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

If $y=\tan ^{-1}\left(\frac{12 x-64 x^3}{1-48 x^2}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

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)

In L.P.P., the maximum value of objective function $\mathrm{Z}=6 x+3 y$ subject to constraints $x+y \leq 5, x+2 y \geq 4,4 x+y \leq 12, x, y \geq 0$ is