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Question

The solution of (1+ex)ydy=exdx is:

A
y2=logc(ex+1)
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B
y22=logcex
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C
y22=logc(ex+1)
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D
2y=logc(ex+1)
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Solution

The correct option is D y22=logc(ex+1)
(1+ex)ydy=exdx

ydy=(ex1+ex)dx+c

y22=log(1+ex)c

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