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Question

The solution of (ex+ex)dydx=exex is:

A
y=log|exex|+c
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B
y=sinhx+c
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C
y=log|ex+ex|+c
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D
y=secx+c
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Solution

The correct option is B y=log|ex+ex|+c
dydx=exexex+ex

let v=ex+ex

dv=(exex)dx
dy=1vdv+c

y=log v+c

y=log(ex+ex)+c

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