Question: Use geometric arguments for the model of a negative feedback system,$$\frac{dx}{dt}=\frac{A\theta^2}{\theta^2+x^2}-\gamma x$$where $A, \theta$ and $\gamma$ are positive constants.
$(a)$ How many steady states are there in this system, and what is the stability of these states?
$(b)$ starting from an initial condition of $x=100$ what happens in the limit $t\rightarrow \infty$?
For $(a)$, I try
$$ \begin{align} \frac{A\theta^2}{\theta^2+x^2}-\gamma x&=0\\ \gamma x^3+\gamma \theta^2 x - A\theta^2&=0\\ x^3+\theta^2 x - \frac{A\theta^2}{\gamma}&=0 \end{align} $$
To solve this depressed cubic using closed formula,
$$x=\left\{ \frac{A\theta^2}{2}+\sqrt{\frac{A^2\theta^4}{4}+\frac{\gamma^3\theta^6}{27}} \right\}^{\frac13}+\left\{ \frac{A\theta^2}{2}-\sqrt{\frac{A^2\theta^4}{4}+\frac{\gamma^3\theta^6}{27}} \right\}^{\frac13}$$ With the other two roots found with the cube roots of unity.
It seems very messy to use these steady states to determine the stability of these states. I was wondering what they mean by geometric arguments here. And for $(b)$ I couldn't come up any technique which can help to solve the DE. Any solution or hint will be appreciated. TIA
To simplify the notation, let's rewrite the ODE as $\frac{dx}{dt}=f(x)-g(x)$, where $f(x):=\frac{A\theta^2}{\theta^2+x^2}$ and $g(x):=\gamma x$.
(a) The question asks how many steady states there are, not to explicitly find them. Now, since $f(x)$ is a positive bell-shaped curve and $g(x)$ is a line with positive slope (see figure), there is only one real solution to the equation $f(x)-g(x)=0$. Let's call this solution $x_*$.
If $x>x_*$, $g(x)>f(x)$, hence $\frac{dx}{dt}<0$ and $x(t)$ decreases with $t$, approaching $x_*$. Similarly, if $x<x_*$, $f(x)>g(x)$, so $\frac{dx}{dt}>0$ and $x(t)$ increases with $t$, again approaching $x_*$. Therefore, $x_*$ is a stable fixed point.
(b) According to the answer to (a), starting from any initial condition (in particular $x=100$), the system will tend to the fixed point $x_*$ as $t\to\infty$.