linear systems and equilibrium For a one-parameter family of linear equations
$ \frac{dY}{dt} = \left( \begin{array}{cc}
1 & 3 \\
k &-2 \\
\end{array} \right)\ Y $
Where k is a parameter, determine the equilibrium at origin for all values of k.
Could someone tell me how to find the equilibrium at the origin? 
Thanks..
 A: An equilibrium is defined as the $Y^*$ such that $\left.\frac{dY}{dt}\right|_{Y=Y^*} = 0$. In your case, you have to solve the following:
$$A(k)Y^* = 0$$
where $A(k)$ is the given matrix.
It's clear that $Y^* = (0 ~~ 0)^T$ is always an equilibrium. Anyway, when $\det(A(k)) = 0$, then there are infinite equilibria.
$$\det(A(k)) = 0 \Rightarrow -2 - 3k = 0 \Rightarrow k = -\frac{2}{3}$$
So, when $k \neq -\frac{2}{3}$, the only equilibrium is $Y^* = (0 ~~ 0)^T$. But when $k = -\frac{2}{3}$, then there are infinite equilibria. In fact:
$$A(k)Y^* = 0 \Rightarrow \left\{\begin{array}{l}y_1^* + 3y_2^* = 0\\-\frac{2}{3}y_1^* - 2 y_2^* = 0\end{array}\right. \Rightarrow Y^* = \left(\begin{array}{c}-3h \\ h\end{array}\right) ~\forall h \in \mathbb{R}$$
If you are interested in the stability of the equilibrium $Y^* = (0 ~~ 0)^T$, then you have to evaluate eigenvalues of $A(k)$:
$$\det\left(\lambda I - A(k)\right) = \det\left(\begin{array}{cc}\lambda-1 & -3 \\ -k & \lambda + 2\end{array}\right) = (\lambda-1)(\lambda+2) - 3k = $$
$$=\lambda^2 + \lambda - (2 + 3k) = 0$$
Then:
$$\lambda_{1,2} = \frac{-1 \pm \sqrt{1 + 2 + 3k}}{2} = \frac{-1 \pm \sqrt{3(1 + k)}}{2}$$


*

*If $k < -1$, then $\lambda_{1,2}$ are complex conjugated, and their real part is $-\frac{1}{2}$. In this case we have stable oscillations.  $Y^*$ is asymptotically stable

*If $k = -1$, then $\lambda_1 = \lambda_2 = -\frac{1}{2}$. Oscillations disappears, but stability remains.

*For $-1 < k < -\frac{2}{3}$, both $\lambda_1$ and $\lambda_2$ are real and negative, then we have stability.

*When $k = -\frac{2}{3}$, an eigenvalue is $0$. Then we have only simple stability, not asymptotic. This is clear since we have infinite equilibria for $k = -\frac{2}{3}$, i.e. they are all indifferent.

*Finally, for $k > -\frac{2}{3}$, an eigenvalue is positive. Then we have only instability.

