Let y = y ( x ) be the solution of the differential equation, x 2 + 1 2 d y d x + 2 x ( x 2 + 1 ) y = 1 such that y 0 = 0 . If a y 1 = π 32 , then the value of a is

Step-by-step solution

x 2 + 1 2   d y d x + 2 x x 2 + 1 y = 1 ⇒ d y d x + 2 x x 2 + 1 y = 1 ( x 2 + 1 ) 2 Which is a linear differential equation with integrating factor I.F. = e ∫ 2 x x 2 + 1 d x = e l n ( x 2 + 1 ) = x 2 + 1 The solution of the given differential equation will be, y . x 2 + 1 = ∫ x 2 + 1 . 1 x 2 + 1 2 d x = ∫ d x x 2 + 1 = t a n - 1 x + c Now, y 0 =   0 ⇒ C = 0 and y = 1 ( x 2 + 1 ) t a n - 1 x a y ( 1 ) = π 32 ⇒ a = π / 32 π / 8 = 1 4 ⇒ a = 1 16