Find the equilibrium solution and eigenvalues and eigenvectors of system?

[Norm850]

Hey guys, I need to find the equilibrium solution (critical point) for the given system. Also I need to take the homogeneous equation x' = Ax (matrix notation) and find the eigenvalues and eigenvectors.

system: x' = -x - 4y - 4
y' = x - y - 6


Can you help?

Thanks

 

[tiny-tim]

Hi Norm!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

(start with the eigenvalues and eigenvectors)

 

[Norm850]

Okay so, the critical points are when x' and y' equal zero I believe, so adding the two equations gets -5y - 10 = 0 => y = -2, x = 4. So the critical point is (x,y) = (4,-2).

Now for writing the homogeneous equation in matrix form, by using change of variables:

x_1 = x
x_2 = y
x_2 = y'

Gives equations:

(x_1)' = -x_1 - 4x_2 - 4
(x_2)' = x_1 - x_2 - 6

And that gives the matrix form x' = Ax + b, which would be (follow link)
http://i39.tinypic.com/35858ye.png

but we want the homogeneous, so we have x' = Ax, which would be (follow link)
http://i41.tinypic.com/53r0hi.png

Now to find eigenvalues, det(A - λI) = λ^2 + 2λ + 5.

This is where I'm confused so far. I can do the quadratic but we haven't had to do quadratic equation yet so I want to make sure i have everything correct so far?

Thanks

 

[tiny-tim]

Hi Norm850!

(just got up :zzz:)

… I want to make sure i have everything correct so far?

Fine so far.

 

[HallsofIvy]

You will get a homogeneous set of equations if you let $x_1= x- 4$ and $x_2= y+ 2$. Of course, $x_1'= x'$ and $x_2'= y'$ and $x= x_1+ 4$, $y= x_2- 2$. Putting those into the equation.

$x_1'= -(x_1+ 4)-4(x_2- 2)- 4= -x_1- 4x_2$
$x_2'= (x_1+ 4)- (x_2- 2)- 6= x_1- x_2$

Determine the eigenvalues of
$$\begin{bmatrix}-1 & 4 \\ 1 & -1\end{bmatrix}$$