I have two ordinary differential equations equations:
$$ \dot{x}=1+x^{2}y-(1+A)x $$
$$ \dot{y}=Ax-yx^{2} $$
I need to find the single equilibrium point in terms of $A$. So set $\dot{x}$ and $\dot{y}$ to zero.
I have rearranged Equation 2 to $Ax=yx^{2}$ and then substituted this into Equation 1. This allows me to solve for $x$, giving $x=1$.
I then solve for $y$ and get the point $(1,A)$.
Is this the correct method?
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$\begingroup$We have:
$$ \dot{x}=1+x^{2}y-(1+A)x $$
$$ \dot{y}=Ax-yx^{2} $$
So, to find the equilibrium points, we need to simultaneously find $x' = y' = 0$.
We have:
$\dot{x}=1+x^{2}y-(1+A)x = 0$ and $\dot{y}=Ax-yx^{2} = 0$.
From the second equation, we get that $x^2 y = Ax$ and substitute that back in to the first equation, yielding:
$\dot{x}=1+x^{2}y-(1+A)x = 1 + Ax -(1+A) x = 1 + Ax -Ax -x = 0$, so
$1 - x = 0$, yielding $x = 1$.
So, we have a fixed point at $(1, A)$.
Yes, your method is correct.
If you wanted to analyze the behaviors, you would use this fixed point and see if it provides information on the behavior with the Jacobian evaluated at it (if these are not borderline points).
Here are the direction fields and phase portraits for $A = -1$, $A = 0$ and $A=1$.
Regards
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