The equation of the curve passing through origin and satisfying $\left(1+x^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y=4 x^2$ is

The order of the differential equation whose general solution is given by $y=\left(\mathrm{C}_1+\mathrm{C}_2\right) \sin \left(x+\mathrm{C}_3\right)-\mathrm{C}_4 \mathrm{e}^{x+\mathrm{C}_5}$ is (where $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3, \mathrm{C}_4, \mathrm{C}_5$ are arbitrary constants)

If $x=\operatorname{sint}$ and $y=\sin p t$, then the value of

$$ \left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{~d} y}{\mathrm{~d} x}+\mathrm{p}^2 y= $$

The general solution of differential equation $\left(y^2-x^2\right) \mathrm{d} x=x y \mathrm{~d} y(x \neq 0)$ is