JEE Main20262 April 2026Morning ShiftMathematicsDifferential EquationsActual
Let y = y(x) be the solution curve of the differential equation (1 + x) dy dx + (y+1) x = 0 , y(0) = 0 . If the curve y = y(x) passes through the point ( , -1 2 ) , then a value of is :
Options
- A6
- B4
- C3
- D2
Correct answer
D. 2
Step-by-step solution
The given differential equation can be written as: (1 + x)dy + (y+1) x dx = 0 d((y+1)(1 + x)) = 0 Integrating both sides: (y+1)(1 + x) = C Given y(0) = 0 , substituting x = 0 and y = 0 : (0+1)(1 + 0) = C C = 1 The equation of the curve is (y+1)(1 + x) = 1 Since the curve passes through ( , -1 2 ) , substituting x = and y = -1 2 : ( -1 2 + 1 )(1 + ) = 1 1 2 (1 + ) = 1 1 + = 2 = 1 = 2 Answer: 2