JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation is dy dx = (1 + x + x^2)(1 - y + y^2). Separating the variables: dy y^2 - y + 1 = (x^2 + x + 1) dx Integrating both sides: dy (y - 1 2 )^2 + 3 4 = (x^2 + x + 1) dx 2 3 ^ -1 ( y - 1 2 3 2 ) = x^3 3 + x^2 2 + x + C 2 3 ^ -1 ( 2y - 1 3 ) = x^3 3 + x^2 2 + x + C Given y(0) = 1 2 , substituting x = 0 and y = 1 2 : 2 3 ^ -1 (0) = C C = 0 Substituting x = 1 to find y(1): 2 3 ^ -1 ( 2y(1) - 1 3 ) = 1^3 3 + 1^2 2 + 1 2 3 ^ -1 ( 2y(1) - 1 3 ) = 11 6 ^ -1 ( 2y(1) - 1 3 ) = 11 3 12 2y(1) - 1 = 3 ( 11 3 12 ) Answer: 3 ( 11 3 12 )