Question

Solve. You should show all work to find eigenvalues and eigenvectors. What sort of equilibrium do we have here and where does it occur? Does the type of equilibrium surprise you? x' =-y 6. x(0) = 0, y(0) = 2 y' = 10x – 2y

Answer 1

The above linear system of O.D.E.s can be expressed as
  $\dot{z} = \begin{bmatrix} 0 & -1\\ 10 & 2 \end{bmatrix} z$
where, $z = \begin{bmatrix} x\\y \end{bmatrix}$   


Let us call the associated matrix to be A.

Observe that Az = 0 implies that -y = 0 and 10 - 2y = 0. These two equations give (x,y) = (0,0)
Thus, the only critical point(or equillibrium point) of the system is (0,0).

Now, the characteristic polynomial of A is det(xI - A) = x(x-2) + 10 = x² - 2x + 10.
Thus, the eigenvalues of A are the roots of this polynomial and hence, the eigen values of A are 1+3i and 1-3i.

Since both the eigenvalues of A have non-zero real part, hence the equillibrium point (0,0) is a HYPERBOLIC equillibruim point.
Further, since both the eigenvalues of A have positive real part, hence the hyperbolic equillibrium point (0,0) is a SOURCE.