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

The given differential equation is:

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

with initial condition $$y(2) = \frac{\pi^2 - 8}{4}$$.

To solve, substitute $$t = \sin^{-1}\left(\frac{x}{2}\right)$$. Then, $$\frac{dt}{dx} = \frac{1}{\sqrt{4 - x^2}}$$, so $$\frac{dx}{\sqrt{4 - x^2}} = dt$$.

Rewriting the differential equation:

$$\sqrt{4 - x^2} \frac{dy}{dx} = t(t^2 - y)$$

Dividing both sides by $$\sqrt{4 - x^2}$$:

$$\frac{dy}{dx} = \frac{t(t^2 - y)}{\sqrt{4 - x^2}}$$

Multiplying by $$dx$$:

$$dy = t(t^2 - y) \cdot \frac{dx}{\sqrt{4 - x^2}} = t(t^2 - y) dt$$

Thus, the equation becomes:

$$\frac{dy}{dt} = t^3 - t y$$

Rearranging into standard linear form:

$$\frac{dy}{dt} + t y = t^3$$

Here, $$P(t) = t$$ and $$Q(t) = t^3$$. The integrating factor is:

$$IF = e^{\int P(t) dt} = e^{\int t dt} = e^{\frac{t^2}{2}}$$

Multiplying both sides by the integrating factor:

$$e^{\frac{t^2}{2}} \frac{dy}{dt} + t e^{\frac{t^2}{2}} y = t^3 e^{\frac{t^2}{2}}$$

The left side is the derivative of $$y e^{\frac{t^2}{2}}$$:

$$\frac{d}{dt} \left( y e^{\frac{t^2}{2}} \right) = t^3 e^{\frac{t^2}{2}}$$

Integrating both sides with respect to $$t$$:

$$y e^{\frac{t^2}{2}} = \int t^3 e^{\frac{t^2}{2}} dt + C$$

To solve the integral, set $$w = \frac{t^2}{2}$$, so $$dw = t dt$$ and $$t^2 = 2w$$. Then:

$$\int t^3 e^{\frac{t^2}{2}} dt = \int t^2 \cdot t e^{\frac{t^2}{2}} dt = \int 2w e^w dw$$

Using integration by parts with $$u = w$$, $$dv = 2e^w dw$$, so $$du = dw$$, $$v = 2e^w$$:

$$\int 2w e^w dw = 2w e^w - \int 2e^w dw = 2w e^w - 2e^w + C_1 = 2e^w (w - 1) + C_1$$

Substituting back $$w = \frac{t^2}{2}$$:

$$2e^{\frac{t^2}{2}} \left( \frac{t^2}{2} - 1 \right) + C_1 = e^{\frac{t^2}{2}} (t^2 - 2) + C_1$$

Thus:

$$y e^{\frac{t^2}{2}} = e^{\frac{t^2}{2}} (t^2 - 2) + C$$

Solving for $$y$$:

$$y = t^2 - 2 + C e^{-\frac{t^2}{2}}$$

Replacing $$t = \sin^{-1}\left(\frac{x}{2}\right)$$:

$$y(x) = \left( \sin^{-1}\left(\frac{x}{2}\right) \right)^2 - 2 + C e^{-\frac{1}{2} \left( \sin^{-1}\left(\frac{x}{2}\right) \right)^2 }$$

Applying the initial condition $$y(2) = \frac{\pi^2 - 8}{4}$$. When $$x = 2$$, $$t = \sin^{-1}(1) = \frac{\pi}{2}$$:

$$y(2) = \left( \frac{\pi}{2} \right)^2 - 2 + C e^{-\frac{1}{2} \left( \frac{\pi}{2} \right)^2 } = \frac{\pi^2}{4} - 2 + C e^{-\frac{\pi^2}{8}}$$

Set equal to the given value:

$$\frac{\pi^2}{4} - 2 + C e^{-\frac{\pi^2}{8}} = \frac{\pi^2 - 8}{4} = \frac{\pi^2}{4} - 2$$

Thus:

$$C e^{-\frac{\pi^2}{8}} = 0$$

Since $$e^{-\frac{\pi^2}{8}} \neq 0$$, it follows that $$C = 0$$. Therefore:

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

Now, evaluate at $$x = 0$$:

$$y(0) = \left( \sin^{-1}\left(\frac{0}{2}\right) \right)^2 - 2 = (\sin^{-1}(0))^2 - 2 = 0^2 - 2 = -2$$

Then:

$$y^2(0) = (-2)^2 = 4$$

The value of $$y^2(0)$$ is 4.