+ytanx=secx which is a linear differential equation
∴I.F=e∫tanxdx=elogsecx=secx
∴ The solution is given by y⋅secx=∫secx⋅secxdx+C ysecx=tanx+C.....(i) Now, y=0, when x=0 ∴0=0+C [From Eq. (i)] ⇒c=0 Putting c=0 in Eq. (i), we get ysecx=tanx