The solution of dydx+2ytanx=sin⁡x , given that y=0,x=π/3 is

I.F=e∫2ttax⁡dx=e2log⁡|sec⁡x|=sec2⁡x Solution of the differential equation is ysec2⁡x==∫sin⁡xcos2⁡xdx=∫sec⁡xtan⁡xdx=sec⁡x+c If x=π/3→4y=2+cy=0⇒c=−2⇒ysec2x=secx−2y=cos⁡x−2cos2⁡x