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Question: Solution of the equation \(\left( e ^ { x } + 1 \right) y d y = ( y + 1 ) e ^ { x } d x\) is...

Solution of the equation (ex+1)ydy=(y+1)exdx\left( e ^ { x } + 1 \right) y d y = ( y + 1 ) e ^ { x } d x is

A

c(y+1)(ex+1)+ey=0c ( y + 1 ) \left( e ^ { x } + 1 \right) + e ^ { y } = 0

B

c(y+1)(ex−1)+ey=0c ( y + 1 ) \left( e ^ { x } - 1 \right) + e ^ { y } = 0

C

c(y+1)(ex−1)−ey=0c ( y + 1 ) \left( e ^ { x } - 1 \right) - e ^ { y } = 0

D

c(y+1)(ex+1)=eyc ( y + 1 ) \left( e ^ { x } + 1 \right) = e ^ { y }

Answer

c(y+1)(ex+1)=eyc ( y + 1 ) \left( e ^ { x } + 1 \right) = e ^ { y }

Explanation

Solution

(ex+1)ydy=(y+1)exdx\left( e ^ { x } + 1 \right) y d y = ( y + 1 ) e ^ { x } d x

⇒ (yy+1)dy=(exex+1)dx\left( \frac { y } { y + 1 } \right) d y = \left( \frac { e ^ { x } } { e ^ { x } + 1 } \right) d x⇒ [1−1y+1]dy=(exex+1)dx\left[ 1 - \frac { 1 } { y + 1 } \right] d y = \left( \frac { e ^ { x } } { e ^ { x } + 1 } \right) d x

⇒ ∫{1−1y+1}dy=∫exex+1dx\int \left\{ 1 - \frac { 1 } { y + 1 } \right\} d y = \int \frac { e ^ { x } } { e ^ { x } + 1 } d x

⇒ y=log⁡(y+1)+log⁡(ex+1)+log⁡cy = \log ( y + 1 ) + \log \left( e ^ { x } + 1 \right) + \log c or ey=c(y+1)(ex+1)e ^ { y } = c ( y + 1 ) \left( e ^ { x } + 1 \right) .