Systems of ODE's double-zero eigenvalues

Click For Summary
SUMMARY

The discussion focuses on solving systems of ordinary differential equations (ODEs) with double-zero eigenvalues using the variation of parameters method. The participant successfully identified the first eigenvalue solution as a constant multiplied by c1, represented by the matrix ([1]) * c1. To find the second solution corresponding to the double-zero eigenvalue, it is necessary to introduce a function of t, such as t, to ensure linear independence of the solutions. The linear dependence of the column vectors is also highlighted, indicating that x1' equals x2' for all t.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors in linear algebra
  • Familiarity with the variation of parameters method for solving ODEs
  • Knowledge of linear dependence and independence of vectors
  • Basic concepts of ordinary differential equations (ODEs)
NEXT STEPS
  • Study the variation of parameters method in detail for ODEs
  • Learn about the implications of double-zero eigenvalues in system stability
  • Explore techniques for finding linearly independent solutions in ODE systems
  • Investigate the role of the Wronskian in determining linear independence of solutions
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations, linear algebra, and systems analysis. This discussion is beneficial for anyone tackling problems involving eigenvalues in ODEs.

gabriels-horn
Messages
92
Reaction score
0

Homework Statement


Image.jpg
I put a triangle around the problem of interest.


Homework Equations





The Attempt at a Solution


I solved for the eigenvalues, resulting in double-zero values. My question is, using the variation of parameters method, which is what (14) refers to in the question. How do I solve for the second zero. For the first I get the matrix

([1]) *c1 for the general solution.
([1])

Do I need to add a guess such as t to the next zero eigenvalue?
 
Physics news on Phys.org
yeah so as you say, a constant value is your 1st general solution, i would try multplying by t and whether you can find a 2nd...

note also, that the column vectors of the matrix are the linearly dependent, and that x1' = x2' for all t
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K