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