If y = y ( x ) is the solution of the differential equation d y d x + 4 x x 2 - 1 y = x + 2 x 2 - 1 5 2 , x > 1 such that y ( 2 ) = 2 9 log e 2 + 3 and y 2 = α log e α + β + β - γ , α , β , γ ∈ ℕ , then α β γ is equal to

Step-by-step solution

Given differential equation is d y d x + 4 x x 2 - 1 y = x + 2 x 2 - 1 5 2 , x > 1    Now IF = e ∫ 4 x x 2 - 1 d x = x 2 - 1 2 The required equation will be ⇒ y · x 2 - 1 2 = ∫ x + 2 x 2 - 1 1 / 2 d x ⇒ y · x 2 - 1 2 = 1 2 ∫ 2 x x 2 - 1 1 / 2 d x + 2 ∫ d x x 2 - 1 1 / 2 ⇒ y · x 2 - 1 2 = 2 ln x 2 - 1 + x + x 2 - 1 + C Now using, at x = 2 , y ( 2 ) = 2 9 log e 2 + 3 we get, ⇒ 9 · 2 9 ln 2 + 3 = 2 ln 2 + 3 + 3 + C ⇒ C = - 3 Now fi