If y = y(x) is the solution of the differenti | vedclass

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(A) The given differential equation is $\sqrt{4-x^2} \frac{dy}{dx} + y \sin^{-1}\left(\frac{x}{2}\right) = \left(\sin^{-1}\left(\frac{x}{2}\right)\rig

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$4$

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(A) The given differential equation is $\sqrt{4-x^2} \frac{dy}{dx} + y \sin^{-1}\left(\frac{x}{2}\right) = \left(\sin^{-1}\left(\frac{x}{2}\right)\right)^3$. Dividing by $\sqrt{4-x^2}$,we get $\frac{dy}{dx} + \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} y = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}}$ and $Q(x) = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} dx}$. Let $u = \sin^{-1}(x/2)$,then $du = \frac{1}{\sqrt{1-(x/2)^2}} \cdot \frac{1}{2} dx = \frac{1}{\sqrt{4-x^2}} dx$. So,$IF = e^{\int u du} = e^{u^2/2} = e^{\frac{(\sin^{-1}(x/2))^2}{2}}$. The solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$. $y e^{\frac{(\sin^{-1}(x/2))^2}{2}} = \int u^3 e^{u^2/2} du + C$. Let $t = u^2/2$,then $dt = u du$. The integral becomes $\int 2t e^t dt = 2(t e^t - e^t) + C = 2e^t(t-1) + C$. Substituting back,$y e^{u^2/2} = 2e^{u^2/2} (\frac{u^2}{2} - 1) + C = e^{u^2/2} (u^2 - 2) + C$. $y = u^2 - 2 + C e^{-u^2/2}$. Given $y(2) = \frac{\pi^2-8}{4}$,$u = \sin^{-1}(1) = \pi/2$. $\frac{\pi^2-8}{4} = \frac{\pi^2}{4} - 2 + C e^{-\pi^2/8} \Rightarrow C = 0$. Thus,$y = u^2 - 2 = (\sin^{-1}(x/2))^2 - 2$. At $x=0$,$y(0) = (\sin^{-1}(0))^2 - 2 = -2$. Therefore,$y^2(0) = (-2)^2 = 4$.

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

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