Let y=yx be the solution of the differential equation  cosx3sinx+cosx+3dy=1+ysinx3sinx+cosx+3dx,0≤x≤π2,y0=0. Then, yπ3 is equal to:

The given differential equation is cosx3sinx+cosx+3dy=1+ysinx3sinx+cosx+3dxRewrite the above equation as dydx=1+ysinx3sinx+cosx+3cosx3sinx+cosx+3=1cosx3sinx+cosx+3+ytanxdydx−ytanx=1cosx3sinx+cosx+3This is a linear differential equation  Integrating factor is e−∫tanxdx=e−logsecx=cosx Hence, the solution of the given differenital equation is as belowycosx=∫1cosx3sinx+cosx+3cosxdx=∫13sinx+cosx+3dxSubstitute tanx2=t,dx=2dt1+t2,sinx=2t1+t2,cosx=1-t21+t2ycosx=∫13⋅2t1+t2+1−t21+t2+32dt1+t2=∫26t+1−t2+3+3t2dt=∫22t2+6t+4dt=∫1t2+3t+2dt=∫1t+1t+2dt=∫1t+1−1t+2dt=lnt+1−lnt+2=lntanx2+1tanx2+2+csubstitute x=0,y=0 it gives c=ln2Therefore, the solution for the given differential equation is ycosx=ln1+tanx22+tanx2when x=π3y12=loge1+tanπ62+tanπ6+ln2simplifying the above expression  we get y=2loge10+2311