The solution of the differential equation `(e^x+e^(-x))dy-(e^x-e^(-x))dx`=0 is

Explanation

Given,

`=>(dy)/(dx)​=(e^x-e^(−x))/(e^x+e^(−x))​`

let `v=e^x+e^(−x)`

`⇒dv=(e^x−e^(−x))dx`

`⇒∫dy=∫(1​dv)/v+c`

`⇒y=log v+c`

`=>y=log(e^x+e^(−x))+c`