If y=y(x) is the solution of the differential equation, dydx+2ytanx=sinx,yπ3=0, then the maximum value of the function y (x) over R is equal to :

The given differential equation is dydx+2ytanx=sinx,yπ3=0This is the linear differential eqution, so that the integrating factor is I.F=e∫2tanxdx=elogsec2x=sec2xTherefore, the solution of the above differential equation is y.sec2x=∫sinxcos2xdx⇒ysec2x=secx+c Substitute the initial condition, yπ3=0, it gives 0(4)=2+c⇒c=-2 Hence the solution is y=cosx-2cos2x It has maximum value when cosx=14 Therefore, y=18