Question 14 Marks
If $y=y(x)$ is the solution of the differential equation,$\sqrt{4-x^{2}} \frac{d y}{d x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^{2}-y\right) \sin ^{-1}\left(\frac{x}{2}\right)$,$-2 \leq x \leq 2, y(2)=\left(\frac{\pi^{2}-8}{4}\right)$, then $\mathrm{y}^{2}(0)$ is equal to _______________ .
Answer
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$\frac{d y}{d x}+\frac{\left(\sin ^{-1} \frac{x}{2}\right)}{\sqrt{4-x^{2}}} y-\frac{\left(\sin ^{-3} \frac{x}{2}\right)^{3}}{\sqrt{4-x^{2}}}$
$y e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}}=\int \frac{\left(\sin ^{-3} \frac{x}{2}\right)^{3}}{4-x^{2}} e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}} d x$
$y=\left(\sin ^{-1} \frac{x}{2}\right)^{2}-2+c . e^{\frac{-\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}}$
$y(2)=\frac{\pi^{2}}{4}-2 \Rightarrow c=0$
$y(0)=-2$
$\frac{d y}{d x}+\frac{\left(\sin ^{-1} \frac{x}{2}\right)}{\sqrt{4-x^{2}}} y-\frac{\left(\sin ^{-3} \frac{x}{2}\right)^{3}}{\sqrt{4-x^{2}}}$
$y e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}}=\int \frac{\left(\sin ^{-3} \frac{x}{2}\right)^{3}}{4-x^{2}} e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}} d x$
$y=\left(\sin ^{-1} \frac{x}{2}\right)^{2}-2+c . e^{\frac{-\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}}$
$y(2)=\frac{\pi^{2}}{4}-2 \Rightarrow c=0$
$y(0)=-2$
