The solution of differential equation 2ysin⁡xdydx=2sin⁡xcos⁡x−y2cos⁡x at x=π2,y=1 is

Given equation can be rewrite as 2ydydx+y2cot⁡x=dvdx Put y2=v=2ydydx=dvdxdvdx+vcot⁡x=2cos⁡x If e∫cos⁡xsin⁡x=dx=elog⁡sin⁡x=sin⁡x Solution is vsin⁡x+∫2cos⁡xsin⁡xdx+cy2sin⁡x=sin2⁡x+c ; when x=π2,y=11=1+c⇒c=0; y2sin⁡x=sin2x⇒y2=sinx