NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation d y d x + y x = 1 ( 1 + ln x + ln y ) 2 is (where, c is an arbitrary constant)
Options
- Ax y 1 + ln ( x y 2 ] = x 2 2 + c
- B1 + ln x y 2 = x 2 2 + y + c
- Cx y 1 + ln x y = x 2 2 + c
- Dx y 1 + ln x y = x 2 + c
Correct answer
A. x y 1 + ln ( x y 2 ] = x 2 2 + c
Step-by-step solution
d y d x + y x = 1 ( 1 + ln x y ) 2 Let x y = u so that d u d x = x ( 1 + ln   u ) 2 ∴       ∫ ( 1 + ln u ) 2 d u = ∫ x   d x + c ⇒ u ( 1 + ln u ) 2 - ∫ 2 1 + ln   u u · u d u = x 2 2 + c ⇒ u 1 + 2   ln ⁡ u + 2 u ( ln ⁡ u ) 2 - 2 u ln ⁡ u = x 2 2 + c ∴ x y ( 1 + ln x y ) 2 = x 2 2 + c