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2e^(x)dx-sec^(2)ydy=0...

`2e^(x)dx-sec^(2)ydy=0`

A

`e^(2x)-tany=c`

B

`2e^(x)-tany=c`

C

`2e^(x)cosy-siny=c`

D

`2e^(2x)+tany=c`

Text Solution

Verified by Experts

The correct Answer is:
B
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