# Find the equilibrium solution and eigenvalues and eigenvectors of system?

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

Last edited:

tiny-tim
Homework Helper
Hi Norm!!! Show us what you've tried, and where you're stuck, and then we'll know how to help! 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
Homework Helper
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.
$x_1'= -(x_1+ 4)-4(x_2- 2)- 4= -x_1- 4x_2$
$x_2'= (x_1+ 4)- (x_2- 2)- 6= x_1- x_2$
$$\begin{bmatrix}-1 & 4 \\ 1 & -1\end{bmatrix}$$