The solution of dydx+2ytanx=sinx , given that y=0,x=π/3 is
y=cosx-2cos2x
y=cosx+2cos2x
y=cosx-cos2x
y=2cosx-2cos2x
I.F=e∫2ttaxdx=e2log|secx|=sec2x Solution of the differential equation is ysec2x==∫sinxcos2xdx=∫secxtanxdx=secx+c If x=π/3→4y=2+cy=0⇒c=−2⇒ysec2x=secx−2y=cosx−2cos2x