Find the particul

The given differential equation is … (1)

Separating given differential equation, we get

On integrating, we get

Let 2 – e^y = t

- e^y dy = dt

⇒ - log (2 – e^y) = log (x + 1) + C ... (2)

Putting y = 0, when x = 0 in (2), we get

⇒ - log (2 – 1) = log (0 + 1) + C

⇒ 0 = 0 + C

∴ C = 0

Equation (2) becomes - log (2 – e^y) = log (x + 1)

⇒ log (x + 1) + log (2 – e^y) = 0

⇒ log (x + 1) (2 – e^y) = 0

⇒ (x + 1) (2 – e^y) = e⁰ = 1

∴ The solution is (x + 1) (2 – e^y) = 1