Solving second order linear homogeneous differential equation HELP?

In summary: Is this for a college course in linear algebra? In that case, your book should have a good discussion of Jordan form and how to find it. If this is for a high school course- sorry- I thought this was a college course.In summary, solving a second order linear homogeneous differential equation with constant coefficients involves rewriting it as a system of two first order linear differential equations and finding the coefficient matrix. If the matrix is not similar to a diagonal matrix, it can be similar to a Jordan matrix. In order to find the matrix P so that A = PJP^-1, generalized eigenvectors can be used to form the Jordan normal form.
  • #1
davidson89
8
0
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 similar to the diagonal matrix, but is similar 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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  • #2


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


hunt_mat said:
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 similar to a Jordan matrix. Maybe I'm not understanding the Jordan matrix...
 
  • #4


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


Not too sure what went wrong with my tex...
 
  • #6


Yeah, I can't really tell what you put there ahahah
 
  • #7


(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


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


hunt_mat said:
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


Cheers
 
  • #12


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


vela said:
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


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


vela said:
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


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


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?
 

1. What is a second order linear homogeneous differential equation?

A second order linear homogeneous differential equation is a mathematical equation that involves the second derivative of an unknown function, along with the function itself and its first derivative. The equation is considered "homogeneous" because all the terms in the equation involve the unknown function and its derivatives, without any constant terms.

2. How is a second order linear homogeneous differential equation solved?

To solve a second order linear homogeneous differential equation, we first need to find the roots of the characteristic equation associated with the equation. Then, based on the nature of the roots (real, complex, or repeated), we can determine the form of the solution and use initial conditions to find the specific values of the constants in the solution.

3. What are the main methods for solving second order linear homogeneous differential equations?

The main methods for solving second order linear homogeneous differential equations are the method of undetermined coefficients, the method of variation of parameters, and the method of reduction of order. Each method has its own advantages and is suitable for different types of equations.

4. What are some real-world applications of second order linear homogeneous differential equations?

Second order linear homogeneous differential equations are commonly used in physics, engineering, and other scientific fields to model and analyze various natural processes. Some examples of real-world applications include modeling the motion of a spring, the oscillations of a pendulum, and the growth of a population.

5. Are there any tools or software that can help with solving second order linear homogeneous differential equations?

Yes, there are various tools and software available that can help with solving second order linear homogeneous differential equations. Some popular options include WolframAlpha, MATLAB, and Maple. These tools use advanced algorithms to quickly and accurately solve differential equations and provide a visual representation of the solution.

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