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e^(x) tan y dx + (1 - e^(x))sec^(2)y dy...

`e^(x) tan y dx + (1 - e^(x))sec^(2)y dy = 0`

Text Solution

Verified by Experts

The correct Answer is:
`tany=C(1-e^(x))`
`e^(x)tanydx+(1-e^(x))sec^(2)ydy=0`
or `(1-e^(x))sec^(2)ydy=-e^(x)tanydx`
or `int(sec^(2)y)(tany)dy=int(-e^(x))/(1-e^(x))`dx
or `log(tany)=log(1-e^(x))+logC`
or `log(tany)=log[C(1-e^(x))]`
or `tany=C(1-e^(x))`
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