Let y = y ( x ) be the solution of the differential equation 1 - x 2 d y = x y + x 3 + 2 3 1 - x 2 d x , - 1 < x < 1 , y ( 0 ) = 0 . If y 1 2 = m n , m and n are coprime numbers, then m + n is equal to __________.

Step-by-step solution

Given, 1 - x 2 d y = x y + x 3 + 2 3 1 - x 2 d x ⇒ d y d x - x y 1 - x 2 = x 3 + 2 3 1 - x 2 1 - x 2 The above equation is a linear differential equation, So, IF = e - ∫ x 1 - x 2 d x = e 1 2 ln 1 - x 2 = 1 - x 2 Now, the solution of the differential equation is given by, y 1 - x 2 = ∫ x 3 + 2 3 1 - x 2 1 - x 2 × 1 - x 2 d x ⇒ y 1 - x 2 = 3 ∫ x 3 + 2 d x ⇒ y 1 - x 2 = 3 x 4 4 + 2 x + c Now, using y ( 0 ) = 0 we get, c = 0 Now, finding y 1 2 y 1 - 1 2 2 = 3 1 2 4 · 4 + 2 · 1 2 ⇒ y = 1 32 + 2 ⇒ y 1 2 = 65 32 = m n He