The solution to the differential equation sin⁡xdydxcos⁡y=dydx+sin⁡ycos⁡xdydx is

Given equation is sin⁡xdydx−y=dydx⇒y=xdydx−sin−1⁡dydxAgain differentiating we getdydx=dydx+xd2ydx2−11−dydx2d2ydx2⇒d2ydx2=0 or x=11−dydx2⇒dydx=c or dydx2=1−1x2using, dydx=c   in given equation   we get    y=cx−sin−1⁡cAlso for particular value of c = 0,y = 0 is also a solution.Finally using dydx2=1−1x2 in (i) we get i.e., y=x2−1−sin−1⁡x2−1x