## 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?
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 Recognitions: Homework Help 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.

 Quote by hunt_mat 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...

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

So the System you have is:
$$\left( \begin{array}{c} z' \\ y' \end{array}\right) =\left( \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array}\right)\left( \begin{array}{c} z \\ y \end{array}\right)$$
That is your system is matrix form. Now I think the idea is to diagonalise this by computing the eigenvectors and eigenvalues.
 Recognitions: Homework Help Not too sure what went wrong with my tex...
 Yeah, I can't really tell what you put there ahahah
 Recognitions: Homework Help (z)' = (2 1)(z) (y)' (1 0)(y)
 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?
 Recognitions: Homework Help 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.

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 Quote by hunt_mat 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.
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 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.

 Quote by vela 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!
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus The generalized way works for this kind of matrix. 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.

 Quote by vela 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?
 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?
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.

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