JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Divide the equation by ^2 y: x ^2 y dy dx - 2 y = x^3(2-x^3). Let v = y, so ^2 y dy dx = dv dx . The equation becomes x dv dx - 2v = x^3(2-x^3). Dividing by x: dv dx - 2v x = x^2(2-x^3). Using integrating factor = x^ -2 : d dx (x^ -2 v) = 2 - x^3. Integrating: x^ -2 v = 2x - x^4 4 + C. From y(2) = 0, we have 0 = 0 = 16 - 16 + 4C, giving C = 0. Therefore y = 2x^3 - x^6 4 . At x = 1: (y(1)) = 2 - 1 4 = 7 4 .