How to Solve a System with Repeated Eigenvalues?

[jrsweet]

Homework Statement

Solve the system.

dx/dt=[1 -4; 4 -7]*x with x(0)=[3; 2]

Homework Equations

The Attempt at a Solution

I am apparently not getting this at all. Can someone walk me through it? I konw I have to first find the eigenvalues and eigenvectors:

(1-λ)(-7-λ)+16=0
λ²+6λ+9=0
λ=-3,-3

So, (A-3I)C1 = 0
(A-3I)= [4 -4; 4 -4]

So, eigenvector = [1; 1]

(A-3I)C2=C1

Eigenvector = [1; 0]And, x1= [1; 1] e^-3t
x2 = ([1; 1]t + [1; 0])e^-3t

So, using fundamental matrices...

F = [ e^-3t (t+3) e^-3t; e^-3t t e^-3t]
F(0) = [1 3; 1 0]
(F(0))'= [0 1; 1/3 -1/3]

So,
x(t)=F*(F(0))'*X₀ = [X₁ ; X₂]

Is there anything wrong with my method?
The homework asks for two answers: X₁ and X₂ and I'm not exactly sure what that is asking for. Thanks! Any help is appreciated.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[lanedance]

hopefully the follwing will help, with a have a more detialed look when i get a chance
http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx

 

[HallsofIvy]

Homework Statement

Solve the system.

dx/dt=[1 -4; 4 -7]*x with x(0)=[3; 2]

Homework Equations

The Attempt at a Solution

I am apparently not getting this at all. Can someone walk me through it? I konw I have to first find the eigenvalues and eigenvectors:

(1-λ)(-7-λ)+16=0
λ²+6λ+9=0
λ=-3,-3

So, (A-3I)C1 = 0
(A-3I)= [4 -4; 4 -4]

So, eigenvector = [1; 1]

(A-3I)C2=C1

Eigenvector = [1; 0]


And, x1= [1; 1] e^-3t
x2 = ([1; 1]t + [1; 0])e^-3t

So, using fundamental matrices...

F = [ e^-3t (t+3) e^-3t; e^-3t t e^-3t]
F(0) = [1 3; 1 0]
(F(0))'= [0 1; 1/3 -1/3]

So,
x(t)=F*(F(0))'*X₀ = [X₁ ; X₂]

Is there anything wrong with my method?
The homework asks for two answers: X₁ and X₂ and I'm not exactly sure what that is asking for. Thanks! Any help is appreciated.

The initial value problem has, of course, a single solution. Perhaps the "X₁" and "X₂" are the two independent solutions to the equation without the initial values.