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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: $C$

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Equilibrium points for any curve y=f(x)

points where $\frac{\mathrm{d}y}{\mathrm{d} x}=0$ (which is a function of y) are called Equilibrium points.

// y'=f(y)=0

Now Equilibrium points can be of two types

1) Stable :- from stable points if we move little bit right or left , in both cases we will come closer to equilibrium. in more simple words you can think like some ⋃ kind bowl , in which you places some glass pebbles , then you can think they will always reach to equilibrium so thins is stable point of equilibrium.

Mathematically stable points are those where $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be -ve if you move bit right from point of equilibrium and $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be +ve if you move bith left from the point of equilibrium.

2) Unstable :-From unstable points if we move lil bit right or left , we're going away from equilibrium in both cases.Now think any bowl of shape ⋂ , place any glass pebble , it will always move away from equilibrium so this is unstable point fo equilibrium.Mathematically unstable points are those where $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be +ve if you move bit right from point of equilibrium and $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be -ve if you move bith left from the point of equilibrium.

Now from this theory we can easily see x=2 is an unstable point of equilibrium while x=1 and x=3 are both stable points of equilibrium.

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