Obtain a particular solution of the problem: (1+e^-x) dy/dx = y-1?

Question

given that as x goes to minus infinity, y = 2

Answer 1

Just use separation of variables
(1+e^-x) dy = (y-1) dx
1/(y-1) dy = 1/(1+e^-x) dx
ln |y-1| = ln |1+e^x| + C
y - 1 = C(1+e^x)
y = C(1+e^x) + 1
as x goes to minus infinity, e^x goes to 0
2 = C(1+0) + 1
2 = C + 1
C = 1
answer:
y = 2 + e^x

Answer 2

(1+e^-x) dy/dx = y - 1

dy/(y - 1) = dx/(1 + e^-x)

ln(y - 1) = x + ln(1 + e^-x) + C .......... Integral Tables

As x gets very large 1 + e^(-x) -> e^(-x)
And ln (e ^-x) = -x giving
ln(y - 1) = x - x + C = C = ln(2 - 1) = 0
So C= 0 and the solution is

ln(y - 1) = x + ln(1 + e^-x)

Answer 3

∫1 / (y - 1) dy = ∫ 1 / [1 + e^(-x) ] dx
LHS
ln (y - 1)

RHS
∫ 1 / [1 + 1 / e^x] dx
∫ e^x / (e^x + 1) dx
ln (e^x + 1)

ln(y - 1) = ln K(e^x + 1)
ln(2 - 1) = ln K(0 + 1)
ln 1 = ln K
K = 1
ln(y - 1) = ln(e^x + 1)
y - 1 = e^x + 1
y = e^x + 2