Mathematics Question from JEE Main 2025 · 28 January, Shift 2

If $y = y(x)$ is the solution of the differential equation,

$$\sqrt{4 - x^2} \frac{dy}{dx} = \left( \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - y \right) \sin^{-1} \left( \frac{x}{2} \right)$$

where $-2 \leq x \leq 2$, and $y(2) = \frac{\pi^2 - 8}{4}$, then $y^2(0)$ is equal to: