NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation xdy = tan ⁡ y + e 1 / x 2 x sec ⁡ y dx is (where C is the constant of integration)
Options
- Asin y = e 1 x 2 + C
- B2 sin y x + e 1 x 2 = C
- Csin y x - e 1 x 2 = C
- Dsin ⁡ y - x e 1 x 2 = C
Correct answer
B. 2 sin y x + e 1 x 2 = C
Step-by-step solution
d y d x = tan y x + e 1 x 2 x 2 sec y cos y d y d x - sin y x = e 1 x 2 x 2 Let, sin y = t cos y d y d x = d t d x ⇒ d t d x - t x = e 1 x 2 x 2 I.F. = e - ∫ 1 x d x = e ln 1 x = 1 x ⇒ t x = ∫ e 1 x 2 x 3 d x ⇒ t x = - 1 2 ∫ - 2 x 3 e 1 x 2 d x ⇒ sin y x = - 1 2 e 1 x 2 + C