Repeated eigenvalues+ differential equation

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Homework Help Overview

The discussion revolves around solving a system of differential equations involving repeated eigenvalues. The original poster presents a specific problem with initial conditions and attempts to find the solutions for x(t) and y(t).

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to find eigenvalues and eigenvectors, and applies a method for repeated eigenvalues. Some participants question the calculations and the setup of the equations, particularly regarding the formulation of the matrix equation.

Discussion Status

There is an ongoing exploration of the calculations involved in finding the matrix P and the eigenvalues. Some guidance has been offered regarding potential errors in the original poster's approach, and a correction has been acknowledged.

Contextual Notes

The discussion includes references to specific calculations and methods, as well as a link to an external resource for further clarification on repeated eigenvalues.

jeffy
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Homework Statement



dx/dt= -4x -y

dy/dt= x-2y

x(0)=4 y(0)=1

x(t)=?
y(t)=?

Homework Equations




The Attempt at a Solution



1) find eigenvalues
(x+4)(X+2)+1
X=-3,-3

2)eigenvectors:
(-3-A)(x,y)=(0,0)

eignvector=(-1,1)

3)using the P from this page http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx

i found P to be (-1,0)

4) so i plugged the eigenvector and the P into the solution for repeated eigenvalues in the link above and got C1=1 and C2=-5

5) plugging those values in i got x=4e^(-3t)+(5t)e^(-3t) and y=e^(-3t)-(5t)e^-3t


however when i entered this into the program(igot all the syntax right) my answer is wrong and I've tried it many times
 
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Recheck your calculation of ρ. I think you accidentally flipped a sign.
 
Thanks for the reply
This is how i calculated P

(lambda-AI)=(-1,1)

1 1 = -1
-1 -1 = 1

addded (1) to (2) to get (2)'

1 1 = -1
0 0 = 0
therefore P1+P2=-1
P1= -P2-1

P=[ -1-P2,P2]

setting P2 to zero i get

(-1,0)
 
The equation you want to solve is

(A-\lambda I) \vec{\rho} = \vec{\eta}

not

(\lambda I-A) \vec{\rho} = \vec{\eta}
 
Its works Thanks for your help
 

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