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 $y^{2}(0)$ is equal to ____.

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Differential Equations

y = y (x)

is the solution of the differential equation,

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 ____.

Answer: 4

Let

t = sin^{- 1} (\frac{x}{2})

, so

\frac{d t}{d x} = \frac{1}{4 - x ^{2}}

. Dividing the given equation by

4 - x^{2}

and switching to

t

as the independent variable (using

\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x}

\frac{d y}{d t} = t (t^{2} - y) = t^{3} - t y ⟹ \frac{d y}{d t} + t y = t^{3}

The integrating factor is

e^{\int t d t} = e^{t^{2} /2}

. Multiplying through:

\frac{d}{d t} (y e^{t^{2} /2}) = t^{3} e^{t^{2} /2}

With

s = t^{2} /2

(so

d s = t d t

t^{2} = 2 s

\int t^{3} e^{t^{2} /2} d t = \int 2 s e^{s} d s = 2 e^{s} (s - 1) = e^{t^{2} /2} (t^{2} - 2)

y e^{t^{2} /2} = e^{t^{2} /2} (t^{2} - 2) + C ⟹ y = t^{2} - 2 + C e^{- t^{2} /2}

x = 2

t = sin^{- 1} (1) = π /2

, and

y (2) = \frac{π ^{2} - 8}{4} = \frac{π ^{2}}{4} - 2

. Substituting:

\frac{π ^{2}}{4} - 2 + C e^{- π^{2} /8} = \frac{π ^{2}}{4} - 2 ⟹ C = 0

y = t^{2} - 2

. At

x = 0

t = sin^{- 1} (0) = 0

, giving

y (0) = - 2

y^{2} (0) = 4

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Integer answer