Let y=y(x) be the solution of the differential equation sinxdydx+ycosx=4x,∈(0,π). If yπ2=0, then yπ6 is equal to

Given that sinxdydx+ycosx=4x⇒ddxysinx=4xIntegration on both sides∫dysinx=∫4xdx⇒ysinx=4x22+C⇒ysinx=2x2+C Given yπ2=0C=−π22 Thus ysinx=2x2-π22 Now ysinπ6=2π236-π22⇒y·12=π218−π22⇒y2=−8π218∴y=−8π29