Q9 of 118 Page 419

Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.

It is given that (1 + e2x) dy + (1 + y2) ex dx = 0


On integrating both sides, we get,


------(1)


Let ex = t


⇒ e2x = t2




⇒ exdx = dt


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



⇒ tan-1 y + tan-1 t = C


⇒ tan-1 y + tan-1 (ex) = 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 (ex) =


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