Linear differential equationsHard

Solving Inverse Trigonometric Differential Equation

If y = y(x) is the solution of the differential equation, $\sqrt {4 - {x^2}} \frac{{dy}}{{dx}} = \left( {{{\left( {{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right)} \right)}^2} - y} \right){\sin ^{ - 1}}\left( {\frac{x}{2}} \right)$ , –2 ≤ x ≤ 2, $y(2) = \left( {\frac{{{\pi ^2} - 8}}{4}} \right)$ , then y²(0) is equal to ____.

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