The solution of differential equation e−x(y+1)dy+cos2⁡x−sin⁡2xydx=0 , if y(0)=1,is

Given that e−x(y+1)dy+cos2⁡x−sin⁡2xydx=0⇒y+1ydy=−excos2⁡x−sin⁡2xdx Integrating on both sides ∫1+1ydy=−∫excos2⁡x−sin⁡2xdx⇒y+log⁡y=−excos2⁡x+c…….. 1)  ∵∫exf(x)+f′(x)dx=exf(x) As y(0)=11+log⁡(1)=−e0cos2⁡0+c⇒y+log⁡y+excos2⁡x=2