Find the particular solution of the differential equation (1 + e^2x) dy + (1 + y²) e^x dx = 0, given that y = 1 when x = 0.

It is given that (1 + e^2x) dy + (1 + y²) e^x dx = 0

On integrating both sides, we get,

------(1)

Let e^x = t

⇒ e^2x = t²

⇒ e^xdx = dt

Substituting the value in equation (1), we get,

⇒ tan^-1 y + tan^-1 t = C

⇒ tan^-1 y + tan^-1 (e^x) = C -------(2)

Now, y =1 at x = 0

Therefore, equation (2) becomes:

tan^-1 1 + tan^-1 1 = C

Substituting in (2), we get,

tan^-1 y + tan^-1 (e^x) =