Question

Y-y' = 1/(e^-x-1) y=?

Answer 1

`y=(ln(e^-x-1)+C)/e^-x`

Explanation:

`y-y'=1/(e^-x-1)`
`y'-y=-1/(e^-x-1)`

The integrating factor is `e^(int-1dx)=e^-x`

`e^-x[y'-y=-1/(e^-x-1)]`
`e^-xy'-e^-xy=-(e^-x)/(e^-x-1)`

The left hand side of the equation is just the product rule for derivatives.

`[e^-xy]'=-(e^-x)/(e^-x-1)`

Integrate both sides:

`e^-xy=int-(e^-x)/(e^-x-1)dx`

Let `u=e^-x` then `(du)/dx=-e^-x`

`e^-xy=int 1/(u-1)du`

`e^-xy=ln(u-1)+C`

Undo substitution:

`e^-xy=ln(e^-x-1)+C`

`y=(ln(e^-x-1)+C)/e^-x`