JEE Main 2025 Jan 28 Shift 2, Mathematics Q25: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Jan 28, Shift 2Mathematics

If $y = y (x)$ 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 ____.

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Solution

Let $u = sin^{- 1} (x /2)$ , so $d u = \frac{d x}{4 - x ^{2}}$ . The equation becomes

$\frac{d y}{d u} = (u^{2} - y) u \Rightarrow \frac{d y}{d u} + u y = u^{3}$ .

This is linear with integrating factor $μ = e^{u^{2} /2}$ . Then

$\frac{d}{d u} (y e^{u^{2} /2}) = u^{3} e^{u^{2} /2}$ .

Let $v = u^{2} /2$ , so $u^{3} d u = 2 v d v$ and the RHS integrates to $\int 2 v e^{v} d v = 2 (v - 1) e^{v} = (u^{2} - 2) e^{u^{2} /2}$ .

Hence $y e^{u^{2} /2} = (u^{2} - 2) e^{u^{2} /2} + C$ , i.e. $y = u^{2} - 2 + C e^{- u^{2} /2}$ .

At $x = 2$ : $u = π /2$ and $y = \frac{π ^{2}}{4} - 2 = \frac{π ^{2} - 8}{4}$ , which forces $C = 0$ .

So $y (x) = (sin^{- 1} (x /2))^{2} - 2$ . At $x = 0$ : $y (0) = 0 - 2 = - 2$ , and $y^{2} (0) = 4$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q25: Methods of Solving First Order, First Degree Differential Equation

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