Solve the equation `(dy)/(dx)=1/x=(e^y)/(x^2)`

Answer

`e^(-y)/(x) = 1/(2x^(2))+c`

Explanation

Dividing by `e^(y)`, we get
`e^(-y)(dy)/(dx)+e^(-y)1/x=1/x^(2)`
Putting `e^(-y)=v`, we get
`-(dv)/(dx)+v/x=1/x^(2)`
or `(dv)/(dx) -1/xv=-1/x^(2)`(linear)
I.F. `e^(-int(1//x)dx)=e^(-logx)=1//x`
Therefore, solution is
`v/x = int-1/x^(2)1/xdx+c=x^(-2)/2+c`
`therefore (e^(-y))/x=x^(-2)/2+c`