Verified Solution • 2025–26 Session moderate

QJEEM25Mab908: JEE Main 2025 Mathematics - Class 12 JEEM25Mab908 Solution: If is the solution of the differential equation, , then is…

If y=y(x)y=y(x) is the solution of the differential equation, 4−x2 dy dx=((sin⁡−1(x2))2−y)sin⁡−1(x2),−2≤x≤2,y(2)=π2−84\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 y2(0)y^2(0) is equal to ___________.

Step-by-Step Verified Solution

amp;dydx+(sin⁡−1x2)4−x2y=(sin⁡−3x2)34−x2amp;ye(sin⁡−1x2)22=∫(sin⁡−3x2)34−x2e(sin⁡−1x2)22dxamp;y=(sin⁡−1x2)2−2+c⋅e−(sin⁡−1x2)22amp;y(2)=π24−2⇒c=0amp;y(0)=−2\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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