The solution of the differential equation `3e^x tan y dx+(1+e^x)sec^2 y dy=0` is

Explanation

`3e^(x)*tanydx+(1-e^(x))sec^(2)ydy=0`
`implies (3e^(x))/(1-e^(x))dx+(sec^(2)y)/(tany)dy=0`
`implies int(3e^(x))/(1-e^(x))dx+int(sec^(2)y)/(tany)dy=logc`
Let `1-e^(x)=t`
`implies e^(x)dx=-dt`
and `tany=z`
`implies int(-3dt)/(t)+int(1)/(z)dz=logc`
`implies-3logt+logz=logc`
`implies (z)/(t^(3))=c`
`implies z=ct^(3)`
`implies tany=c*(1-e^(x))^(3)`.