Let y = y ( x ) be the solution of the differential equation x log e x d y d x + y = x 2 log e x , ( x > 1 ) . If y ( 2 ) = 2 , then y ( e ) is equal to

Step-by-step solution

Given, x ln x d y d x + y = x 2 ln x ⇒ d y d x + y x ln x = x This is a linear differential equation. Now, finding integrating factor we get I . F = e ∫ 1 xlnx dx = e ln lnx = lnx Now, solution of the differential equation is given by y × I . F = ∫ x × I . F d x ⇒ y × ln x = ∫ xlnx   d x ⇒ y × ln x = ln x · x 2 2 - x 2 4 + c Now given y 2 = 2 , 2 × ln 2 = ln 2 · 2 2 2 - 2 2 4 + c ⇒ c = 1 So, equation will be y × ln x = ln x &#183