Find the general solution of each of the following differential equations:
`e^(x) tan y dx +(1-e^(x))sec^(2)y dy =0`

Answer

`tany=C(1-e^(x))`

Explanation

`-int (-e^(x))/((1-e^(x)))dx + int (sec^(2)y)/(tan y) dy = log |C_(1)| rArr logh|tan y|-log |1-e^(x)|=log |C_(1)|.`
` therefore log |(tany)/((1-e^(x)))|=log |C_(1)| rArr (tan y)/((1-e^(x))) = pm C_(1) = C.`