Here, $dx/dt$ is neither constant nor function of variable $t.$ So, on solving for $x,$ we get exponential decay function something similar to radioactive decay $dN/dt = -KN ;K>0.$
For any equilibrium point, we have $dx/dt = 0 \implies x = 1,2,3.$
Now if we consider sign scheme of $dx/dt$ for different ranges of $x,$ we have :
- $dx/dt > 0$ for $x < 1;$
- $dx/dt < 0$ for $1 < x < 2;$
- $dx/dt > 0$ for $2 < x < 3;$
- $dx/dt < 0$ for $x > 3;$
For $x \to 2, dx/dt$ changes from negative to positive. So, $x = 2$ is a point of unstable equilibrium.
Since, here $dx/dt$ is a function of $x$ not $t,$ so condition for checking stable / unstable points (aka local minima / local maxima) is opposite to what we apply in case of $dy/dx = f(x).$ Here we have $dy/dx = f(y)$ which is actually exponential function in $y$ and $x.$
We can understand equilibrium in terms of radioactive decay.
Let $dN/dt = -KN ;K>0$ its significance is that an element is loosing energy so it is getting stability because we know more energy an element gets,more de-stability it gains and vice versa.
Correct Answer: $C$