If y=y(x) is the solution of the differential equation, 4-x^2 ~d y ~d x =(( ^ -1 ((x)/(2)))^2-y) ^ -1 ((x)/(2)),-2 ≤ x ≤ 2, y(2)=(π^2-8)/(4) , then y^2(0) is equal to

Correct answer: 0 aligned & (d y)/(d x)+ ( ^ -1 (x)/(2)) 4-x^2 y= ( ^ -3 (x)/(2))^3 4-x^2 & y e^ ( ^ -1 (x)/(2))^2 2 =∫ ( ^ -3 (x)/(2))^3 4-x^2 e^ ( ^ -1 (x)/(2))^2 2 d x & y=( ^ -1 (x)/(2))^2-2+c · e^ -( ^ -1 (x)/(2))^2 2 & y(2)=(π^2)/(4)-2 c=0 & y(0)=-2 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Differential Equations

If y=y(x) is the solution of the differential equation, 4-x^2 ~d y...

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

Correct answer

Step-by-step solution

\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}

Practice Differential Equations on Quantrex Academy →

More from Differential Equations