JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 23)
If $y=y(x)$ is the solution of the differential equation,
$\sqrt{4-x^2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^2-y\right) \sin ^{-1}\left(\frac{x}{2}\right),-2 \leq x \leq 2, y(2)=\frac{\pi^2-8}{4}$, then $y^2(0)$ is equal to ___________.
Answer
4
Explanation
$$\begin{aligned}
& \frac{d y}{d x}+\frac{\left(\sin ^{-1} \frac{x}{2}\right)}{\sqrt{4-x^2}} y=\frac{\left(\sin ^{-3} \frac{x}{2}\right)^3}{\sqrt{4-x^2}} \\
& y e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}}=\int \frac{\left(\sin ^{-3} \frac{x}{2}\right)^3}{4-x^2} e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}} d x \\
& y=\left(\sin ^{-1} \frac{x}{2}\right)^2-2+c \cdot e^{\frac{-\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}} \\
& y(2)=\frac{\pi^2}{4}-2 \Rightarrow c=0 \\
& y(0)=-2
\end{aligned}$$
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