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Question

The general solution of the differential equation (ex+1) y dy=(y+1) ex dx is (where c is a constant of integration)

A
(y1)|ex1|=cey
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B
|y1|(ex+1)=2cey
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C
(y+1)(ex1)=4ce|y|
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D
|y+1|(ex+1)=ey|c|
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Solution

The correct option is D |y+1|(ex+1)=ey|c|
The given differential equation is (ex+1) y dy=(y+1) ex dx

ydy(y+1)=exdx(ex+1)

Integrating both sides, we get

yln|y+1|=ln|ex+1|+ln|k|y=ln|(y+1)(ex+1)k|

|y+1|(ex+1)=ey|c|; where c=1k

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