JEE Main20206 Sep 2020Morning ShiftMathematicsDifferential EquationsActual
The general solution of the differential equation 1 + x 2 + y 2 + x 2 y 2 + x y d y d x = 0 (where C is a constant of integration)
Options
- A1 + y 2 + 1 + x 2 = 1 2 log e 1 + x 2 - 1 1 + x 2 + 1 + C
- B1 + y 2 - 1 + x 2 = 1 2 log e 1 + x 2 - 1 1 + x 2 + 1 + C
- C1 + y 2 + 1 + x 2 = 1 2 log e 1 + x 2 + 1 1 + x 2 - 1 + C
- D1 + y 2 - 1 + x 2 = 1 2 log e 1 + x 2 + 1 1 + x 2 - 1 + C
Correct answer
C. 1 + y 2 + 1 + x 2 = 1 2 log e 1 + x 2 + 1 1 + x 2 - 1 + C
Step-by-step solution
1 + x 2 1 + y 2 + x y d y d x = 0 Integrating, ⇒ ∫ 2 y 2 1 + y 2 d y = - ∫ 1 + x 2 x 1 + x 2 d x ⇒ 1 + y 2 = - ∫ x 1 + x 2 d x - ∫ x x 2 1 + x 2 d x Put 1 + x 2 = t ⇒ x 2 = t 2 - 1 to solve RHS 2nd integration, ⇒ 1 + y 2 = - 1 + x 2 - ∫ t t 2 - 1 t d t ⇒ 1 + y 2 + 1 + x 2 = - ∫ 1 t 2 - 1 d t ⇒ 1 + y 2 + 1 + x 2 = 1 2 ln 1 + x 2 + 1 1 + x 2 - 1 + C