So my question is to calculate the (range of) values for which x* (fixed point) = 0 and I want to investigate the type of stability.
$$ \frac{d}{du} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-5&a\\2&1\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}$$
In solving this question - I have calculated the determinant to be (-5 - 2a).
So then to show the values that satisfy the fixed point being a saddle:
The range of values for -5-2a < 0 $\Rightarrow$ -2.5 > a.
To show that the values that satisfy the fixed point being a node:
This occurs if the eigenvalues are real and negative, and I have calculated the eigenvalues to be:
λ = -2 $\pm$ $\sqrt{5+2a}$
So for the equilibrium point at the origin, then the eigenvalues must be real and negative so then:
$\sqrt{5+2a}$ > 0 $\Rightarrow$ a>= -2.5 where -2<= λ <0
And for the case of focus:
Then the equilibrium position at the origin is a focus when the eigenvalues are complex conjugate numbers with a positive real part so then $\sqrt{5+2a}$ < 0, so then a < -2.5
Can anyone verify if this correct, I am trying my hand at other questions, from where I begun to learn the theory and practice from this site https://www.math24.net/linear-autonomous-systems-equilibrium-points-page-2/
Thank you for any help.
I consider this to be correct - although I am unsure if your phrasing for node is right.