JEE Main20205 Sep 2020Morning ShiftMathematicsDifferential EquationsActual
If y = y x is the solution of the differential equation 5 + e x 2 + y ⋅ d y d x + e x = 0 satisfying y 0 = 1 then value of y ( log e 13 ) is
Options
- A1
- B- 1
- C0
- D2
Correct answer
B. - 1
Step-by-step solution
Given d y 2 + y = - e x d x 5 + e x ⇒ ∫ d y 2 + y = - ∫ e x d x 5 + e x ⇒ log e ( 2 + y ) = - log e 5 + e x + log e C ⇒ log e ( 2 + y ) = log e C 5 + e x ⇒ y = C 5 + e x - 2 ∵ y ( 0 ) = 1 ∴   c = 18 y = 18 5 + e x - 2 y ( log e 13 ) = 18 5 + e log e 13 - 2 ⇒ y ( log e 13 ) = 18 5 + 13 - 2 ∴    y log e 13 = - 1