It is given that

On integrating both sides, we get,

----------------(1)

Let

⇒ e^ydt = -dt

Substituting value in equation (1), we get,

⇒ -log|t| = log|C(x+1)|

⇒ -log|2 – e^y| = log|C(x + 1)|

------------------(2)

Now, at x = 0 and y = 0, equation (2) becomes,

⇒ C = 1

Now, substituting the value of C I equation (2), we get,

Therefore, the required particular solution of the given differential equation is