(extany)dx+(1+ex)sec2y)dy=0 ...(i) Its of the form Mdx+Ndy=0 Now,
∂M
∂y
=exsec2y and
∂N
∂x
=exsec2y ⇒ Eq. (i) is exact. Thus, its solution will be ∫Mdx+∫Ndy=C ∫extanydx+∫sec2ydy=C extany+tany=C tany(1+ex)=C ...(ii) Since, Eq. (ii) passes through (0,
