in the line above. You didn't multiply (2 - λ)(3 - λ) correctly.
=

15-15λ-3λ2-5λ+5λ23-3+λ

=

λ3+2λ2-19λ-12...

Thanks
 
Last edited:

Related Threads on Using eigenvalues and eigenvectors to solve system of ODEs

Replies
1
Views
5K
Replies
0
Views
2K
Replies
2
Views
5K
Replies
8
Views
18K
Replies
4
Views
2K
Replies
6
Views
4K
Replies
0
Views
2K
Replies
1
Views
2K
Top