If y = y(x) is the solution of the differential equation , , then is equal to _________. [2025]

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Solution

Q.
If y = y(x) is the solution of the differential equation $\sqrt{4 - x^{2}} \frac{d y}{d x} = ((\sin^{- 1} {(\frac{x}{2}))}^{2} - y) \sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2, y (2) = \frac{π^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to _________. [2025]

Ans.
(4)

From the given D.E.

$\frac{d y}{d x} + \frac{(\sin^{- 1} \frac{x}{2})}{\sqrt{4 - x^{2}}} y = \frac{{(\sin^{- 1} \frac{x}{2})}^{3}}{\sqrt{4 - x^{2}}}$

$I . F . = e^{\int \frac{\sin^{- 1} \frac{x}{2}}{\sqrt{4 - x^{2}}} d x} = e^{\frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}}$

$∴ y e^{\frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}} = \int \frac{{(\sin^{- 1} \frac{x}{2})}^{3}}{\sqrt{4 - x^{2}}} e^{\frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}} d x$

$\Rightarrow y = {(\sin^{- 1} \frac{x}{2})}^{2} - 2 + c e^{- \frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}}$

Now, $y (2) = \frac{π^{2}}{4} - 2 + c e^{- \frac{π^{2}}{8}}$

On comparing with $y (2) = \frac{π^{2} - 8}{4}$ , we get c = 0

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

Hence, $y (0) = - 2 \Rightarrow y^{2} (0) = 4$

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