Find the particular solution of the following differential equation: (x+1)dy/dx = 2e-y, y=0 when x=0

Given (x+1)dy/dx = 2e-y ⇒ dy/2e-y = dx/(x+1)
Both the sides, we get
∫dy/2e-y = ∫dx/(x+1)
⇒∫eydy/2 = log|x+1| + c
⇒ey/2 = log|x+1| + c
Let z = ey, then dz = eydy = zdy
⇒∫dz/2z = log|x+1| + c
⇒(1/2)log|z| = log|x+1| + c
⇒log|z1/2| = log|x+1| + c
⇒log|√z| = log|x+1| + c
Since z = ey, we have
⇒log|√ey| = log|x+1| + c
⇒log|ey/2| = log|x+1| + c
⇒y/2 = log|x+1| + c
⇒y = 2log|x+1| + 2c
Substituting y=0 and x=0, we get
0 = 2log(1) + 2c
0 = 2(0) + 2c
2c = 0
c = 0
Therefore, the required particular solution is y = 2log|x+1|

Given (x+1)dy/dx = 2e-y
Separating variables, we get
eydy = 2dx/(x+1)
Integrating both sides,
∫eydy = ∫2dx/(x+1)
ey = 2ln|x+1| + C
When x=0, y=0, so
1 = 2ln(1) + C
C = 1
Therefore, the particular solution is ey = 2ln|x+1| + 1