Let y = y x be the solution curve of the differential equation d y d x = y x 1 + x 2 1 + log e x , x > 0 , y ( 1 ) = 3 . Then y 2 ( x ) 9 is equal to :

Step-by-step solution

Given differential equation can be rewritten as, d y d x - y x = y 3 1 + log e x 1 y 3 dy dx - 1 xy 2 = 1 + log e x Let - 1 y 2 = t ⇒ 2 y 3 dy dx = dt dx ∴ dt 2 dx + t x = 1 + log e x ⇒ dt dx + 2 t x = 2 1 + log e x   . . . . . . . . . . ( 1 ) We know the solution of the differential equation d y d x + P y = Q is given by, ye ∫ Pdx = ∫ Qe ∫ Pdx + C     ( Where   I . F . = e ∫ Pdx ) Therefore, on solving equation(1) we get, I.F. = e ∫ 2 x d x = x 2