Question

e^ x cot y
dx + (1-e^ x) cosec^2y dy =0​

Answer 1

Given,

3e

x

cos

2

ydx+(1−e

x

)cotydy=0

3e

x

cos

2

ydx=(e

x

−1)cotydy

e

x

−1

3e

x

dx=

cos

2

y

coty

dy

integrating on both sides, we get,

∫

e

x

−1

3e

x

dx=∫

cos

2

y

coty

dy

∫

e

x

−1

3e

x

dx=∫

siny

secy

dy

3log(e

x

−1)=log(tany)+c