JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation is: x 1-x^2 \,dy + (y 1-x^2 - x ^ -1 x )dx = 0 Rearranging the terms, we get: x 1-x^2 \,dy + y 1-x^2 \,dx = x ^ -1 x\,dx Dividing the entire equation by 1-x^2 : x\,dy + y\,dx = x ^ -1 x 1-x^2 \,dx The left side is the exact differential of xy: d(xy) = x ^ -1 x 1-x^2 \,dx Integrating both sides: xy = x ^ -1 x 1-x^2 \,dx + C To evaluate the integral, use integration by parts. Let u = ^ -1 x and dv = x 1-x^2 \,dx. du = - 1 1-x^2 \,dx and v = - 1-x^2 u\,dv = uv - v\,du x ^ -1 x 1-x^2 \,dx = - 1-x^2 ^ -1 x - (- 1-x^2 ) (- 1 1-x^2 )\,dx = - 1-x^2 ^ -1 x - 1\,dx = - 1-x^2 ^ -1 x - x Substituting this back into the equation: xy = - 1-x^2 ^ -1 x - x + C It is given that _ x 1^- y(x) = 1. Substituting x = 1 and y = 1: 1(1) = - 1-1^2 ^ -1 (1) - 1 + C 1 = 0 - 1 + C C = 2 The particular solution is: xy = 2 - x - 1-x^2 ^ -1 x To find y ( 1 2 ), substitute x = 1 2 : 1 2 y ( 1 2 ) = 2 - 1 2 - 1 - ( 1 2 )^2 ^ -1 ( 1 2 ) 1 2 y ( 1 2 ) = 3 2 - 3 2 ( 3 ) 1 2 y ( 1 2 ) = 3 2 - 2 3 Multiplying by 2: y ( 1 2 ) = 3 - 3