The curve that satisfies the differential equation $x y d y-\left(1+y^2\right) d x=0$ passes through $(1,0)$ and intersects the curve $x^2+3 y^2=3$ at an angle $\theta$. Then, $\frac{2 \theta}{\pi}=$

The order and degree of the differential equation $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0$, are respectively.

The general solution of the differential equation $\frac{d y}{d x}=\frac{2 x-3 y+4}{3 x+2 y-7}$ is

The general solution of $\frac{d y}{d x}=\frac{x+y+1}{y-x+1}$ is