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

1 Answer
Jun 11, 2017

#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#