dwsmith said:
Since I no longer own a calculus book anymore, I have a simple question.
\begin{align}
\dot{x} =& -x + ay + xy^2\notag\\
\dot{y} =& b - ay - x^2y\notag
\end{align}
How can parameterize those equations for a and b?
So I found an easy example which I understand but applying it to this problem is tough.
The example is
\begin{align}
\dot{u} &= a - u + u^2v = f(u,v)\\
\dot{v} &= b - u^2v = g(u,v)
\end{align}
\begin{align}
f(u_0,v_0) &= a - u + u^2v = 0\\
g(u_0,v_0) &= b - u^2v 0
\end{align}
So this is relatively easy to find $u_0 = b+a$ and $v_0 = \dfrac{b}{(a+b)^2}$.
Then using the Jacobian, we get $\text{tr}(A) = f_u+g_v = \dfrac{b-a}{a+b}-(a+b)^2$ and $\text{det}(A)=f_ug_v-f_vg_u = (a+b)^2$.
Then
$\text{tr}(A) > 0, \ |A| > 0, \quad (\text{tr}(A))^2\begin{cases} >\\<\end{cases} \ 4|A|\Rightarrow\text{unstable}\begin{cases} \text{node} \\ \text{spiral}\end{cases}$.
The equations then are $a-b=(a+b)^3$ and $a+b=0$.
However, when I solved for my equation I obtained:
$$
y_0 = \frac{b}{a+x_0^2}
$$
which lead to $-x_0+\dfrac{ab}{a+x_0^2}+\dfrac{x_0^2b}{a+x_0^2}=0$.
This didn't lead to fruit solutions for $x_0$ and $y_0$.
I did I overlook something when I solved mine: a mistake or technique to avoid this?