The solution of 2ycos⁡y2dydx−2x+1sin⁡y2=(x+1)3 is

Let sin⁡y2=t⇒2ycos⁡y2dydx=dtdx⇒ dtdx−2t(x+1)=(x+1)3 Solve the linear equation I. F=1(x+1)2G.S is t I.F.=∫Q(x) IFdxt(x+1)2=∫(x+1)dxsiny2(x+1)2=(x+1)22+c