If y=y(x) is the solution of the differential equation, sqrt 4-x 2 d y/d x= ( (sin -1 (x/2 ) ) 2 -y ) sin -1 (x/2 ) , -2 leq x leq 2, y(2)= (frac π 2 -8 4 ) , then y 2 (0) is equal to

Mathematics · Differential Equations

JEE Main 2025 — 28 January, Evening Shift — Question 25

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

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$\frac{d y}{d x}+\frac{\left(\sin ^{-1} \frac{x}{2}\right)}{\sqrt{4-x^{2}}} y=\frac{\left(\sin ^{-3} \frac{x}{2}\right)^{3}}{\sqrt{4-x^{2}}}$

$y e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}}=\int \frac{\left(\sin ^{-1} \frac{x}{2}\right)^{3}}{\sqrt{4-x^{2}}} e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}} d x$

$y=\left(\sin ^{-1} \frac{x}{2}\right)^{2}-2+c \cdot e^{\frac{-\left(\sin ^{-1} \frac{x}{2}\right)^{2}}{2}}$

$y(2)=\frac{\pi^{2}}{4}-2 \Rightarrow c=0$

$y(0)=-2$

$y^{2}(0)=4$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential