JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation is 2y^2 dx dy - 2xy + x^2 = 0. Dividing by 2y^2, we get: dx dy - x y + x^2 2y^2 = 0 This is a homogeneous differential equation. Let x = vy, then dx dy = v + y dv dy . Substituting these into the equation: v + y dv dy - v + v^2 2 = 0 y dv dy = - v^2 2 Separating the variables: dv v^2 = - dy 2y Integrating both sides: v^ -2 dv = - 1 2 dy y - 1 v = - 1 2 y - C Substituting v = x y : y x = 1 2 y + C Given x(e) = e, substituting x = e and y = e: e e = 1 2 e + C 1 = 1 2 + C C = 1 2 The particular solution is: y x = 1 2 y + 1 2 = y + 1 2 x = 2y y + 1 To find x(e^2), substitute y = e^2: x(e^2) = 2e^2 (e^2) + 1 = 2e^2 2 + 1 = 2 3 e^2 Answer: 2 3 e^2