BITSAT 2019 Paper

(c) Since, (e‌x + 1) ydy = (y + 1 )e‌x dx
⇒

dx

dy

=

y

1+y

+

y

(1+y)ex

⇒

dx

dy

= (

y

1+y

)+(

ex+1

ex

)
⇒ (

y

1+y

)dy = (

ex+1

ex

)dx
After integrating on both sides, we have
∫

y

1+y

dy= ∫

ex

ex+1

dx
⇒ ∫1 dy - ∫

1

1+y

dy = ∫

ex

ex+1

dx
⇒ y - log |(1 + y)|= log |(1 + e‌x) + log k
Hence y = log [k (1 + y) (1 + e‌x)]