Explanation:

Let’s break it down step by step:

- We start from the curve: y² = 2cx + 2cvc.

- c is a parameter. To get its differential equation, differentiate w.r.t. x, and eliminate c.

- Differentiating both sides: 2y(dy/dx) = 2c.

- So, c = y(dy/dx).

- Substitute c back: y² = 2(y dy/dx)x + 2(y dy/dx)v(y dy/dx).

- Rearranging gives: y² = 2x y(dy/dx) + 2y(dy/dx)v(y dy/dx).

- Notice that the highest order derivative (just dy/dx, which is first-order) is raised to power 3/2 because of the square root and multiplied by itself.

- When you clear the radicals and express in polynomial form, the highest power to which dy/dx is raised is 3.

Now, for the options:

- Option 1: 2 – that would be if the highest power was 2

- Option 2: 3 – this matches what we found

- Option 3: 4 – nope, not this high

- Option 4: Degree does not exist – that only happens if the equation has non-polynomial forms (e.g., powers which are not integer and can’t be cleared), but here it can be expressed as a polynomial in dy/dx.

So, the correct answer is Option 2: 3