If the general solution of $\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$ is $x=f(y)+c e^{-\tan ^{-1} y}$, then $f(y)=$

If $y=f(x)$ is the solution of the differential equation $\left(1+\cos ^2 x\right) f^{\prime}(x)-4 \sin 2 x-f(x) \sin 2 x=0$ when $f(0)=0$, then $f\left(\frac{\pi}{3}\right)=$

The differential equation corresponding to the family of ellipses $\frac{x^2}{a^2}+\frac{y^2}{4}=1$, where ' $a$ ' is an arbitrary constant is

The general solution of the differential equation $\frac{d y}{d x}+(\sec x \operatorname{cosec} x) y=\cos ^2 x$