Q. 9

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

Answer :

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

On integrating both sides, we get,

------(1)

Let e^{x} = t

⇒ e^{2x} = t^{2}

⇒ e^{x}dx = 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}) =

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