The correct option is A 1−√32
cosxdydx+ysinx=1⇒dydx+ytanx=secx
which is a linear differential equation.
I.F.=e∫tanx dx=secx
The general solution is
y×secx=∫sec2x dx⇒ysecx=tanx+c
Given y(π4)=√2
⇒2=1+c⇒c=1
Hence, the solution of the given equation is ysecx=tanx+1
Putting x=−π3, we get
2y=−√3+1⇒y=1−√32