Solving second order linear homogeneous differential equation! HELP?

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  • #1
Solving second order linear homogeneous differential equation! HELP!?

Solve the second order linear homogeneous differential equation with constant coefficients by reqriting as a system of two first order linear differential equations. Show that the coefficient matrix is not similiar to the diagonal matrix, but is similiar to a Jordan matrix, J. Determine the matrix P so that A = PJP^-1. y'' + 2y' + y = 0

I'm not sure how to go on about solving this question. Can someone help me get to the answer?
 

Answers and Replies

  • #2
hunt_mat
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You write y'=z, so y''=z' and you have the system:

z'+2z+y=0
y'=z

You write a matrix W=(y,z)^T and then the system is W=AW for some A which you have to find.
 
  • #3


You write a matrix W=(y,z)^T and then the system is W=AW for some A which you have to find.

Do you think you can explain more as to why I need to find the matrix W = (y,z)^T? And also when finding A for system W = AW, how is it similiar to a Jordan matrix. Maybe I'm not understanding the Jordan matrix...
 
  • #4
hunt_mat
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So the System you have is:
[tex]
\left(
\begin{array}{c}
z' \\
y'
\end{array}\right) =\left(
\begin{array}{cc}
2 & 1 \\
1 & 0
\end{array}\right)\left(
\begin{array}{c}
z \\
y
\end{array}\right)
[/tex]
That is your system is matrix form. Now I think the idea is to diagonalise this by computing the eigenvectors and eigenvalues.
 
Last edited:
  • #5
hunt_mat
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Not too sure what went wrong with my tex...
 
  • #6


Yeah, I can't really tell what you put there ahahah
 
  • #7
hunt_mat
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(z)' = (2 1)(z)
(y)' (1 0)(y)
 
  • #8


So, when after finding the eigenvalues and eigenvectors, do I form a matrix out of the eigenvectors? Would that be the answer? How is that a Jordan matrix?
 
  • #9
hunt_mat
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The matrix of eigenvectors, P can be used to solve the system. i don't know why they are referring to a jordan matrix, you can swap the rows around and that will be a jordan matrix.
 
  • #10
vela
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Not too sure what went wrong with my tex...
You have a typo in the second \begin{array}. Instead of a closing brace, you used a parenthesis.
 
  • #11
hunt_mat
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Cheers
 
  • #12
vela
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If you try to solve for eigenvectors the regular way, you'll find you can only find one independent vector, so you can't diagonalize the matrix. You can get to Jordan normal form, however, using generalized eigenvectors.
 
  • #13


If you try to solve for eigenvectors the regular way, you'll find you can only find one independent vector, so you can't diagonalize the matrix. You can get to Jordan normal form, however, using generalized eigenvectors.

So what is the difference between solving for eigenvectors the regular way and the generalized way? Also what is Jordan normal form? Thanks a lot for your help!
 
  • #14
vela
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The generalized way works for this kind of matrix. :smile: It allows you do deal with certain situations that arise when you have repeated eigenvalues.

The first thing you need to do when you approach a problem is know what it's talking about. You presumably have a textbook. Don't let all that money you spent on it go to waste! Look up what Jordan normal form is.
 
  • #15


If you try to solve for eigenvectors the regular way, you'll find you can only find one independent vector, so you can't diagonalize the matrix. You can get to Jordan normal form, however, using generalized eigenvectors.

So, after getting the generalized eigenvectors, how do i get it to Jordan normal form? My book doesn't explain this part. Is there some steps I need to follow?
 
  • #16


So I got the Jordan form to be :
[ -1 1 ]
[ 0 -1 ]

and A is:
[ -2 -1 ]
[ 1 0 ]

But I need to find a matrix P so that A = PJP^-1. Any ideas on how to find P?
 
  • #17
vela
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When you diagonalize a matrix, you form the matrix by using the eigenvectors as its columns. To get Jordan form, you do the same thing except you use the generalized eigenvectors.
 
  • #18
HallsofIvy
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In order to find the eigenvalues, you had to solve the characteristic equation x^2+ 2x+ 1= (x+ 1)^2= 0. Now every matrix satisfies its own characteristice equation. That is, (A^2+ 2A+ I)v= (A+ I)v= 0 for every vector v. You have already determined that any eigenvector corresponding to eigenvalue -1 is a multiple of < 1, -1>, a one dimensional subspace. But that means that there exist other vectors, v, such that Av is not equal to -v, (A+ I)v is not 0, but we still must have (A+ I)^2v= 0. That is the same as saying that (A+ I)[(A+I)v]= 0 which means that (A+ I)v must be an eigenvalue.

A "generalized eigenvector" is such a vector- a vector v such that Av is not equal to -v but such that (A+ I)v is an eigenvcctor.

Now, A+ I=
[-1 -1]
[1 1]
so you are looking for a vector, <x, y>, such that x- y= 1 and x+ y= -1.

Use <1, -1> as the first column of P and the "generalized eigenvector" as the second column.

I hope you don't mind my asking- if you really have never seen "Jordan form" etc. before, why have you been assigned such a problem?
 

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