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JEE Mains · Maths · STD 12 - 9. differential equations

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

  1. A 4
  2. B 6
  3. C 8
  4. D 10
Verified Solution

Answer & Solution

Correct Answer

(A) 4

Step-by-step Solution

Detailed 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}…

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