(x + 1)

dy

dx

= 2e−y - 1
⇒

dy

2e−y−1

=

dx

x+1

⇒

ey

2−ey

=

dx

x+1

Integrating both sides, we get:
∫

eydy

2−ey

= log |x + 1| + log C ... (1)
Let 2 - ey = t
∴

d

dy

(2−ey)={dt}/{dy}$$
⇒ - ey =

dt

dy

⇒ ey dy = - dt
Substituting this value in equation 1 , we get:
∫ -

dt

t

= log |x + 1| + log C
⇒ - log |t| = log |C (x + 1)|
⇒ - log |2 - ey| = log |C (x + 1)|
⇒

1

2−ey

= C (x + 1)
⇒ 2 - ey =

1

C(x+1)

... (2)
Now, at x = 0 and y = 0, equation 2 becomes:
⇒ 2 - 1 =

1

C

⇒ C = 1
Substituting C = 1 in equation 2 , we get:
2 - ey =

1

x+1

⇒ ey = 2 -

1

x+1

⇒ ey =

2x+2−1

x+1

⇒ ey =

2x+1

x+1

⇒ y = log |

2x+1

x+1

| , (x ≠ - 1)
This is the required particular solution of the given differential equation.