# Multiple eigenvalue solutions

## Homework Statement

Solve the system.

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

## 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.

HallsofIvy
Homework Helper

## Homework Statement

Solve the system.

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

## 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.