The general solution of the differential equation $$\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$$

The solution of the differential equation $$\frac{d y}{d x}+y \cos x=\frac{1}{2} \sin 2 x$$

The sum of the degree and order of the following differential equation $$\left[1-\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}=k x \frac{d^2 y}{d x^2}$$

The solution of the differential equation $${{{d^2}y} \over {d{x^2}}} = 0$$ represents