If dydx+xsin⁡2y=x3cos2⁡y,y(0)=0 then tan⁡y(1)=

sec2⁡ydydx+2xtan⁡y=x3 I. F=e∫2xdx=ex2 Solution is tan⁡yex2=∫x3⋅ex2dx=12x2−1ex2+c⇒tan⁡y(x)=12x2−1+c⋅e−x2x=0,y=0⇒c=12⇒tan⁡y(x)=12x2−1+e−x2tan⁡y(y(1))=12e