Given Equation:

Re-arranging, we get,

Let 2 – e^y = t

-e^ydy = dt

Therefore,

Integrating both sides, we get,

log t = log(x + 1) + C

log (2 – e^y) = log (x + 1) + C

At x = 0, y = 0.

Therefore,

log(2) = log(1) + C

Therefore,

C = log 2

Now, we have,

log (2 – e^y) – log (x + 1) – log 2 = 0