JEE Advanced2014MathematicsDifferential EquationsActual
The function y = f x is the solution of the differential equation d y d x + x y x 2 - 1 = x 4 + 2 x 1 - x 2 in (- 1 , 1) satisfying f 0 = 0 . Then ∫ - 3 2 3 2 f x d x is
Options
- Aπ 3 - 3 2
- Bπ 3 - 3 4
- Cπ 6 - 3 4
- Dπ 6 - 3 2
Correct answer
B. π 3 - 3 4
Step-by-step solution
d y d x + x x 2 - 1 y = x 4 + 2 x 1 - x 2 This is a linear differential equation I.F. = e ∫ x x 2 - 1 d x = e 1 2 I n x 2 - 1 = 1 - x 2 ⇒ s o l u t i o n i s y 1 - x 2 = ∫ x x 3 + 2 1 - x 2 . 1 - x 2 d x Or y 1 - x 2 = ∫ x 4 + 2 x d x = x 5 5 + x 2 + c f 0 = 0 ⇒ c = 0 ⇒ f x 1 - x 2 = x 5 5 + x 2 Now, ∫ - 3 2 3 2 f x d x = ∫ - 3 2 3 2 x 2 1 - x 2 d x (Using property) = 2 ∫ 0 3 2 x 2 1 - x 2 d x = 2 ∫ 0 π 3 sin 2 θ cos θ cos θ d θ (Taking x = sin θ ) = 2 ∫ 0 π 3 sin 2 θ d θ = 2 θ 2 - sin 2 θ 4 0 π 3 = 2 π