The potential energy of a conservative force field is given by
`U=ax^(2)-bx`
where, a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral.

Explanation

To solve the problem, we need to find the equilibrium position and analyze the stability of that position based on the given potential energy function \( U = ax^2 - bx \).

### Step 1: Find the Force
The force \( F \) associated with a potential energy \( U \) is given by:
\[ F = -\frac{dU}{dx} \]
First, we will differentiate the potential energy function \( U \) with respect to \( x \).

### Step 2: Differentiate the Potential Energy
Given:
\[ U = ax^2 - bx \]
Differentiating \( U \) with respect to \( x \):
\[ \frac{dU}{dx} = \frac{d}{dx}(ax^2 - bx) = 2ax - b \]

### Step 3: Set the Force to Zero for Equilibrium
For equilibrium, the force must be zero:
\[ F = -\frac{dU}{dx} = 0 \]
This implies:
\[ 2ax - b = 0 \]
Solving for \( x \):
\[ 2ax = b \implies x = \frac{b}{2a} \]

### Step 4: Determine the Stability of the Equilibrium
To analyze the stability of the equilibrium position, we need to find the second derivative of the potential energy function:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(2ax - b) = 2a \]

### Step 5: Analyze the Second Derivative
The sign of the second derivative tells us about the stability:
- If \( \frac{d^2U}{dx^2} > 0 \), the equilibrium is stable.
- If \( \frac{d^2U}{dx^2} < 0 \), the equilibrium is unstable.
- If \( \frac{d^2U}{dx^2} = 0 \), the equilibrium is neutral.

Since \( a \) is a positive constant, we have:
\[ \frac{d^2U}{dx^2} = 2a > 0 \]
This indicates that the equilibrium position \( x = \frac{b}{2a} \) is stable.

### Final Answer
The equilibrium position is \( x = \frac{b}{2a} \) and it is a stable equilibrium.

---