Given dx + xdy = e^–ysec²ydy

This is a first order linear differential equation of the form

Here, P = 1 and e^–ysec²y

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^y [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ xe^y = tan y + c

∴ x = (tan y + c)e^–y

Thus, the solution of the given differential equation is x = (tan y + c)e^–y