Given, differential equation is

Also, given that y = 1, when x = 0
On putting y = 1 and x = 0 in Eq, (i), we get
\(\sqrt{1-1}\) = 0 - e0 + C
⇒ c = 1 [∵ e0 = 1]
On substituting the value of C in Eq. (i), we get
\(\sqrt{1-y^2}\) = xeX - ex + 1
which is the required particular solution of given differential equation.