MHT CET2016MathematicsDifferential Equations
The particular solution of the differential equation y 1 + log x d x d y - x log x = 0 when y ( e ) = e 2 is
Options
- Ay = e x log x
- Be y = x log x
- Cx y = e log x
- Dy log x = e x
Correct answer
A. y = e x log x
Step-by-step solution
y 1 + log x d x d y - x log x = 0 ⇒ 1 + log x x log x d x = d y y Integrating on both side ∫ 1 + log x x log x d x = ∫ d y y log x log x = log y + log C ⇒ log x log x = log y . c ∴ x log x = y . c ...... (i) As x = e , y = e 2 ∴ e = e 2 . c ∴ c = 1 e Putting c = 1 e in equation (i) we get x log x = y e ⇒ y = e x log x