JEE Main20262 April 2026Evening ShiftMathematicsDifferential EquationsActual
Let x = x(y) be the solution of the differential equation 2y^2 dx dy - 2xy + x^2 = 0 , y > 1 , x(e) = e . Then x(e^2) is equal to:
Options
- A3 2 e^2
- B2 3 e^2
- Ce^2
- D2e^2
Correct answer
B. 2 3 e^2
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